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values | problem_index
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stringclasses 5
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float64 0
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float64 15
4.26k
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float64 0
271M
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stringlengths 7
9
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stringlengths 10
27.7k
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float64 2
360
| correct_timeConsumedMillis
int64 30
8.06k
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int64 0
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stringlengths 7
9
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stringlengths 28
21.2k
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stringclasses 3
values | contest_start_year
int64 2.01k
2.02k
| time_limit
float64 0.5
15
| memory_limit
float64 64
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| title
stringlengths 2
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| description
stringlengths 35
3.16k
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stringlengths 67
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⌀ | interaction_format
null | note
stringclasses 840
values | examples
stringlengths 34
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| rating
int64 800
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stringclasses 533
values | testset_size
int64 2
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stringlengths 44
19.7M
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class | input_mode
stringclasses 1
value | generated_checker
stringclasses 231
values | executable
bool 1
class |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1006/D
|
1006
|
D
|
PyPy 3
|
TESTS
| 2 | 139 | 5,120,000 |
109034614
|
from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi,inf
from collections import deque,defaultdict
from bisect import bisect,bisect_left
from itertools import permutations as per
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(str(x)+'\n')
nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
n=I()
x=S()
y=S()
ans=0
for i in range(n//2):
a=x[i]
b=x[-i-1]
c=y[i]
d=y[-i-1]
if a==b==c==d or a==b!=c==d or a==c!=b==d or a==d!=c==b:continue
elif c!=a==b!=d!=c or a!=c==d!=b!=a or len(set([a,b,c,d]))==4:ans+=2
else:ans+=1
print(ans+(1 if n%2 and x[n//2]!=y[n//2] else 0))
| 23 | 93 | 3,379,200 |
206735302
|
n=int(input())
s=input()
t=input()
ans=0
for i in range((n+1)//2):
if 2*i==n-1:
if s[i]!=t[i]:
ans+=1
continue
S=[s[i],s[n-1-i],t[i],t[n-i-1]]
S.sort()
if S[0]==S[1] and S[2]==S[3]:
continue
if len(set(S))==4:
ans+=2
continue
if len(set(S))==2:
ans+=1
continue
if s[i]==s[n-1-i]:
ans+=2
else:
ans+=1
print(ans)
|
Codeforces Round 498 (Div. 3)
|
ICPC
| 2,018 | 2 | 256 |
Two Strings Swaps
|
You are given two strings $$$a$$$ and $$$b$$$ consisting of lowercase English letters, both of length $$$n$$$. The characters of both strings have indices from $$$1$$$ to $$$n$$$, inclusive.
You are allowed to do the following changes:
- Choose any index $$$i$$$ ($$$1 \le i \le n$$$) and swap characters $$$a_i$$$ and $$$b_i$$$;
- Choose any index $$$i$$$ ($$$1 \le i \le n$$$) and swap characters $$$a_i$$$ and $$$a_{n - i + 1}$$$;
- Choose any index $$$i$$$ ($$$1 \le i \le n$$$) and swap characters $$$b_i$$$ and $$$b_{n - i + 1}$$$.
Note that if $$$n$$$ is odd, you are formally allowed to swap $$$a_{\lceil\frac{n}{2}\rceil}$$$ with $$$a_{\lceil\frac{n}{2}\rceil}$$$ (and the same with the string $$$b$$$) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in $$$a$$$ with another character. In other words, in a single preprocess move you can choose any index $$$i$$$ ($$$1 \le i \le n$$$), any character $$$c$$$ and set $$$a_i := c$$$.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings $$$a$$$ and $$$b$$$ equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string $$$b$$$ or make any preprocess moves after the first change is made.
|
The first line of the input contains one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of strings $$$a$$$ and $$$b$$$.
The second line contains the string $$$a$$$ consisting of exactly $$$n$$$ lowercase English letters.
The third line contains the string $$$b$$$ consisting of exactly $$$n$$$ lowercase English letters.
|
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string $$$a$$$ equal to string $$$b$$$ with a sequence of changes from the list above.
| null |
In the first example preprocess moves are as follows: $$$a_1 := $$$'b', $$$a_3 := $$$'c', $$$a_4 := $$$'a' and $$$a_5:=$$$'b'. Afterwards, $$$a = $$$"bbcabba". Then we can obtain equal strings by the following sequence of changes: $$$swap(a_2, b_2)$$$ and $$$swap(a_2, a_6)$$$. There is no way to use fewer than $$$4$$$ preprocess moves before a sequence of changes to make string equal, so the answer in this example is $$$4$$$.
In the second example no preprocess moves are required. We can use the following sequence of changes to make $$$a$$$ and $$$b$$$ equal: $$$swap(b_1, b_5)$$$, $$$swap(a_2, a_4)$$$.
|
[{"input": "7\nabacaba\nbacabaa", "output": "4"}, {"input": "5\nzcabd\ndbacz", "output": "0"}]
| 1,700 |
["implementation"]
| 23 |
[{"input": "7\r\nabacaba\r\nbacabaa\r\n", "output": "4\r\n"}, {"input": "5\r\nzcabd\r\ndbacz\r\n", "output": "0\r\n"}, {"input": "1\r\na\r\nb\r\n", "output": "1\r\n"}, {"input": "5\r\nahmad\r\nyogaa\r\n", "output": "3\r\n"}]
| false |
stdio
| null | true |
1006/D
|
1006
|
D
|
PyPy 3
|
TESTS
| 2 | 139 | 7,680,000 |
95549583
|
n = int(input())
A = list(str(input()))
B = list(str(input()))
ans = 0
if n%2 == 0:
for i in range(n//2):
sa = {A[i], A[n-1-i]}
sb = {B[i], B[n-1-i]}
#print(sa, sb)
if len(sa) == 1 and len(sb) == 1:
continue
ans += max(len(sa-sb), len(sb-sa))
else:
for i in range(n//2):
sa = {A[i], A[n-1-i]}
sb = {B[i], B[n-1-i]}
#print(sa, sb)
if len(sa) == 1 and len(sb) == 1:
continue
ans += max(len(sa-sb), len(sb-sa))
if A[n//2] != B[n//2]:
ans += 1
print(ans)
| 23 | 93 | 3,686,400 |
170137185
|
from collections import *
from heapq import *
from bisect import *
from itertools import *
from functools import *
from math import *
from string import *
import operator
import sys
input = sys.stdin.readline
def solve():
n = int(input())
A = input().strip()
B = input().strip()
ans = 0
for i in range(n // 2):
w, x, y, z = A[i], A[~i], B[i], B[~i]
ans += w != x if y == z else len({y, z} - {w, x})
if n % 2 and A[n//2] != B[n//2]:
ans += 1
print(ans)
def main():
tests = 1
for _ in range(tests):
solve()
if __name__ == "__main__":
main()
|
Codeforces Round 498 (Div. 3)
|
ICPC
| 2,018 | 2 | 256 |
Two Strings Swaps
|
You are given two strings $$$a$$$ and $$$b$$$ consisting of lowercase English letters, both of length $$$n$$$. The characters of both strings have indices from $$$1$$$ to $$$n$$$, inclusive.
You are allowed to do the following changes:
- Choose any index $$$i$$$ ($$$1 \le i \le n$$$) and swap characters $$$a_i$$$ and $$$b_i$$$;
- Choose any index $$$i$$$ ($$$1 \le i \le n$$$) and swap characters $$$a_i$$$ and $$$a_{n - i + 1}$$$;
- Choose any index $$$i$$$ ($$$1 \le i \le n$$$) and swap characters $$$b_i$$$ and $$$b_{n - i + 1}$$$.
Note that if $$$n$$$ is odd, you are formally allowed to swap $$$a_{\lceil\frac{n}{2}\rceil}$$$ with $$$a_{\lceil\frac{n}{2}\rceil}$$$ (and the same with the string $$$b$$$) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in $$$a$$$ with another character. In other words, in a single preprocess move you can choose any index $$$i$$$ ($$$1 \le i \le n$$$), any character $$$c$$$ and set $$$a_i := c$$$.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings $$$a$$$ and $$$b$$$ equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string $$$b$$$ or make any preprocess moves after the first change is made.
|
The first line of the input contains one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of strings $$$a$$$ and $$$b$$$.
The second line contains the string $$$a$$$ consisting of exactly $$$n$$$ lowercase English letters.
The third line contains the string $$$b$$$ consisting of exactly $$$n$$$ lowercase English letters.
|
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string $$$a$$$ equal to string $$$b$$$ with a sequence of changes from the list above.
| null |
In the first example preprocess moves are as follows: $$$a_1 := $$$'b', $$$a_3 := $$$'c', $$$a_4 := $$$'a' and $$$a_5:=$$$'b'. Afterwards, $$$a = $$$"bbcabba". Then we can obtain equal strings by the following sequence of changes: $$$swap(a_2, b_2)$$$ and $$$swap(a_2, a_6)$$$. There is no way to use fewer than $$$4$$$ preprocess moves before a sequence of changes to make string equal, so the answer in this example is $$$4$$$.
In the second example no preprocess moves are required. We can use the following sequence of changes to make $$$a$$$ and $$$b$$$ equal: $$$swap(b_1, b_5)$$$, $$$swap(a_2, a_4)$$$.
|
[{"input": "7\nabacaba\nbacabaa", "output": "4"}, {"input": "5\nzcabd\ndbacz", "output": "0"}]
| 1,700 |
["implementation"]
| 23 |
[{"input": "7\r\nabacaba\r\nbacabaa\r\n", "output": "4\r\n"}, {"input": "5\r\nzcabd\r\ndbacz\r\n", "output": "0\r\n"}, {"input": "1\r\na\r\nb\r\n", "output": "1\r\n"}, {"input": "5\r\nahmad\r\nyogaa\r\n", "output": "3\r\n"}]
| false |
stdio
| null | true |
515/E
|
515
|
E
|
PyPy 3-64
|
TESTS
| 3 | 62 | 2,662,400 |
213981380
|
"""
We use a ST to encode vertex information:
if vertex represents [l, r]
then we keep (lbm, rbm, ism, sl, sr) where
lbm: maximum lbm(x) where lbm(x) is distance from left boundary to x + 2*h(x)
rbm: similarly for right
ism: best intra-segment we can do
sl: segment left index of arr
sr: segment right index of arr
During building phase, we combine appropriately.
Suppose we have range_query: [l, r] where l <= r, a normal interval
Then we again combine appropriately.
Now suppose we have query amounting to [0, r] and [l, n-1]
We query each subinterval like normal, then see if we can form an even better answer
by combining lbm1 and rbm2
"""
from math import inf
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
print = lambda p: sys.stdout.write(str(p) + "\n")
n, m = [int(component) for component in input().split(" ")]
d = [int(component) for component in input().split(" ")]
h = [int(component) for component in input().split(" ")]
### prefix work
def get_prefix_sums(arr):
prefix = [0] * (len(arr)+1)
for index in range(1,len(arr)+1):
prefix[index] = prefix[index-1] + arr[index-1]
return prefix
prefix = get_prefix_sums(d)
def get_dist(l, r):
# print(f" called with {(l, r)}")
return prefix[r] - prefix[l]
def make_segment_tree(d, h):
tree = [None] * (2*len(d)+1)
build(tree, d, h)
return tree
def merge(s1, s2):
# print(f" merge called with {s1} and {s2}")
if s1 == None:
return s2
if s2 == None:
return s1
if s1[3] > s2[3]:
s1, s2 = s2, s1
llbm, lrbm, lism, ll, lr = s1
rlbm, rrbm, rism, rl, rr = s2
nlbm = max(llbm, get_dist(ll, rl) + rlbm)
nrbm = max(lrbm + get_dist(lr, rr), rrbm)
nism = max(lism, lrbm + get_dist(lr, rl) + rlbm, rism)
return nlbm, nrbm, nism, ll, rr
def build(tree, d, h):
for i in range(len(d)):
tree[i + len(d)] = (2*h[i], 2*h[i], -inf, i, i)
# print(f"tree so far: {tree}")
for i in range(len(d)-1, 0, -1):
# print(f"i = {i}, about to merge {tree[i*2]}, with {tree[i*2+1]}")
tree[i] = merge(tree[i << 1], tree[(i << 1) | 1])
# print(f" result = {tree[i]}")
return tree
def range_query(tree, l, r):
# print(f"range_query received {(l, r)}")
res = None
l += (len(tree)-1) >> 1
r += ((len(tree)-1) >> 1) + 1
while l < r:
# print(f" l = {l}, r = {r}, res = {res}")
if l & 1:
# print(" l & 1")
# print(f" merging {tree[l]} and {res}")
res = merge(tree[l], res)
# print(f" result = {res}")
l += 1
if r & 1:
# print(" r & 1")
r -= 1
# print(f" merging {res} and {tree[r]}")
res = merge(res, tree[r])
# print(f" result = {res}")
l >>= 1
r >>= 1
# print(f"final res = {res}")
return res
tree = make_segment_tree(d, h)
# print(tree)
def complex_query(l, r):
# print(f"received {(l, r)}")
if l <= r:
# print("l <= r")
if l > 0 and r < n-1:
llbm, _, lism, _, _ = range_query(tree, 0, l-1)
_, rrbm, rism, _, _ = range_query(tree, r+1, n-1)
return max(lism, rism, llbm + d[-1] + rrbm)
elif l == 0:
# print("l == 0")
_, _, ism, _, _ = range_query(tree, r+1, n-1)
return ism
else:
# print(f"r == {n-1}")
_, _, ism, _, _ = range_query(tree, 0, l-1)
return ism
else:
_, _, ism, _, _ = range_query(tree, r+1, l-1)
return ism
# print(m, m, d, h)
for _ in range(m):
a, b = [int(component) for component in input().split(" ")]
print(complex_query(a-1, b-1))
| 27 | 967 | 57,036,800 |
214984801
|
from math import inf
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
fast_print = lambda p: sys.stdout.write(str(p) + "\n")
n, m = [int(component) for component in input().split(" ")]
d = [int(component) for component in input().split(" ")]
h = [int(component) for component in input().split(" ")]
### prefix work
def get_prefix_sums(arr):
prefix = [0] * (len(arr)+1)
for index in range(1,len(arr)+1):
prefix[index] = prefix[index-1] + arr[index-1]
return prefix
prefix = get_prefix_sums(d)
def get_dist(l, r):
# print(f" called with {(l, r)}")
return prefix[r] - prefix[l]
def merge(s1, s2):
# print(f" merge called with {s1} and {s2}")
if s1 == None:
return s2
if s2 == None:
return s1
# if s1[3] > s2[3]:
# s1, s2 = s2, s1
# print(f" reordered: {s1} and {s2}")
llbm, lrbm, lism, ll, lr = s1
rlbm, rrbm, rism, rl, rr = s2
nlbm = max(llbm, get_dist(ll, rl) + rlbm)
nrbm = max(lrbm + get_dist(lr, rr), rrbm)
nism = max(lism, lrbm + get_dist(lr, rl) + rlbm, rism)
return nlbm, nrbm, nism, ll, rr
def make_segment_tree(d, h):
N = len(d)
size = 1 << (N-1).bit_length()
tree = [None] * (2*size)
for k in range(N):
tree[size+k] = (2*h[k], 2*h[k], -inf, k, k)
for k in range(size-1,0,-1):
tree[k] = merge(tree[2*k], tree[2*k+1])
return tree
def range_query(tree, l, r):
# print(f"\nrange_query received {(l, r)}")
lres = None
rres = None
l += len(tree) >> 1
r += (len(tree) >> 1) + 1
while l < r:
if l & 1:
# print(f" l & 1")
lres = merge(lres, tree[l])
# print(f" now lres = {lres}")
l += 1
if r & 1:
# print(f" r & 1")
r -= 1
rres = merge(tree[r], rres)
# print(f" now rres = {rres}")
l >>= 1
r >>= 1
res = merge(lres, rres)
# print(f" final res = {res}")
return res
tree = make_segment_tree(d, h)
# print()
# print(tree)
def complex_query(l, r):
# print(f"received {(l, r)}")
if l <= r:
# print("l <= r")
if l > 0 and r < n-1:
# print("l > 0, r < n-1")
llbm, _, lism, _, _ = range_query(tree, 0, l-1)
# print(f"llbm = {llbm}, lism = {lism}")
_, rrbm, rism, _, _ = range_query(tree, r+1, n-1)
# print(f"rrbm = {rrbm}, rism = {rism}")
return max(lism, rism, llbm + d[-1] + rrbm)
elif l == 0:
# print("l == 0")
_, _, ism, _, _ = range_query(tree, r+1, n-1)
return ism
else:
# print(f"r == {n-1}")
_, _, ism, _, _ = range_query(tree, 0, l-1)
return ism
else:
_, _, ism, _, _ = range_query(tree, r+1, l-1)
return ism
# print(m, m, d, h)
for _ in range(m):
a, b = [int(component) for component in input().split(" ")]
fast_print(complex_query(a-1, b-1))
|
Codeforces Round 292 (Div. 2)
|
CF
| 2,015 | 2 | 512 |
Drazil and Park
|
Drazil is a monkey. He lives in a circular park. There are n trees around the park. The distance between the i-th tree and (i + 1)-st trees is di, the distance between the n-th tree and the first tree is dn. The height of the i-th tree is hi.
Drazil starts each day with the morning run. The morning run consists of the following steps:
- Drazil chooses two different trees
- He starts with climbing up the first tree
- Then he climbs down the first tree, runs around the park (in one of two possible directions) to the second tree, and climbs on it
- Then he finally climbs down the second tree.
But there are always children playing around some consecutive trees. Drazil can't stand children, so he can't choose the trees close to children. He even can't stay close to those trees.
If the two trees Drazil chooses are x-th and y-th, we can estimate the energy the morning run takes to him as 2(hx + hy) + dist(x, y). Since there are children on exactly one of two arcs connecting x and y, the distance dist(x, y) between trees x and y is uniquely defined.
Now, you know that on the i-th day children play between ai-th tree and bi-th tree. More formally, if ai ≤ bi, children play around the trees with indices from range [ai, bi], otherwise they play around the trees with indices from $${ \left[ a _ { i }, n \right] } \cup { \left[ 1, b _ { i } \right] }$$.
Please help Drazil to determine which two trees he should choose in order to consume the most energy (since he wants to become fit and cool-looking monkey) and report the resulting amount of energy for each day.
|
The first line contains two integer n and m (3 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting number of trees and number of days, respectively.
The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 109), the distances between consecutive trees.
The third line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 109), the heights of trees.
Each of following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) describing each new day. There are always at least two different trees Drazil can choose that are not affected by children.
|
For each day print the answer in a separate line.
| null | null |
[{"input": "5 3\n2 2 2 2 2\n3 5 2 1 4\n1 3\n2 2\n4 5", "output": "12\n16\n18"}, {"input": "3 3\n5 1 4\n5 1 4\n3 3\n2 2\n1 1", "output": "17\n22\n11"}]
| 2,300 |
["data structures"]
| 27 |
[{"input": "5 3\r\n2 2 2 2 2\r\n3 5 2 1 4\r\n1 3\r\n2 2\r\n4 5\r\n", "output": "12\r\n16\r\n18\r\n"}, {"input": "3 3\r\n5 1 4\r\n5 1 4\r\n3 3\r\n2 2\r\n1 1\r\n", "output": "17\r\n22\r\n11\r\n"}, {"input": "10 10\r\n8477 33103 38654 6582 27496 1106 15985 3644 29818 8849\r\n88745 72099 87767 85285 73517 94562 87214 63194 83791 77619\r\n2 8\r\n1 5\r\n9 5\r\n7 8\r\n6 9\r\n8 1\r\n6 1\r\n4 9\r\n8 10\r\n5 10\r\n", "output": "383739\r\n394915\r\n364658\r\n509685\r\n428294\r\n439157\r\n386525\r\n394604\r\n480926\r\n428294\r\n"}, {"input": "9 9\r\n1 1 1 1 1 1 1 1 1\r\n1 1 1 1 1000000000 1 1 1 1\r\n9 1\r\n9 9\r\n1 1\r\n8 9\r\n7 9\r\n9 2\r\n8 2\r\n1 1\r\n1 2\r\n", "output": "2000000005\r\n2000000006\r\n2000000006\r\n2000000006\r\n2000000006\r\n2000000005\r\n2000000004\r\n2000000006\r\n2000000006\r\n"}, {"input": "10 10\r\n91616899 35356500 87449167 31557462 21778951 474730484 302870359 398428048 174667839 183336304\r\n955685310 810816265 348361987 966143351 883722429 699134978 928163574 775129554 873615248 931808862\r\n4 4\r\n7 1\r\n2 7\r\n8 9\r\n6 2\r\n8 4\r\n9 10\r\n8 8\r\n3 10\r\n3 6\r\n", "output": "5234627463\r\n3731289022\r\n4218061919\r\n4645770639\r\n3731289022\r\n4120281441\r\n4510187231\r\n4704051250\r\n3624620049\r\n4827000318\r\n"}]
| false |
stdio
| null | true |
247/D
|
250
|
D
|
Python 3
|
TESTS
| 11 | 840 | 13,619,200 |
13535353
|
__author__ = 'Michael Ilyin'
import math
# debug = True
debug = False
def dist(x1, y1, x2, y2):
return math.sqrt(math.pow(x1 - x2, 2) + math.pow(y1 - y2, 2))
def get_y(x1, y1, x2, y2, x):
return (((x - x1) * (y2 - y1)) / (x2 - x1)) + y1
if debug:
with open("input.txt", "r") as inp:
firstLine = inp.readline()
secondLine = inp.readline()
thirdLine = inp.readline()
fourthLine = inp.readline()
else:
firstLine = input()
secondLine = input()
thirdLine = input()
fourthLine = input()
first = firstLine.split()
n = int(first[0])
m = int(first[1])
a = int(first[2])
b = int(first[3])
A = [int(x) for x in secondLine.split()]
B = [int(x) for x in thirdLine.split()]
L = [int(x) for x in fourthLine.split()]
optimalLen = float("inf")
optimalBIdx = -1
for i, bi in enumerate(B):
d = dist(0, 0, b, bi) + L[i]
if d < optimalLen:
optimalLen = d
optimalBIdx = i
if debug:
print(optimalBIdx + 1, optimalLen)
intersectY = get_y(0, 0, b, B[optimalBIdx], a)
if debug:
print(intersectY)
pointDist = float("inf")
optimalAIdx = -1
for i, ai in enumerate(A):
d = dist(a, ai, a, intersectY)
if d < pointDist:
pointDist = d
optimalAIdx = i
if debug:
print(optimalAIdx + 1, pointDist)
print(optimalAIdx + 1, optimalBIdx + 1)
| 33 | 794 | 14,745,600 |
106349764
|
import sys
def pro():
return sys.stdin.readline().strip()
def rop():
return map(int, pro().split())
def main():
s = list(rop())
a = list(rop())
q = list(rop())
o = list(rop())
p = -1
t = (1e100, -1, -1)
for i in range(s[1]):
while not((p == - 1 or s[2] * q[i] - s[3] * a[p] >= 0)
and (p + 1 == s[0] or s[2] * q[i] - s[3] * a[p+1] < 0)):
p += 1
if p != -1:
t = min(t, (o[i] + (s[2] ** 2 + a[p] ** 2) ** 0.5 + ((s[3] - s[2]) ** 2 + (q[i] - a[p]) ** 2) ** 0.5, p, i))
if p + 1 != s[0]:
t = min(t, (o[i] + (s[2] ** 2 + a[p + 1] ** 2) ** 0.5 + ((s[3] - s[2]) ** 2 + (q[i] - a[p + 1]) ** 2) ** 0.5, p + 1, i))
print(t[1] + 1, t[2] + 1)
main()
|
CROC-MBTU 2012, Final Round
|
CF
| 2,012 | 1 | 256 |
Building Bridge
|
Two villages are separated by a river that flows from the north to the south. The villagers want to build a bridge across the river to make it easier to move across the villages.
The river banks can be assumed to be vertical straight lines x = a and x = b (0 < a < b).
The west village lies in a steppe at point O = (0, 0). There are n pathways leading from the village to the river, they end at points Ai = (a, yi). The villagers there are plain and simple, so their pathways are straight segments as well.
The east village has reserved and cunning people. Their village is in the forest on the east bank of the river, but its exact position is not clear. There are m twisted paths leading from this village to the river and ending at points Bi = (b, y'i). The lengths of all these paths are known, the length of the path that leads from the eastern village to point Bi, equals li.
The villagers want to choose exactly one point on the left bank of river Ai, exactly one point on the right bank Bj and connect them by a straight-line bridge so as to make the total distance between the villages (the sum of |OAi| + |AiBj| + lj, where |XY| is the Euclidean distance between points X and Y) were minimum. The Euclidean distance between points (x1, y1) and (x2, y2) equals $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$.
Help them and find the required pair of points.
|
The first line contains integers n, m, a, b (1 ≤ n, m ≤ 105, 0 < a < b < 106).
The second line contains n integers in the ascending order: the i-th integer determines the coordinate of point Ai and equals yi (|yi| ≤ 106).
The third line contains m integers in the ascending order: the i-th integer determines the coordinate of point Bi and equals y'i (|y'i| ≤ 106).
The fourth line contains m more integers: the i-th of them determines the length of the path that connects the eastern village and point Bi, and equals li (1 ≤ li ≤ 106).
It is guaranteed, that there is such a point C with abscissa at least b, that |BiC| ≤ li for all i (1 ≤ i ≤ m). It is guaranteed that no two points Ai coincide. It is guaranteed that no two points Bi coincide.
|
Print two integers — the numbers of points on the left (west) and right (east) banks, respectively, between which you need to build a bridge. You can assume that the points on the west bank are numbered from 1 to n, in the order in which they are given in the input. Similarly, the points on the east bank are numbered from 1 to m in the order in which they are given in the input.
If there are multiple solutions, print any of them. The solution will be accepted if the final length of the path will differ from the answer of the jury by no more than 10 - 6 in absolute or relative value.
| null | null |
[{"input": "3 2 3 5\n-2 -1 4\n-1 2\n7 3", "output": "2 2"}]
| 1,900 |
[]
| 33 |
[{"input": "3 2 3 5\r\n-2 -1 4\r\n-1 2\r\n7 3\r\n", "output": "2 2"}, {"input": "1 1 10 20\r\n5\r\n-5\r\n1\r\n", "output": "1 1"}, {"input": "2 2 1 2\r\n-1 10\r\n8 9\r\n3 7\r\n", "output": "1 1"}, {"input": "10 20 50 60\r\n-96 -75 32 37 42 43 44 57 61 65\r\n-95 -90 -86 -79 -65 -62 -47 -11 -8 -6 1 8 23 25 32 51 73 88 94 100\r\n138 75 132 116 49 43 96 166 96 161 146 112 195 192 201 186 251 254 220 227\r\n", "output": "2 6"}]
| false |
stdio
| null | true |
175/B
|
175
|
B
|
Python 3
|
TESTS
| 0 | 154 | 0 |
53660628
|
n = int(input())
d = {}
for _ in '0' * n:
s, v = input().split()
d[s] = max(d.get(s, 0), int(v))
v = d.values()
res = str(len(d)) + '\n'
for k in d:
a, b = 0, 0
for x in v:
a += x <= d[k]
b += x > d[k]
s = 'noob'
a /= n
b /= n
if a >= .5 and b >= .2: s = 'random'
if a >= .8 and b >= .1: s = 'average'
if a >= .9 and b >= .01: s = 'hardcore'
if a >= .99: s='pro'
res += k + ' ' + s + '\n'
print(res)
| 46 | 434 | 204,800 |
214357322
|
# LUOGU_RID: 116285050
n=int(input())
d={}
c=[]
for _ in range(n):
a,b=input().split()
b=int(b)
if a not in c:
c.append(a)
if a not in d:
d[a]=b
else:
if d[a]<b:
d[a]=b
print(len(d))
m=len(c)
x=[0]*m
for i in range(m):
for j in range(m):
if d[c[i]]<d[c[j]]:
x[i]+=1
print(c[i],end=' ')
if x[i]>0.50*m:
print('noob')
elif x[i]>0.20*m:
print('random')
elif x[i]>0.10*m:
print('average')
elif x[i]>0.01*m:
print('hardcore')
else:
print('pro')
|
Codeforces Round 115
|
CF
| 2,012 | 2 | 256 |
Plane of Tanks: Pro
|
Vasya has been playing Plane of Tanks with his friends the whole year. Now it is time to divide the participants into several categories depending on their results.
A player is given a non-negative integer number of points in each round of the Plane of Tanks. Vasya wrote results for each round of the last year. He has n records in total.
In order to determine a player's category consider the best result obtained by the player and the best results of other players. The player belongs to category:
- "noob" — if more than 50% of players have better results;
- "random" — if his result is not worse than the result that 50% of players have, but more than 20% of players have better results;
- "average" — if his result is not worse than the result that 80% of players have, but more than 10% of players have better results;
- "hardcore" — if his result is not worse than the result that 90% of players have, but more than 1% of players have better results;
- "pro" — if his result is not worse than the result that 99% of players have.
When the percentage is calculated the player himself is taken into account. That means that if two players played the game and the first one gained 100 points and the second one 1000 points, then the first player's result is not worse than the result that 50% of players have, and the second one is not worse than the result that 100% of players have.
Vasya gave you the last year Plane of Tanks results. Help Vasya determine each player's category.
|
The first line contains the only integer number n (1 ≤ n ≤ 1000) — a number of records with the players' results.
Each of the next n lines contains a player's name and the amount of points, obtained by the player for the round, separated with a space. The name contains not less than 1 and no more than 10 characters. The name consists of lowercase Latin letters only. It is guaranteed that any two different players have different names. The amount of points, obtained by the player for the round, is a non-negative integer number and does not exceed 1000.
|
Print on the first line the number m — the number of players, who participated in one round at least.
Each one of the next m lines should contain a player name and a category he belongs to, separated with space. Category can be one of the following: "noob", "random", "average", "hardcore" or "pro" (without quotes). The name of each player should be printed only once. Player names with respective categories can be printed in an arbitrary order.
| null |
In the first example the best result, obtained by artem is not worse than the result that 25% of players have (his own result), so he belongs to category "noob". vasya and kolya have best results not worse than the results that 75% players have (both of them and artem), so they belong to category "random". igor has best result not worse than the result that 100% of players have (all other players and himself), so he belongs to category "pro".
In the second example both players have the same amount of points, so they have results not worse than 100% players have, so they belong to category "pro".
|
[{"input": "5\nvasya 100\nvasya 200\nartem 100\nkolya 200\nigor 250", "output": "4\nartem noob\nigor pro\nkolya random\nvasya random"}, {"input": "3\nvasya 200\nkolya 1000\nvasya 1000", "output": "2\nkolya pro\nvasya pro"}]
| 1,400 |
["implementation"]
| 78 |
[{"input": "5\r\nvasya 100\r\nvasya 200\r\nartem 100\r\nkolya 200\r\nigor 250\r\n", "output": "4\r\nartem noob\r\nigor pro\r\nkolya random\r\nvasya random\r\n"}, {"input": "3\r\nvasya 200\r\nkolya 1000\r\nvasya 1000\r\n", "output": "2\r\nkolya pro\r\nvasya pro\r\n"}, {"input": "1\r\nvasya 1000\r\n", "output": "1\r\nvasya pro\r\n"}, {"input": "5\r\nvasya 1000\r\nvasya 100\r\nkolya 200\r\npetya 300\r\noleg 400\r\n", "output": "4\r\nkolya noob\r\noleg random\r\npetya random\r\nvasya pro\r\n"}, {"input": "10\r\na 1\r\nb 2\r\nc 3\r\nd 4\r\ne 5\r\nf 6\r\ng 7\r\nh 8\r\ni 9\r\nj 10\r\n", "output": "10\r\na noob\r\nb noob\r\nc noob\r\nd noob\r\ne random\r\nf random\r\ng random\r\nh average\r\ni hardcore\r\nj pro\r\n"}, {"input": "10\r\nj 10\r\ni 9\r\nh 8\r\ng 7\r\nf 6\r\ne 5\r\nd 4\r\nc 3\r\nb 2\r\na 1\r\n", "output": "10\r\na noob\r\nb noob\r\nc noob\r\nd noob\r\ne random\r\nf random\r\ng random\r\nh average\r\ni hardcore\r\nj pro\r\n"}, {"input": "1\r\ntest 0\r\n", "output": "1\r\ntest pro\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_path):
# Read input data
with open(input_path) as f:
n = int(f.readline())
players = {}
for _ in range(n):
name, points = f.readline().strip().split()
points = int(points)
if name not in players or points > players[name]:
players[name] = points
m = len(players)
best_scores = list(players.values())
# Compute correct categories
correct = {}
for name in players:
score = players[name]
cnt = sum(1 for s in best_scores if s > score)
percentage = (cnt * 100.0) / m
if percentage > 50:
cat = 'noob'
elif percentage > 20:
cat = 'random'
elif percentage > 10:
cat = 'average'
elif percentage > 1:
cat = 'hardcore'
else:
cat = 'pro'
correct[name] = cat
# Read submission output
with open(submission_path) as f:
lines = f.read().splitlines()
if not lines:
print(0)
return
try:
m_sub = int(lines[0])
except:
print(0)
return
if m_sub != m:
print(0)
return
submission = {}
for line in lines[1:]:
parts = line.strip().split()
if len(parts) != 2:
print(0)
return
name, cat = parts
if name in submission:
print(0)
return
submission[name] = cat
# Check all players are present and correct
if submission.keys() != correct.keys():
print(0)
return
for name, cat in submission.items():
if correct.get(name) != cat:
print(0)
return
print(1)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
412/D
|
412
|
D
|
Python 3
|
PRETESTS
| 2 | 77 | 0 |
6407699
|
n, p = map(int, input().split())
arr = [i + 1 for i in range(n)]
operations = []
for i in range(p):
t, y = map(int, input().split())
operations.append([t, y])
for elem in operations:
a, b = elem[0], elem[1]
if abs(arr.index(a) - arr.index(b)) > 1:
continue
else:
k, l = arr.index(a), arr.index(b)
if k == l - 1:
arr[k] = b
arr[l] = a
can = 1
for elem in operations:
a, b = elem[0], elem[1]
if abs(arr.index(a) - arr.index(b)) > 1:
continue
else:
k, l = arr.index(a), arr.index(b)
if k == l - 1:
can = 0
break
if (can == 1):
print(' '.join(map(str, arr)))
else:
print(-1)
| 33 | 327 | 9,830,400 |
197318855
|
n,m = map(int,input().split())
g = [set() for i in range(n)]
for i in range(m):
a,b = map(int,input().split())
g[a-1].add(b-1)
c = [0]*n
for i in range(n):
c[i]=i
for i in range(n):
j=i
while j>0 and c[j] in g[c[j-1]]:
c[j],c[j-1]=c[j-1],c[j]
j-=1
for i in c:
print(i+1,end=' ')
|
Coder-Strike 2014 - Round 1
|
CF
| 2,014 | 1 | 256 |
Giving Awards
|
The employees of the R1 company often spend time together: they watch football, they go camping, they solve contests. So, it's no big deal that sometimes someone pays for someone else.
Today is the day of giving out money rewards. The R1 company CEO will invite employees into his office one by one, rewarding each one for the hard work this month. The CEO knows who owes money to whom. And he also understands that if he invites person x to his office for a reward, and then immediately invite person y, who has lent some money to person x, then they can meet. Of course, in such a situation, the joy of person x from his brand new money reward will be much less. Therefore, the R1 CEO decided to invite the staff in such an order that the described situation will not happen for any pair of employees invited one after another.
However, there are a lot of employees in the company, and the CEO doesn't have a lot of time. Therefore, the task has been assigned to you. Given the debt relationships between all the employees, determine in which order they should be invited to the office of the R1 company CEO, or determine that the described order does not exist.
|
The first line contains space-separated integers n and m $$(2\leq n\leq3\cdot10^{4};1\leq m\leq min(10^{5},\frac{n(n-1)}{2}))$$ — the number of employees in R1 and the number of debt relations. Each of the following m lines contains two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi), these integers indicate that the person number ai owes money to a person a number bi. Assume that all the employees are numbered from 1 to n.
It is guaranteed that each pair of people p, q is mentioned in the input data at most once. In particular, the input data will not contain pairs p, q and q, p simultaneously.
|
Print -1 if the described order does not exist. Otherwise, print the permutation of n distinct integers. The first number should denote the number of the person who goes to the CEO office first, the second number denote the person who goes second and so on.
If there are multiple correct orders, you are allowed to print any of them.
| null | null |
[{"input": "2 1\n1 2", "output": "2 1"}, {"input": "3 3\n1 2\n2 3\n3 1", "output": "2 1 3"}]
| 2,000 |
["dfs and similar"]
| 33 |
[{"input": "2 1\r\n1 2\r\n", "output": "2 1 \r\n"}, {"input": "3 3\r\n1 2\r\n2 3\r\n3 1\r\n", "output": "2 1 3 \r\n"}, {"input": "10 45\r\n10 5\r\n10 7\r\n6 1\r\n5 8\r\n3 5\r\n6 5\r\n1 2\r\n6 10\r\n2 9\r\n9 5\r\n4 1\r\n7 5\r\n1 8\r\n6 8\r\n10 9\r\n7 2\r\n7 9\r\n4 10\r\n7 3\r\n4 8\r\n10 3\r\n10 8\r\n2 10\r\n8 2\r\n4 2\r\n5 2\r\n9 1\r\n4 5\r\n1 3\r\n9 6\r\n3 8\r\n5 1\r\n6 4\r\n4 7\r\n8 7\r\n7 1\r\n9 3\r\n3 4\r\n6 2\r\n3 2\r\n6 7\r\n6 3\r\n4 9\r\n8 9\r\n1 10\r\n", "output": "2 3 1 5 7 8 4 6 9 10 \r\n"}, {"input": "4 6\r\n3 4\r\n4 1\r\n4 2\r\n2 1\r\n3 2\r\n3 1\r\n", "output": "1 2 4 3 \r\n"}, {"input": "5 10\r\n4 2\r\n4 5\r\n3 1\r\n2 1\r\n3 4\r\n3 2\r\n5 1\r\n5 2\r\n4 1\r\n5 3\r\n", "output": "1 2 4 3 5 \r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m = map(int, f.readline().split())
edges = set()
for _ in range(m):
a, b = map(int, f.readline().split())
edges.add((a, b))
# Read submission output
with open(submission_path) as f:
line = f.readline().strip()
if line == '-1':
# Check if it's actually impossible
# To determine this, we'd need to know if any valid order exists, which is complex.
# However, the problem requires the checker to verify the submission's output.
# Since the submission claims it's impossible, but how can the checker know?
# This suggests that the checker cannot handle the -1 case properly.
# Wait, the problem says if the order does not exist, output -1.
# But how can the checker verify if the submission's -1 is correct?
# This is a problem because the checker would need to solve the problem itself.
# However, the checker can't do that. So for this problem, the checker can't validate a submission that outputs -1.
# Therefore, in the checker code, if the submission outputs -1, we have to check if a valid order exists.
# But that's equivalent to solving the problem. Which is impossible for a checker to do without the correct algorithm.
# Hence, this problem's checker is incomplete. However, given the problem statement, perhaps the checker is expected to handle cases where the submission outputs -1 correctly.
# So, the approach here is that the checker cannot fully validate submissions that output -1. Thus, in practice, such cases would have to be handled by the contest system's test cases, which would have the correct answer (either a permutation or -1).
# However, in the problem statement, the output is either a permutation or -1. So the checker must check two possibilities:
# 1. The submission outputs a valid permutation (so -1 is wrong)
# 2. The submission outputs -1, and no valid permutation exists.
# But how can the checker distinguish between the two? It can't unless it solves the problem.
# Therefore, the checker code here can't handle the -1 case. Thus, the problem may have test cases where the correct answer is -1, and the checker would need to check that the submission's output is -1 and that no valid permutation exists. But how?
# This is a problem. Therefore, perhaps the correct approach is that in this problem, the checker can only check the permutation case. If the submission outputs -1, the checker can't validate it unless it knows whether such a permutation exists (which requires solving the problem).
# Given that the checker's code is supposed to be written without solving the problem, this is impossible. Therefore, the problem is not a 'checker' type but 'diff'? But the problem allows multiple correct outputs, so it's a checker. However, the -1 case complicates things.
# However, according to the problem statement, the correct answer is either -1 or a permutation. So if the submission outputs -1, the checker must check whether the correct answer is -1. But how?
# This is a contradiction. Therefore, in practice, perhaps the test cases where the answer is -1 would be evaluated as 'diff' (exact match), and the rest as 'checker'.
# However, the problem statement's output allows any permutation. So if the reference output is a permutation, then the submission must also be a permutation. If the reference output is -1, then the submission must be -1.
# Therefore, the checker must compare the submission's output with the reference output in the case where the reference output is -1. Otherwise, check the permutation.
# So, the checker must read the reference output. If the reference output is -1, then the submission must also output -1. Else, the submission must output a valid permutation.
# Therefore, the checker logic is:
# Read the reference output. If it's -1, then submission must be -1.
# Otherwise, submission must be a valid permutation.
# So, the code must compare the submission against the reference output in the case where the reference is -1.
# So, the code should proceed as follows:
# 1. Read the reference output (output_path).
with open(output_path) as ref_f:
ref_line = ref_f.readline().strip()
if ref_line == '-1':
# The correct answer is -1; submission must be -1
if line == '-1':
print(1)
return
else:
print(0)
return
else:
# The correct answer is a permutation. The submission must be a valid permutation.
# So proceed with permutation checks.
pass
# If submission is -1 but reference is not, invalid.
if line == '-1':
print(0)
return
# Parse submission's permutation
perm = list(map(int, line.split()))
if len(perm) != n:
print(0)
return
if set(perm) != set(range(1, n+1)):
print(0)
return
# Check consecutive pairs
for i in range(n-1):
x = perm[i]
y = perm[i+1]
if (x, y) in edges:
print(0)
return
print(1)
if __name__ == '__main__':
main()
| true |
247/D
|
250
|
D
|
PyPy 3-64
|
TESTS
| 4 | 61 | 2,150,400 |
154037124
|
import math
n,m,a,b = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
C = list(map(int,input().split()))
tar = float('inf')
i = 0
dx2 = (a-b)*(a-b)
for j in range(m):
pdy = float('inf')
for k in range(i,n):
dy = abs(A[k]-B[j])
if dy>=pdy:
k -= 1
break
tot = math.sqrt(a*a+A[k]*A[k])+math.sqrt(dx2+dy*dy)+C[j]
pdy = dy
tar = min(tar,tot)
if tot==tar: ans = [k+1,j+1]
i = k
print(*ans)
| 33 | 1,122 | 11,878,400 |
13858446
|
from math import sqrt,fabs
def dist(x1, y1, x2, y2):
return sqrt(pow(abs(x1 - x2), 2) + pow(abs(y1 - y2), 2))
def calcOptimumRightPoint(startX, startY):
l = float("inf")
idx = -1
for i in range(len(B)):
d = dist(startX, startY, b, B[i]) + L[i]
if d <= l:
l = d
idx = i
return idx
n,m,a,b = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]
L = [int(x) for x in input().split()]
optimumRightPoint = calcOptimumRightPoint(0,0)
intersectLeft = (a * B[optimumRightPoint]) / b
l = float("inf")
optimumLeftPoint = -1
for i in range(len(A)):
if fabs(intersectLeft-A[i]) < l:
l = fabs(intersectLeft-A[i])
optimumLeftPoint = i
optimumRightPoint = calcOptimumRightPoint(a, A[optimumLeftPoint])
print(optimumLeftPoint + 1, optimumRightPoint + 1)
|
CROC-MBTU 2012, Final Round
|
CF
| 2,012 | 1 | 256 |
Building Bridge
|
Two villages are separated by a river that flows from the north to the south. The villagers want to build a bridge across the river to make it easier to move across the villages.
The river banks can be assumed to be vertical straight lines x = a and x = b (0 < a < b).
The west village lies in a steppe at point O = (0, 0). There are n pathways leading from the village to the river, they end at points Ai = (a, yi). The villagers there are plain and simple, so their pathways are straight segments as well.
The east village has reserved and cunning people. Their village is in the forest on the east bank of the river, but its exact position is not clear. There are m twisted paths leading from this village to the river and ending at points Bi = (b, y'i). The lengths of all these paths are known, the length of the path that leads from the eastern village to point Bi, equals li.
The villagers want to choose exactly one point on the left bank of river Ai, exactly one point on the right bank Bj and connect them by a straight-line bridge so as to make the total distance between the villages (the sum of |OAi| + |AiBj| + lj, where |XY| is the Euclidean distance between points X and Y) were minimum. The Euclidean distance between points (x1, y1) and (x2, y2) equals $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$.
Help them and find the required pair of points.
|
The first line contains integers n, m, a, b (1 ≤ n, m ≤ 105, 0 < a < b < 106).
The second line contains n integers in the ascending order: the i-th integer determines the coordinate of point Ai and equals yi (|yi| ≤ 106).
The third line contains m integers in the ascending order: the i-th integer determines the coordinate of point Bi and equals y'i (|y'i| ≤ 106).
The fourth line contains m more integers: the i-th of them determines the length of the path that connects the eastern village and point Bi, and equals li (1 ≤ li ≤ 106).
It is guaranteed, that there is such a point C with abscissa at least b, that |BiC| ≤ li for all i (1 ≤ i ≤ m). It is guaranteed that no two points Ai coincide. It is guaranteed that no two points Bi coincide.
|
Print two integers — the numbers of points on the left (west) and right (east) banks, respectively, between which you need to build a bridge. You can assume that the points on the west bank are numbered from 1 to n, in the order in which they are given in the input. Similarly, the points on the east bank are numbered from 1 to m in the order in which they are given in the input.
If there are multiple solutions, print any of them. The solution will be accepted if the final length of the path will differ from the answer of the jury by no more than 10 - 6 in absolute or relative value.
| null | null |
[{"input": "3 2 3 5\n-2 -1 4\n-1 2\n7 3", "output": "2 2"}]
| 1,900 |
[]
| 33 |
[{"input": "3 2 3 5\r\n-2 -1 4\r\n-1 2\r\n7 3\r\n", "output": "2 2"}, {"input": "1 1 10 20\r\n5\r\n-5\r\n1\r\n", "output": "1 1"}, {"input": "2 2 1 2\r\n-1 10\r\n8 9\r\n3 7\r\n", "output": "1 1"}, {"input": "10 20 50 60\r\n-96 -75 32 37 42 43 44 57 61 65\r\n-95 -90 -86 -79 -65 -62 -47 -11 -8 -6 1 8 23 25 32 51 73 88 94 100\r\n138 75 132 116 49 43 96 166 96 161 146 112 195 192 201 186 251 254 220 227\r\n", "output": "2 6"}]
| false |
stdio
| null | true |
436/C
|
436
|
C
|
Python 3
|
TESTS
| 2 | 61 | 0 |
6888989
|
n, m, k, w = map(int, input().split())
m *= n
x, y, z = range(n), range(m), range(k)
t = [''.join(input() for i in x) for j in z]
def f(a, b): return sum(a[i] != b[i] for i in y)
u = [[w * f(t[i], t[j]) for j in range(i + 1, k)] for i in z]
def g(i, j): return min(m, u[i][j - i - 1] if i < j else u[j][i - j - 1])
d, q = [m] * k, [-1] * k
a, b = [-1] * k, list(z)
for i in z:
q[b[0]], k = a[b[0]], b[0]
for j in b:
if d[j] < d[k]: q[j], k = a[j], j
b.remove(k)
for i in b:
if d[i] > g(i, k): d[i], a[i] = g(i, k), k
print(sum(d))
print('\n'.join([str(i + 1) + ' ' + str(q[i] + 1) for i in z]))
| 30 | 1,466 | 70,041,600 |
87968659
|
def put():
return map(int, input().split())
def diff(x,y):
ans = 0
for i in range(n*m):
if s[x][i]!= s[y][i]:
ans+=1
return ans
def find(i):
if i==p[i]:
return i
p[i] = find(p[i])
return p[i]
def union(i,j):
if rank[i]>rank[j]:
i,j = j,i
elif rank[i]==rank[j]:
rank[j]+=1
p[i]= j
def dfs(i,p):
if i!=0:
print(i,p)
for j in tree[i]:
if j!=p:
dfs(j,i)
n,m,k,w = put()
s = ['']*k
for i in range(k):
for j in range(n):
s[i]+=input()
edge = []
k+=1
rank = [0]*(k)
p = list(range(k))
cost = 0
tree = [[] for i in range(k)]
for i in range(k):
for j in range(i+1,k):
if i==0:
z=n*m
else:
z = diff(i-1,j-1)*w
edge.append((z,i,j))
edge.sort()
for z,i,j in edge:
u = find(i)
v = find(j)
if u!=v:
union(u,v)
cost+= z
tree[i].append(j)
tree[j].append(i)
print(cost)
dfs(0,-1)
|
Zepto Code Rush 2014
|
CF
| 2,014 | 2 | 256 |
Dungeons and Candies
|
During the loading of the game "Dungeons and Candies" you are required to get descriptions of k levels from the server. Each description is a map of an n × m checkered rectangular field. Some cells of the field contain candies (each cell has at most one candy). An empty cell is denoted as "." on the map, but if a cell has a candy, it is denoted as a letter of the English alphabet. A level may contain identical candies, in this case the letters in the corresponding cells of the map will be the same.
When you transmit information via a network, you want to minimize traffic — the total size of the transferred data. The levels can be transmitted in any order. There are two ways to transmit the current level A:
1. You can transmit the whole level A. Then you need to transmit n·m bytes via the network.
2. You can transmit the difference between level A and some previously transmitted level B (if it exists); this operation requires to transmit dA, B·w bytes, where dA, B is the number of cells of the field that are different for A and B, and w is a constant. Note, that you should compare only the corresponding cells of levels A and B to calculate dA, B. You cannot transform the maps of levels, i.e. rotate or shift them relatively to each other.
Your task is to find a way to transfer all the k levels and minimize the traffic.
|
The first line contains four integers n, m, k, w (1 ≤ n, m ≤ 10; 1 ≤ k, w ≤ 1000). Then follows the description of k levels. Each level is described by n lines, each line contains m characters. Each character is either a letter of the English alphabet or a dot ("."). Please note that the case of the letters matters.
|
In the first line print the required minimum number of transferred bytes.
Then print k pairs of integers x1, y1, x2, y2, ..., xk, yk, describing the way to transfer levels. Pair xi, yi means that level xi needs to be transferred by way yi. If yi equals 0, that means that the level must be transferred using the first way, otherwise yi must be equal to the number of a previously transferred level. It means that you will transfer the difference between levels yi and xi to transfer level xi. Print the pairs in the order of transferring levels. The levels are numbered 1 through k in the order they follow in the input.
If there are multiple optimal solutions, you can print any of them.
| null | null |
[{"input": "2 3 3 2\nA.A\n...\nA.a\n..C\nX.Y\n...", "output": "14\n1 0\n2 1\n3 1"}, {"input": "1 1 4 1\nA\n.\nB\n.", "output": "3\n1 0\n2 0\n4 2\n3 0"}, {"input": "1 3 5 2\nABA\nBBB\nBBA\nBAB\nABB", "output": "11\n1 0\n3 1\n2 3\n4 2\n5 1"}]
| 1,800 |
["dsu", "graphs", "greedy", "trees"]
| 30 |
[{"input": "2 3 3 2\r\nA.A\r\n...\r\nA.a\r\n..C\r\nX.Y\r\n...\r\n", "output": "14\r\n1 0\r\n2 1\r\n3 1\r\n"}, {"input": "1 1 4 1\r\nA\r\n.\r\nB\r\n.\r\n", "output": "3\r\n1 0\r\n2 0\r\n4 2\r\n3 0\r\n"}, {"input": "1 3 5 2\r\nABA\r\nBBB\r\nBBA\r\nBAB\r\nABB\r\n", "output": "11\r\n1 0\r\n3 1\r\n2 3\r\n4 2\r\n5 1\r\n"}, {"input": "2 2 5 1\r\n..\r\nBA\r\n.A\r\nB.\r\n..\r\nA.\r\nAB\r\n.B\r\n..\r\n..\r\n", "output": "12\r\n1 0\r\n2 1\r\n3 1\r\n5 3\r\n4 5\r\n"}, {"input": "3 3 10 2\r\nBA.\r\n..A\r\n.BB\r\nB..\r\n..B\r\n.AA\r\nB..\r\nAB.\r\n..A\r\nBAB\r\n.A.\r\n.B.\r\n..B\r\nA..\r\n...\r\n...\r\n.B.\r\nBA.\r\n..B\r\n.AB\r\n.B.\r\nB.A\r\n.A.\r\n.BA\r\n..B\r\n...\r\n.A.\r\n.AA\r\n..A\r\n.B.\r\n", "output": "67\r\n1 0\r\n10 1\r\n2 1\r\n3 2\r\n4 1\r\n7 4\r\n9 7\r\n5 9\r\n6 9\r\n8 4\r\n"}, {"input": "3 1 5 1\r\nB\r\nA\r\nB\r\nA\r\nA\r\nB\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\n", "output": "5\r\n1 0\r\n2 1\r\n3 2\r\n4 3\r\n5 3\r\n"}, {"input": "3 2 10 1\r\nAB\r\nBA\r\nAB\r\nAA\r\nAA\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBA\r\nAB\r\nAA\r\nBB\r\nAB\r\nBA\r\nBB\r\nBB\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBB\r\nAB\r\nAA\r\n", "output": "16\r\n1 0\r\n3 1\r\n8 3\r\n2 3\r\n4 2\r\n9 4\r\n6 4\r\n7 6\r\n10 6\r\n5 10\r\n"}, {"input": "2 3 10 2\r\nABB\r\nABA\r\nAAB\r\nBAB\r\nAAA\r\nBBA\r\nBBB\r\nBAA\r\nBBB\r\nABB\r\nABA\r\nBBA\r\nBBB\r\nAAB\r\nABA\r\nABB\r\nBBA\r\nBAB\r\nBBB\r\nBBB\r\n", "output": "38\r\n1 0\r\n5 1\r\n7 5\r\n4 7\r\n9 4\r\n10 5\r\n6 1\r\n3 6\r\n8 1\r\n2 0\r\n"}, {"input": "1 1 1 1\r\n.\r\n", "output": "1\r\n1 0\r\n"}]
| false |
stdio
|
import sys
def read_levels(input_path):
with open(input_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip()]
first_line = lines[0].split()
n, m, k, w = map(int, first_line)
levels = [None] # 1-based indexing
idx = 1
for _ in range(k):
level = []
for _ in range(n):
if idx >= len(lines):
return None, None, None, None, None
level.append(lines[idx].strip())
idx += 1
levels.append(level)
return n, m, k, w, levels
def main(input_path, output_path, submission_path):
n, m, k, w, levels = read_levels(input_path)
if levels is None:
return 0
# Read reference output's first line
with open(output_path) as f_ref:
ref_lines = f_ref.read().splitlines()
if not ref_lines:
return 0
ref_total = int(ref_lines[0])
# Read submission output
with open(submission_path) as f_sub:
sub_lines = f_sub.read().splitlines()
if not sub_lines:
return 0
try:
sub_total = int(sub_lines[0])
except:
return 0
# Check sub_total matches reference
if sub_total != ref_total:
return 0
# Parse pairs
tokens = []
for line in sub_lines[1:]:
tokens.extend(line.strip().split())
if len(tokens) != 2 * k:
return 0
pairs = []
try:
for i in range(0, len(tokens), 2):
xi = int(tokens[i])
yi = int(tokens[i+1])
pairs.append((xi, yi))
except:
return 0
# Check all xi are unique and 1..k
seen_xi = set()
for xi, _ in pairs:
if xi < 1 or xi > k or xi in seen_xi:
return 0
seen_xi.add(xi)
if len(seen_xi) != k:
return 0
# Check pairs order and calculate total
transmitted = set()
total = 0
for xi, yi in pairs:
if yi != 0 and yi not in transmitted:
return 0
if yi == 0:
cost = n * m
else:
# Compute d between xi and yi
d = 0
for row in range(n):
for col in range(m):
if levels[xi][row][col] != levels[yi][row][col]:
d += 1
cost = d * w
total += cost
transmitted.add(xi)
if total != sub_total:
return 0
return 1
if __name__ == "__main__":
input_path, output_path, submission_path = sys.argv[1:]
score = main(input_path, output_path, submission_path)
print(score)
| true |
174/C
|
174
|
C
|
PyPy 3
|
TESTS
| 0 | 124 | 20,172,800 |
127712038
|
def process(A):
n = len(A)
if n==1:
return [[1,1] for i in range(A[0])]
current = [1 for i in range(A[0])]
answer = []
for i in range(1, n):
if A[i] < A[i-1]:
while len(current) > A[i]:
a = current.pop()
answer.append([a, i-1])
elif A[i] > A[i-1]:
while len(current) < A[i]:
current.append(i)
while len(current) > 0:
a = current.pop()
answer.append([a, n])
return answer
n = int(input())
A = [int(x) for x in input().split()]
answer = process(A)
print(len(answer))
for a, b in answer:
print(f'{a} {b}')
| 63 | 1,152 | 30,003,200 |
15764001
|
N = 100000
v = []
for i in range(0, N) :
v.append([])
line = input()
n = int(line)
line = input()
lineSplit = line.split()
a = 0
for i in range(0, len(lineSplit)) :
b = int(lineSplit[i])
if b > a :
for j in range(a, b) :
v[j].append([i, -1])
elif b < a :
for j in range(a-1, b-1, -1) :
v[j][-1][1] = i
a = b
if a > 0 :
for j in range(0, a) :
v[j][-1][1] = n
c = 0
for i in range(0, N) :
c += len(v[i])
print(c)
#for i in v :
# print(i)
for i in range(0, N) :
for j in range(0, len(v[i])) :
print(v[i][j][0] + 1, v[i][j][1])
|
VK Cup 2012 Round 3 (Unofficial Div. 2 Edition)
|
CF
| 2,012 | 2 | 256 |
Range Increments
|
Polycarpus is an amateur programmer. Now he is analyzing a friend's program. He has already found there the function rangeIncrement(l, r), that adds 1 to each element of some array a for all indexes in the segment [l, r]. In other words, this function does the following:
Polycarpus knows the state of the array a after a series of function calls. He wants to determine the minimum number of function calls that lead to such state. In addition, he wants to find what function calls are needed in this case. It is guaranteed that the required number of calls does not exceed 105.
Before calls of function rangeIncrement(l, r) all array elements equal zero.
|
The first input line contains a single integer n (1 ≤ n ≤ 105) — the length of the array a[1... n].
The second line contains its integer space-separated elements, a[1], a[2], ..., a[n] (0 ≤ a[i] ≤ 105) after some series of function calls rangeIncrement(l, r).
It is guaranteed that at least one element of the array is positive. It is guaranteed that the answer contains no more than 105 calls of function rangeIncrement(l, r).
|
Print on the first line t — the minimum number of calls of function rangeIncrement(l, r), that lead to the array from the input data. It is guaranteed that this number will turn out not more than 105.
Then print t lines — the descriptions of function calls, one per line. Each line should contain two integers li, ri (1 ≤ li ≤ ri ≤ n) — the arguments of the i-th call rangeIncrement(l, r). Calls can be applied in any order.
If there are multiple solutions, you are allowed to print any of them.
| null |
The first sample requires a call for the entire array, and four additional calls:
- one for the segment [2,2] (i.e. the second element of the array),
- three for the segment [5,5] (i.e. the fifth element of the array).
|
[{"input": "6\n1 2 1 1 4 1", "output": "5\n2 2\n5 5\n5 5\n5 5\n1 6"}, {"input": "5\n1 0 1 0 1", "output": "3\n1 1\n3 3\n5 5"}]
| 1,800 |
["data structures", "greedy"]
| 63 |
[{"input": "6\r\n1 2 1 1 4 1\r\n", "output": "5\r\n2 2\r\n5 5\r\n5 5\r\n5 5\r\n1 6\r\n"}, {"input": "5\r\n1 0 1 0 1\r\n", "output": "3\r\n1 1\r\n3 3\r\n5 5\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n1 1\r\n"}, {"input": "1\r\n100000\r\n", "output": "100000\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 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1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n"}, {"input": "5\r\n1 2 3 4 5\r\n", "output": "5\r\n5 5\r\n4 5\r\n3 5\r\n2 5\r\n1 5\r\n"}, {"input": "12\r\n0 1 1 1 3 4 3 3 3 3 2 2\r\n", "output": "4\r\n6 6\r\n5 10\r\n5 12\r\n2 12\r\n"}, {"input": "2\r\n1 1\r\n", "output": "1\r\n1 2\r\n"}, {"input": "2\r\n2 1\r\n", "output": "2\r\n1 1\r\n1 2\r\n"}, {"input": "2\r\n1 3\r\n", "output": "3\r\n2 2\r\n2 2\r\n1 2\r\n"}, {"input": "2\r\n2 4\r\n", "output": "4\r\n2 2\r\n2 2\r\n1 2\r\n1 2\r\n"}, {"input": "3\r\n1 1 1\r\n", "output": "1\r\n1 3\r\n"}, {"input": "3\r\n0 2 1\r\n", "output": "2\r\n2 2\r\n2 3\r\n"}, {"input": "3\r\n2 2 1\r\n", "output": "2\r\n1 2\r\n1 3\r\n"}, {"input": "3\r\n2 4 2\r\n", "output": "4\r\n2 2\r\n2 2\r\n1 3\r\n1 3\r\n"}, {"input": "5\r\n1 1 0 0 0\r\n", "output": "1\r\n1 2\r\n"}, {"input": "5\r\n0 0 1 1 0\r\n", "output": "1\r\n3 4\r\n"}, {"input": "5\r\n1 0 2 1 0\r\n", "output": "3\r\n1 1\r\n3 3\r\n3 4\r\n"}, {"input": "5\r\n2 1 2 3 3\r\n", "output": "4\r\n1 1\r\n4 5\r\n3 5\r\n1 5\r\n"}, {"input": "20\r\n4 5 4 4 3 2 2 1 2 2 2 3 3 4 2 2 2 1 1 1\r\n", "output": "8\r\n2 2\r\n1 4\r\n1 5\r\n1 7\r\n14 14\r\n12 14\r\n9 17\r\n1 20\r\n"}, {"input": "20\r\n1 6 8 9 10 10 11 11 10 10 9 6 6 6 6 4 3 2 1 0\r\n", "output": "11\r\n7 8\r\n5 10\r\n4 11\r\n3 11\r\n3 11\r\n2 15\r\n2 15\r\n2 16\r\n2 17\r\n2 18\r\n1 19\r\n"}, {"input": "20\r\n4 6 7 8 8 8 9 9 10 12 12 11 12 12 11 9 8 8 5 2\r\n", "output": "13\r\n10 11\r\n13 14\r\n10 15\r\n9 15\r\n7 16\r\n4 18\r\n3 18\r\n2 18\r\n2 19\r\n1 19\r\n1 19\r\n1 20\r\n1 20\r\n"}, {"input": "20\r\n2 2 4 5 5 6 7 6 5 5 7 6 4 3 3 3 3 3 3 1\r\n", "output": "9\r\n7 7\r\n6 8\r\n11 11\r\n11 12\r\n4 12\r\n3 13\r\n3 19\r\n1 19\r\n1 20\r\n"}, {"input": "20\r\n5 9 11 12 13 13 13 13 13 13 13 13 13 13 12 11 11 8 6 4\r\n", "output": "13\r\n5 14\r\n4 15\r\n3 17\r\n3 17\r\n2 17\r\n2 18\r\n2 18\r\n2 19\r\n1 19\r\n1 20\r\n1 20\r\n1 20\r\n1 20\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n = int(f.readline())
a = list(map(int, f.readline().split()))
# Compute minimal t
minimal_t = 0
prev = 0
for num in a:
if num > prev:
minimal_t += num - prev
prev = num
try:
with open(submission_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip()]
except:
print(0)
return
if not lines:
print(0)
return
try:
t = int(lines[0])
if t != minimal_t:
print(0)
return
intervals = []
for line in lines[1:t+1]:
l, r = map(int, line.split())
if l < 1 or r > n or l > r:
print(0)
return
intervals.append((l, r))
except:
print(0)
return
# Apply intervals
diff = [0] * (n + 2)
for l, r in intervals:
diff[l] += 1
if r + 1 <= n:
diff[r + 1] -= 1
current = 0
result = []
for i in range(1, n+1):
current += diff[i]
result.append(current)
if result == a:
print(1)
else:
print(0)
if __name__ == "__main__":
if len(sys.argv) != 4:
print("Usage: checker.py input_path output_path submission_output_path")
sys.exit(1)
main(sys.argv[1], sys.argv[2], sys.argv[3])
| true |
559/E
|
559
|
E
|
Python 3
|
TESTS
| 0 | 15 | 0 |
193545212
|
n = int(input())
a = [0] * (n + 1)
l = [0] * (n + 1)
for i in range(1, n + 1):
a[i], l[i] = map(int, input().split())
a[0] = 0 # add monument at position 0
l[0] = 0 # monument has no illumination
a, l = zip(*sorted(zip(a, l)))
r = 0 # rightmost illuminated position
total_length = 0 # total illuminated length
for i in range(1, n + 1):
if a[i] - l[i] > r: # segment from a[i] - l[i] to a[i] is illuminated
r = a[i] - l[i]
total_length += l[i]
elif a[i] > r: # segment from r to a[i] is illuminated
total_length += a[i] - r
r = a[i]
print(total_length)
| 90 | 373 | 20,684,800 |
163335825
|
import dataclasses
import sys
inf = float('inf')
# change stdout buffer size
buffer = open(1, 'w', 10**6)
# fast printing function
def print(*args, sep=' ', end='\n'):
buffer.write(sep.join(str(arg) for arg in args) + end)
# flush stdout
def flush():
buffer.flush()
def read_ints(index=None):
if index is not None:
return [(int(x), i + index) for i, x in enumerate(sys.stdin.readline().split())]
return [int(x) for x in sys.stdin.readline().split()]
def go(i, j, t):
pass
def solve():
n = int(input())
x = [0] * n
y = [0] * n
vs = set()
for i in range(n):
x[i], y[i] = read_ints()
vs.add(x[i] - y[i])
vs.add(x[i])
vs.add(x[i] + y[i])
v = list(vs)
v.sort()
nv = len(v)
num = dict()
for i in range(nv):
num[v[i]] = i
f = [0] * nv
w = [[] for _ in range(nv)]
for i in range(n):
a = num[x[i] - y[i]]
b = num[x[i]]
c = num[x[i] + y[i]]
f[b] = c
w[a].append(b)
dp = [[[0, 0] for _ in range(nv + 1)] for _ in range(nv + 1)]
for i in range(nv - 1, -1, -1):
for j in range(nv - 1, -1, -1):
for t in range(2):
if j < i:
dp[i][j][t] = dp[i][i][0]
continue
cur = dp[i + 1][j][t]
if (t == 0 or j > i) and f[i] > j:
cur = max(cur, dp[j + 1 if t else i + 1][f[i]][0] + v[f[i]] - v[j])
for k in w[i]:
if k > j:
cur = max(cur, dp[j + 1 if t else i + 1][k][1] + v[k] - v[j])
dp[i][j][t] = cur
print(dp[0][0][0])
flush()
if __name__ == '__main__':
solve()
|
Codeforces Round 313 (Div. 1)
|
CF
| 2,015 | 4 | 256 |
Gerald and Path
|
The main walking trail in Geraldion is absolutely straight, and it passes strictly from the north to the south, it is so long that no one has ever reached its ends in either of the two directions. The Geraldionians love to walk on this path at any time, so the mayor of the city asked the Herald to illuminate this path with a few spotlights. The spotlights have already been delivered to certain places and Gerald will not be able to move them. Each spotlight illuminates a specific segment of the path of the given length, one end of the segment is the location of the spotlight, and it can be directed so that it covers the segment to the south or to the north of spotlight.
The trail contains a monument to the mayor of the island, and although you can walk in either directions from the monument, no spotlight is south of the monument.
You are given the positions of the spotlights and their power. Help Gerald direct all the spotlights so that the total length of the illuminated part of the path is as much as possible.
|
The first line contains integer n (1 ≤ n ≤ 100) — the number of spotlights. Each of the n lines contains two space-separated integers, ai and li (0 ≤ ai ≤ 108, 1 ≤ li ≤ 108). Number ai shows how much further the i-th spotlight to the north, and number li shows the length of the segment it illuminates.
It is guaranteed that all the ai's are distinct.
|
Print a single integer — the maximum total length of the illuminated part of the path.
| null | null |
[{"input": "3\n1 1\n2 2\n3 3", "output": "5"}, {"input": "4\n1 2\n3 3\n4 3\n6 2", "output": "9"}]
| 3,000 |
["dp", "sortings"]
| 90 |
[{"input": "3\r\n1 1\r\n2 2\r\n3 3\r\n", "output": "5\r\n"}, {"input": "4\r\n1 2\r\n3 3\r\n4 3\r\n6 2\r\n", "output": "9\r\n"}, {"input": "5\r\n3 3\r\n4 1\r\n2 2\r\n0 3\r\n9 5\r\n", "output": "13\r\n"}, {"input": "5\r\n3 3\r\n4 3\r\n6 4\r\n2 3\r\n1 5\r\n", "output": "14\r\n"}, {"input": "5\r\n1 2\r\n7 5\r\n9 4\r\n5 1\r\n3 5\r\n", "output": "13\r\n"}, {"input": "5\r\n7 2\r\n3 5\r\n2 4\r\n8 1\r\n9 5\r\n", "output": "15\r\n"}, {"input": "5\r\n7 1\r\n5 5\r\n1 4\r\n4 4\r\n2 2\r\n", "output": "12\r\n"}, {"input": "5\r\n9 5\r\n2 4\r\n3 3\r\n5 2\r\n1 1\r\n", "output": "13\r\n"}, {"input": "3\r\n0 3\r\n3 3\r\n6 3\r\n", "output": "9\r\n"}, {"input": "3\r\n0 3\r\n4 3\r\n7 3\r\n", "output": "9\r\n"}, {"input": "10\r\n78329099 25986078\r\n9003418 30942874\r\n32350045 8429148\r\n78842461 58122669\r\n89820027 42334842\r\n76809240 3652872\r\n77832962 54942701\r\n76760300 50934062\r\n53414406 14348704\r\n3119584 40577983\r\n", "output": "168539695\r\n"}, {"input": "10\r\n7117 86424\r\n87771 51337\r\n12429 34872\r\n53590 17922\r\n54806 13188\r\n8575 11567\r\n73589 76161\r\n71136 14076\r\n85527 6121\r\n83455 12523\r\n", "output": "227599\r\n"}, {"input": "10\r\n228 4\r\n833 58\r\n27 169\r\n775 658\r\n981 491\r\n979 310\r\n859 61\r\n740 324\r\n747 126\r\n785 410\r\n", "output": "1524\r\n"}, {"input": "4\r\n66 61\r\n715 254\r\n610 297\r\n665 41\r\n", "output": "653\r\n"}, {"input": "5\r\n44326737 210514\r\n61758935 9618\r\n34426105 9900632\r\n34195486 5323398\r\n28872088 135139\r\n", "output": "15579301\r\n"}, {"input": "5\r\n44549379 754619\r\n29429248 66713\r\n88414664 12793\r\n37846422 8417174\r\n38662784 5886595\r\n", "output": "15137894\r\n"}, {"input": "1\r\n100 50\r\n", "output": "50\r\n"}, {"input": "20\r\n22164537 5600930\r\n22164533 5600930\r\n22164538 5600930\r\n22164526 5600930\r\n22164527 5600930\r\n22164539 5600930\r\n22164528 5600930\r\n22164542 5600930\r\n22164544 5600930\r\n22164543 5600930\r\n22164530 5600930\r\n22164529 5600930\r\n22164536 5600930\r\n22164540 5600930\r\n22164531 5600930\r\n22164541 5600930\r\n22164535 5600930\r\n22164534 5600930\r\n22164525 5600930\r\n22164532 5600930\r\n", "output": "11201879\r\n"}, {"input": "5\r\n7339431 13372\r\n11434703 8326\r\n9158453 15156\r\n8266053 926622\r\n8286111 872342\r\n", "output": "1835818\r\n"}, {"input": "5\r\n23742227 754619\r\n8622096 66713\r\n37249276 12793\r\n17039270 8417174\r\n17855632 5886595\r\n", "output": "15137894\r\n"}, {"input": "10\r\n200 100\r\n1000100 1000000\r\n1000200 1000000\r\n2000100 89\r\n1000155 13\r\n1000159 1\r\n1000121 12\r\n1000111 1\r\n1000105 3\r\n1000195 13\r\n", "output": "2000089\r\n"}]
| false |
stdio
| null | true |
562/F
|
566
|
A
|
Python 3
|
TESTS
| 2 | 109 | 307,200 |
67346273
|
class Tree():
def __init__(self, value, index):
self.value = value
self.left = None
self.right = None
self.count = 1
self.index = index
def insert(self, value, index):
if value == self.value:
self.count += 1
return
if value < self.value:
if self.left is None:
self.left = Tree(value, index)
return
self.left.insert(value, index)
if value > self.value:
if self.right is None:
self.right = Tree(value, index)
return
self.right.insert(value, index)
@staticmethod
def largest_common_prefix(str1, str2):
common = 0
for i in range(len(str1)):
if str1[i] == str2[i]:
common += 1
else: return common
return len(str1)
def lnr(self):
if self.left is not None:
self.left.lnr()
print (self.value, self.index, self.count)
if self.right is not None:
self.right.lnr()
def search(self, value):
root_prefix = self.largest_common_prefix(self.value, value)
if value == self.value:
self.count -= 1
return len(value), self.index
if value > self.value: #kiem tra ben phai
if self.right: #coi co node phai khong
right_prefix = self.largest_common_prefix(self.right.value, value)
if root_prefix > right_prefix:
if self.count > 0:
self.count -= 1
return root_prefix, self.index
else:
return 0, 0
else:
return self.right.search(value)
else: #khong co thi thang similar nhat la thang root
if self.count > 0:
self.count -= 1
return root_prefix, self.index
else:
return 0, 0
if value < self.value: #kiem tra ben trai
if self.left: #coi co node trai khong
left_prefix = self.largest_common_prefix(self.left.value, value)
if root_prefix > left_prefix:
if self.count > 0:
self.count -= 1
return root_prefix, self.index
else:
return 0, 0
else:
return self.left.search(value)
else:
if self.count > 0:
self.count -= 1
return root_prefix, self.index
else:
return 0, 0
# @staticmethod
def count_nhanh(self):
if self.count == 1:
return True
if self.left:
return self.left.count_nhanh()
if self.right:
return self.right.count_nhanh()
return False
def search_zero(self, value):
if self.right:
check_right = self.right.count_nhanh()
else:
check_right = False
if self.left:
check_left = self.left.count_nhanh()
else:
check_left = False
if check_left == check_right == False and self.count == 1:
return self.index
if value > self.value:
if check_right == True:
return self.right.search_zero(value)
else:
return self.left.search_zero(value)
if value < self.value:
if check_left == True:
return self.left.search_zero(value)
else:
return self.right.search_zero(value)
if __name__ == "__main__":
n = int(input())
tree = Tree(input(), 1)
for i in range(n-1):
tree.insert(input(), i+2)
save = {}
queue = []
for i in range(n):
pseudonyms = input()
value, index = tree.search(pseudonyms)
if value == 0:
queue.append([pseudonyms, i+1])
else:
save["{} {}".format(index, i+1)] = value
for remain in queue:
index = tree.search_zero(remain[0])
save["{} {}".format(index, remain[1])] = 0
print (sum(save.values()))
for i,v in enumerate(save):
if i % 2 == 0:
print (v)
else:
print (v[::-1])
| 38 | 1,372 | 262,553,600 |
12334711
|
import sys
class Node:
def __init__(self, d):
global nodes
self.ch = {}
self.a = [[], []]
self.d = d
nodes += [self]
nodes = []
pairs = []
res = 0
N = int(sys.stdin.readline())
_input = sys.stdin.readlines()
_input = [s[:-1] for s in _input]
A = [_input[:N], _input[N:]]
T = Node(0)
for i, l in enumerate(A):
for j, s in enumerate(l):
t = T
for c in s:
v = ord(c) - ord('a')
if not v in t.ch:
t.ch[v] = Node(t.d + 1)
t = t.ch[v]
t.a[i] += [j + 1]
for n in reversed(nodes):
for i in n.ch:
n.a[0] += n.ch[i].a[0]
n.a[1] += n.ch[i].a[1]
k = min(len(n.a[0]), len(n.a[1]))
for i in range(k):
pairs += [str(n.a[0][-1]) + ' ' + str(n.a[1][-1])]
n.a[0].pop()
n.a[1].pop()
res += n.d
print(res)
print('\n'.join(pairs))
|
VK Cup 2015 - Finals
|
CF
| 2,015 | 2 | 256 |
Matching Names
|
Teachers of one programming summer school decided to make a surprise for the students by giving them names in the style of the "Hobbit" movie. Each student must get a pseudonym maximally similar to his own name. The pseudonym must be a name of some character of the popular saga and now the teachers are busy matching pseudonyms to student names.
There are n students in a summer school. Teachers chose exactly n pseudonyms for them. Each student must get exactly one pseudonym corresponding to him. Let us determine the relevance of a pseudonym b to a student with name a as the length of the largest common prefix a and b. We will represent such value as $$\operatorname{lcp}(a,b)$$. Then we can determine the quality of matching of the pseudonyms to students as a sum of relevances of all pseudonyms to the corresponding students.
Find the matching between students and pseudonyms with the maximum quality.
|
The first line contains number n (1 ≤ n ≤ 100 000) — the number of students in the summer school.
Next n lines contain the name of the students. Each name is a non-empty word consisting of lowercase English letters. Some names can be repeating.
The last n lines contain the given pseudonyms. Each pseudonym is a non-empty word consisting of small English letters. Some pseudonyms can be repeating.
The total length of all the names and pseudonyms doesn't exceed 800 000 characters.
|
In the first line print the maximum possible quality of matching pseudonyms to students.
In the next n lines describe the optimal matching. Each line must have the form a b (1 ≤ a, b ≤ n), that means that the student who was number a in the input, must match to the pseudonym number b in the input.
The matching should be a one-to-one correspondence, that is, each student and each pseudonym should occur exactly once in your output. If there are several optimal answers, output any.
| null |
The first test from the statement the match looks as follows:
- bill → bilbo (lcp = 3)
- galya → galadriel (lcp = 3)
- gennady → gendalf (lcp = 3)
- toshik → torin (lcp = 2)
- boris → smaug (lcp = 0)
|
[{"input": "5\ngennady\ngalya\nboris\nbill\ntoshik\nbilbo\ntorin\ngendalf\nsmaug\ngaladriel", "output": "11\n4 1\n2 5\n1 3\n5 2\n3 4"}]
| 2,300 |
[]
| 38 |
[{"input": "5\r\ngennady\r\ngalya\r\nboris\r\nbill\r\ntoshik\r\nbilbo\r\ntorin\r\ngendalf\r\nsmaug\r\ngaladriel\r\n", "output": "11\r\n4 1\r\n2 5\r\n1 3\r\n5 2\r\n3 4\r\n"}, {"input": "1\r\na\r\na\r\n", "output": "1\r\n1 1\r\n"}, {"input": "2\r\na\r\na\r\na\r\na\r\n", "output": "2\r\n1 1\r\n2 2\r\n"}, {"input": "2\r\na\r\nb\r\na\r\na\r\n", "output": "1\r\n1 1\r\n2 2\r\n"}, {"input": "2\r\nb\r\nb\r\na\r\na\r\n", "output": "0\r\n1 1\r\n2 2\r\n"}, {"input": "2\r\na\r\nb\r\na\r\nb\r\n", "output": "2\r\n1 1\r\n2 2\r\n"}, {"input": "10\r\nbaa\r\na\r\nba\r\naabab\r\naa\r\nbaab\r\nbb\r\nabbbb\r\na\r\na\r\na\r\nba\r\nba\r\nbaabbb\r\nba\r\na\r\naabb\r\nbaa\r\nab\r\nb\r\n", "output": "17\r\n4 7\r\n8 9\r\n2 1\r\n9 6\r\n6 4\r\n1 8\r\n3 2\r\n7 10\r\n10 3\r\n5 5\r\n"}, {"input": "10\r\nabaabbaaa\r\nacccccaacabc\r\nacbaabaaabbca\r\naaccca\r\ncbbba\r\naaba\r\nacab\r\nac\r\ncbac\r\nca\r\nbbbbc\r\nbacbcbcaac\r\nc\r\ncba\r\na\r\nabba\r\nbcabc\r\nabcccaa\r\nab\r\na\r\n", "output": "10\r\n1 9\r\n6 5\r\n4 10\r\n8 6\r\n7 8\r\n9 4\r\n10 3\r\n3 2\r\n2 1\r\n5 7\r\n"}, {"input": "1\r\nzzzz\r\nyyx\r\n", "output": "0\r\n1 1\r\n"}, {"input": "1\r\naa\r\naaa\r\n", "output": "2\r\n1 1\r\n"}, {"input": "1\r\naaa\r\naa\r\n", "output": "2\r\n1 1\r\n"}, {"input": "10\r\nb\r\nb\r\na\r\na\r\na\r\na\r\nb\r\nb\r\na\r\nb\r\nb\r\na\r\na\r\na\r\nb\r\nb\r\nb\r\na\r\nb\r\nb\r\n", "output": "9\r\n3 2\r\n4 3\r\n5 4\r\n6 8\r\n1 1\r\n2 5\r\n7 6\r\n8 7\r\n10 9\r\n9 10\r\n"}, {"input": "10\r\na\r\nb\r\na\r\na\r\nc\r\na\r\na\r\na\r\na\r\na\r\nb\r\nc\r\nc\r\na\r\nc\r\nb\r\na\r\na\r\na\r\nc\r\n", "output": "6\r\n1 4\r\n3 7\r\n4 8\r\n6 9\r\n2 1\r\n5 2\r\n7 6\r\n8 3\r\n9 5\r\n10 10\r\n"}, {"input": "10\r\nw\r\nr\r\na\r\nc\r\nx\r\ne\r\nb\r\nx\r\nw\r\nx\r\nz\r\ng\r\nd\r\ny\r\ns\r\ny\r\nj\r\nh\r\nl\r\nu\r\n", "output": "0\r\n3 3\r\n7 2\r\n4 8\r\n6 7\r\n2 9\r\n1 5\r\n9 10\r\n5 4\r\n8 6\r\n10 1\r\n"}]
| false |
stdio
|
import sys
def lcp(s, t):
min_len = min(len(s), len(t))
cnt = 0
for i in range(min_len):
if s[i] == t[i]:
cnt += 1
else:
break
return cnt
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path, 'r') as f:
n = int(f.readline())
students = [f.readline().strip() for _ in range(n)]
pseudonyms = [f.readline().strip() for _ in range(n)]
# Read reference sum
with open(output_path, 'r') as f:
ref_sum_line = f.readline().strip()
if not ref_sum_line:
print(0)
return
try:
reference_sum = int(ref_sum_line)
except ValueError:
print(0)
return
# Read submission
with open(submission_path, 'r') as f:
# Read submitted sum
sum_line = f.readline().strip()
if not sum_line:
print(0)
return
try:
submitted_sum = int(sum_line)
except ValueError:
print(0)
return
if submitted_sum != reference_sum:
print(0)
return
# Read the pairs
a_list = []
b_list = []
a_set = set()
b_set = set()
for _ in range(n):
line = f.readline().strip()
parts = line.split()
if len(parts) != 2:
print(0)
return
try:
a = int(parts[0])
b = int(parts[1])
except ValueError:
print(0)
return
if a < 1 or a > n or b < 1 or b > n:
print(0)
return
if a in a_set or b in b_set:
print(0)
return
a_set.add(a)
b_set.add(b)
a_list.append(a)
b_list.append(b)
# Check if all a and b are present
if len(a_set) != n or len(b_set) != n:
print(0)
return
# Compute sum
computed_sum = 0
for a, b in zip(a_list, b_list):
s = students[a-1]
p = pseudonyms[b-1]
computed_sum += lcp(s, p)
if computed_sum != submitted_sum:
print(0)
return
# All checks passed
print(1)
if __name__ == "__main__":
main()
| true |
878/E
|
878
|
E
|
Python 3
|
TESTS
| 2 | 93 | 512,000 |
32122313
|
from random import randint
def find_max_ind(arr):
max_el, max_ind = arr[0], 0
for i in range(1, len(arr)):
el = arr[i]
if el >= max_el:
max_el = el
max_ind = i
return max_ind
def calc_max_prod(arr, l):
if l == 1:
return arr[0]
if l == 2:
a = arr[0] + arr[1] * 2
return a
max_ind = find_max_ind(arr)
if max_ind == 0:
a = calc_max_prod([calc_max_prod(arr[:2], 2)] + arr[2:], l - 1)
return a
before = arr[:max_ind - 1]
calc = arr[max_ind - 1:max_ind + 1]
after = arr[max_ind + 1:]
a = calc_max_prod(before + [calc_max_prod(calc, 2)] + after, l - 1)
return a % (10 ** 9 + 7)
def calc_max_prod1(arr, l):
if l == 1:
return arr[0]
if l == 2:
return arr[0] + arr[1] * 2
max_prod = -float("inf")
for i in range(l - 1):
prod = calc_max_prod(arr[:i] + [calc_max_prod(arr[i:i + 2], 2)] + arr[i + 2:], l - 1)
max_prod = max(prod, max_prod)
return max_prod % (10 ** 9 + 7)
def gen_test():
n, q = randint(1, 5), randint(1, 10)
arr1 = []
for i in range(n):
arr1.append(randint(1, 100))
lri = []
for i in range(q):
li = randint(1, n - 2)
ri = randint(li + 1, n)
lri.append([li, ri])
test = [[n, q], arr1, lri]
return test
def test(number_tests):
for i in range(number_tests):
test = gen_test()
n, q = test[0]
arr = test[1]
lri = test[2]
for l in lri:
li, ri = l
res1 = calc_max_prod(arr[li - 1:ri], ri - li + 1)
res2 = calc_max_prod1(arr[li - 1:ri], ri - li + 1)
if res1 != res2:
print("Incorretc answer!!!")
print("Test:")
print(n, q)
print(arr)
print(li, ri)
print("First answer: ")
print(res1)
print("Correct answer: ")
print(res2)
def targ():
n, q = list(map(int, input().split()))
nums = list(map(int, input().split()))
rem = [list(map(int, input().split())) for i in range(q)]
for i in range(q):
li, ri = rem[i]
print(calc_max_prod1(nums[li - 1:ri], ri - li + 1))
DEBUG = False
NUMBER_TESTS = 10
if DEBUG:
test(NUMBER_TESTS)
else:
targ()
| 55 | 999 | 80,281,600 |
303302080
|
import sys
MOD = 10**9 + 7
INF = 2 * 10**9 + 5
def upd(a):
a += (a >> 31) & MOD
return a
def init(n):
_2 = [1] * (n + 1)
for i in range(1, n + 1):
_2[i] = (_2[i-1] << 1) % MOD
return _2
def find(x, fa):
if fa[x] != x:
fa[x] = find(fa[x], fa)
return fa[x]
def merge(x, y, fa, L, s, f, _2):
fa[y] = x
if (L[x] > 30 and s[x] > 0) or (s[x] + (s[y] << L[x]) >= INF):
s[x] = INF
else:
s[x] += s[y] << L[x]
f[x] = (f[x] + f[y] * _2[L[x]]) % MOD
L[x] += L[y]
def calc(l, r, pr, _2):
return (pr[l] - pr[r+1] * _2[r-l+1] % MOD + MOD) % MOD
def main():
input = sys.stdin.read
data = input().split()
idx = 0
n = int(data[idx])
m = int(data[idx+1])
idx += 2
a = [0] * (n + 1)
for i in range(1, n+1):
a[i] = int(data[idx])
idx += 1
_2 = init(n)
pr = [0] * (n + 2)
for i in range(n, 0, -1):
pr[i] = ((pr[i+1] << 1) + a[i] + MOD) % MOD
fa = [i for i in range(n+1)]
L = [1] * (n+1)
f = [0] * (n+1)
s = [0] * (n+1)
for i in range(1, n+1):
f[i] = s[i] = a[i]
vec = [[] for _ in range(n+1)]
for i in range(1, m+1):
l = int(data[idx])
r = int(data[idx+1])
idx += 2
vec[r].append((l, i))
ans = [0] * (m+1)
ps = [0] * (n+1)
for i in range(1, n+1):
while find(i, fa) > 1 and s[find(i, fa)] >= 0:
merge(find(find(i, fa)-1, fa), find(i, fa), fa, L, s, f, _2)
ps[find(i, fa)] = (ps[find(find(i, fa)-1, fa)] + f[find(i, fa)]) % MOD
for q in vec[i]:
l, id = q
ans[id] = ((ps[find(i, fa)] - ps[find(l, fa)] + MOD) * 2 + calc(l, find(l, fa) + L[find(l, fa)] - 1, pr, _2) + MOD) % MOD
for i in range(1, m+1):
print(ans[i])
if __name__ == "__main__":
main()
|
Codeforces Round 443 (Div. 1)
|
CF
| 2,017 | 2 | 512 |
Numbers on the blackboard
|
A sequence of n integers is written on a blackboard. Soon Sasha will come to the blackboard and start the following actions: let x and y be two adjacent numbers (x before y), then he can remove them and write x + 2y instead of them. He will perform these operations until one number is left. Sasha likes big numbers and will get the biggest possible number.
Nikita wants to get to the blackboard before Sasha and erase some of the numbers. He has q options, in the option i he erases all numbers to the left of the li-th number and all numbers to the right of ri-th number, i. e. all numbers between the li-th and the ri-th, inclusive, remain on the blackboard. For each of the options he wants to know how big Sasha's final number is going to be. This number can be very big, so output it modulo 109 + 7.
|
The first line contains two integers n and q (1 ≤ n, q ≤ 105) — the number of integers on the blackboard and the number of Nikita's options.
The next line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the sequence on the blackboard.
Each of the next q lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n), describing Nikita's options.
|
For each option output Sasha's result modulo 109 + 7.
| null |
In the second sample Nikita doesn't erase anything. Sasha first erases the numbers 1 and 2 and writes 5. Then he erases 5 and -3 and gets -1. -1 modulo 109 + 7 is 109 + 6.
|
[{"input": "3 3\n1 2 3\n1 3\n1 2\n2 3", "output": "17\n5\n8"}, {"input": "3 1\n1 2 -3\n1 3", "output": "1000000006"}, {"input": "4 2\n1 1 1 -1\n1 4\n3 4", "output": "5\n1000000006"}]
| 3,300 |
["combinatorics", "dp"]
| 55 |
[{"input": "3 3\r\n1 2 3\r\n1 3\r\n1 2\r\n2 3\r\n", "output": "17\r\n5\r\n8\r\n"}, {"input": "3 1\r\n1 2 -3\r\n1 3\r\n", "output": "1000000006\r\n"}, {"input": "4 2\r\n1 1 1 -1\r\n1 4\r\n3 4\r\n", "output": "5\r\n1000000006\r\n"}, {"input": "4 1\r\n1 1 -3 1\r\n1 4\r\n", "output": "1\r\n"}, {"input": "1 1\r\n1000000000\r\n1 1\r\n", "output": "1000000000\r\n"}, {"input": "1 1\r\n-1000000000\r\n1 1\r\n", "output": "7\r\n"}, {"input": "2 3\r\n0 0\r\n1 2\r\n1 1\r\n2 2\r\n", "output": "0\r\n0\r\n0\r\n"}, {"input": "2 3\r\n1000000000 1000000000\r\n1 2\r\n1 1\r\n2 2\r\n", "output": "999999986\r\n1000000000\r\n1000000000\r\n"}, {"input": "10 10\r\n-7 4 4 -5 2 3 -9 7 -4 -2\r\n8 10\r\n8 9\r\n2 3\r\n2 9\r\n2 3\r\n7 8\r\n1 3\r\n2 6\r\n6 8\r\n3 3\r\n", "output": "1000000002\r\n1000000006\r\n12\r\n208\r\n12\r\n5\r\n17\r\n56\r\n13\r\n4\r\n"}, {"input": "10 10\r\n593536087 56559483 -439122178 -126803734 606390399 -809361217 444436245 71742850 -477364598 -818526589\r\n5 9\r\n5 7\r\n5 5\r\n1 9\r\n5 10\r\n2 2\r\n3 10\r\n4 5\r\n2 4\r\n4 4\r\n", "output": "384626549\r\n765412945\r\n606390399\r\n410699067\r\n747573385\r\n56559483\r\n72910946\r\n85977057\r\n924707673\r\n873196273\r\n"}, {"input": "10 10\r\n-616555628 133372392 -749502876 499498544 927177575 -838173566 -139786799 -676011158 155638259 102225904\r\n1 3\r\n3 3\r\n1 2\r\n2 10\r\n5 7\r\n4 5\r\n6 6\r\n8 9\r\n3 5\r\n7 7\r\n", "output": "151183418\r\n250497131\r\n650189163\r\n270536548\r\n971256859\r\n353853680\r\n161826441\r\n635265367\r\n958204491\r\n860213208\r\n"}, {"input": "10 1\r\n-11 81 -4 79 44 -11 -50 26 -38 13\r\n4 10\r\n", "output": "129\r\n"}, {"input": "10 1\r\n-858350203 62991893 757167826 -643742467 -122005341 210910071 973749788 -554405426 91646398 811009699\r\n1 6\r\n", "output": "208079143\r\n"}, {"input": "101 1\r\n-2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 1000000000\r\n1 101\r\n", "output": "165401054\r\n"}, {"input": "108 1\r\n-1 -1 -1 -2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1000000000\r\n1 108\r\n", "output": "171334758\r\n"}]
| false |
stdio
| null | true |
30/C
|
30
|
C
|
Python 3
|
TESTS
| 4 | 186 | 0 |
7282597
|
def is_reachable(from_state, target):
distance = ((from_state[0]-target[0])**2 + (from_state[1]-target[1])**2)**0.5
time_diff = target[2] - from_state[2]
return time_diff >= distance
num_iter = int(input())
targets = sorted([list(map(float, input().strip().split(' '))) for dummy in range(0, num_iter)], key=lambda single_target: single_target[2])
states = []
max_score = -1
for target in targets:
is_initial_target = True
for i, state in enumerate(states):
if is_reachable(state, target):
is_initial_target = False
score = states[i][3] + target[3]
states[i] = target
states[i][3] = score
if is_initial_target:
states.append(target)
score = target[3]
max_score = (max_score < score) and score or max_score
print(max_score)
| 50 | 810 | 2,560,000 |
56455652
|
F = lambda: map(float, input().split())
n = int(input())
arr = []
for i in range(n):
x, y, t, s = F()
arr.append([int(x), int(y), int(t), s])
arr.sort(key=lambda x: x[2])
dp = [0] * n
for i in range(n):
dp[i] = arr[i][-1]
for j in range(i):
if (arr[i][0] - arr[j][0]) ** 2 + (arr[i][1] - arr[j][1]) ** 2 <= (arr[i][2] - arr[j][2]) ** 2:
dp[i] = max(dp[i], dp[j] + arr[i][-1])
print(max(dp))
|
Codeforces Beta Round 30 (Codeforces format)
|
CF
| 2,010 | 2 | 256 |
Shooting Gallery
|
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize — big pink plush panda. The king is not good at shooting, so he invited you to help him.
The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it.
A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
|
The first line contains integer n (1 ≤ n ≤ 1000) — amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti — integers, - 1000 ≤ xi, yi ≤ 1000, 0 ≤ ti ≤ 109, real number pi is given with no more than 6 digits after the decimal point, 0 ≤ pi ≤ 1). No two targets may be at the same point.
|
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
| null | null |
[{"input": "1\n0 0 0 0.5", "output": "0.5000000000"}, {"input": "2\n0 0 0 0.6\n5 0 5 0.7", "output": "1.3000000000"}]
| 1,800 |
["dp", "probabilities"]
| 50 |
[{"input": "1\r\n0 0 0 0.5\r\n", "output": "0.5000000000\r\n"}, {"input": "2\r\n0 0 0 0.6\r\n5 0 5 0.7\r\n", "output": "1.3000000000\r\n"}, {"input": "1\r\n-5 2 3 0.886986\r\n", "output": "0.8869860000\r\n"}, {"input": "4\r\n10 -7 14 0.926305\r\n-7 -8 12 0.121809\r\n-7 7 14 0.413446\r\n3 -8 6 0.859061\r\n", "output": "1.7853660000\r\n"}, {"input": "5\r\n-2 -2 34 0.127276\r\n5 -5 4 0.459998\r\n10 3 15 0.293766\r\n1 -3 7 0.089869\r\n-4 -7 11 0.772515\r\n", "output": "0.8997910000\r\n"}, {"input": "5\r\n2 5 1 0.955925\r\n9 -9 14 0.299977\r\n0 1 97 0.114582\r\n-4 -2 66 0.561033\r\n0 -10 75 0.135937\r\n", "output": "1.7674770000\r\n"}, {"input": "10\r\n-4 7 39 0.921693\r\n3 -1 50 0.111185\r\n-2 -8 27 0.976475\r\n-9 -2 25 0.541029\r\n6 -4 21 0.526054\r\n-7 2 19 0.488637\r\n-6 -5 50 0.819011\r\n-7 3 39 0.987596\r\n-3 -8 16 0.685997\r\n4 10 1 0.246686\r\n", "output": "3.0829590000\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(output_path, 'r') as f:
correct_line = f.readline().strip()
try:
correct = float(correct_line)
except:
print(0)
return
with open(submission_path, 'r') as f:
sub_line = f.readline().strip()
try:
sub = float(sub_line)
except:
print(0)
return
if abs(correct - sub) <= 1e-6 + 1e-12: # Adding small epsilon for floating point errors
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
1000/D
|
1000
|
D
|
PyPy 3
|
TESTS
| 2 | 93 | 3,584,000 |
147568159
|
import sys
input = sys.stdin.readline
MOD = 998244353
N = 10005
fact = [1]
for i in range(1, N):
fact.append(fact[-1] * i % MOD)
factinv = [0] * N
factinv[N - 1] = pow(fact[N - 1], MOD - 2, MOD)
for j in range(N - 2, -1, -1):
factinv[j] = factinv[j + 1] * (j + 1) % MOD
def nCk(n, k):
if k < 0 or n < k:
return 0
return fact[n] * factinv[n - k] * factinv[k] % MOD
n = int(input())
a = list(map(int, input().split()))
dp = [0] * (n + 1)
dp[n] = 1
for j in range(n - 1, -1, -1):
if a[j] <= 0:
continue
for k in range(j + a[j] + 1, n + 1):
dp[j] += dp[k] * nCk(k - j - 1, a[j]) % MOD
print(sum(dp[:n]))
| 22 | 951 | 1,126,400 |
39719619
|
from copy import deepcopy
import itertools
from bisect import bisect_left
from bisect import bisect_right
import math
from collections import deque
def read():
return int(input())
def readmap():
return map(int, input().split())
def readlist():
return list(map(int, input().split()))
MAX = 10000
MOD = 998244353
fac = [0] * MAX
finv = [0] * MAX
inv = [0] * MAX
def COMinit():
fac[0] = 1
fac[1] = 1
finv[0] = 1
finv[1] = 1
inv[1] = 1;
for i in range(2, MAX):
fac[i] = fac[i - 1] * i % MOD
inv[i] = MOD - inv[MOD%i] * (MOD // i) % MOD
finv[i] = finv[i - 1] * inv[i] % MOD
return None
def COM(n, k):
if n < k:
return 0
if n < 0 or k < 0:
return 0
return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD
N = read()
A = readlist()
for i in range(N):
A[i] = max(0, A[i])
if A[i] > N - i - 1:
A[i] = 0
COMinit()
dp = [0] * N
for n in range(1, N):
dp[n] = dp[n-1]
for i in range(n):
if i == 0:
dp[n] = (dp[n] + COM(n - i - 1, A[i] - 1)) % MOD
else:
dp[n] = (dp[n] + COM(n - i - 1, A[i] - 1) * (dp[i - 1] + 1)) % MOD
print(dp[N - 1])
|
Educational Codeforces Round 46 (Rated for Div. 2)
|
ICPC
| 2,018 | 2 | 256 |
Yet Another Problem On a Subsequence
|
The sequence of integers $$$a_1, a_2, \dots, a_k$$$ is called a good array if $$$a_1 = k - 1$$$ and $$$a_1 > 0$$$. For example, the sequences $$$[3, -1, 44, 0], [1, -99]$$$ are good arrays, and the sequences $$$[3, 7, 8], [2, 5, 4, 1], [0]$$$ — are not.
A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences $$$[2, -3, 0, 1, 4]$$$, $$$[1, 2, 3, -3, -9, 4]$$$ are good, and the sequences $$$[2, -3, 0, 1]$$$, $$$[1, 2, 3, -3 -9, 4, 1]$$$ — are not.
For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.
|
The first line contains the number $$$n~(1 \le n \le 10^3)$$$ — the length of the initial sequence. The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n~(-10^9 \le a_i \le 10^9)$$$ — the sequence itself.
|
In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.
| null |
In the first test case, two good subsequences — $$$[a_1, a_2, a_3]$$$ and $$$[a_2, a_3]$$$.
In the second test case, seven good subsequences — $$$[a_1, a_2, a_3, a_4], [a_1, a_2], [a_1, a_3], [a_1, a_4], [a_2, a_3], [a_2, a_4]$$$ and $$$[a_3, a_4]$$$.
|
[{"input": "3\n2 1 1", "output": "2"}, {"input": "4\n1 1 1 1", "output": "7"}]
| 1,900 |
["combinatorics", "dp"]
| 22 |
[{"input": "3\r\n2 1 1\r\n", "output": "2\r\n"}, {"input": "4\r\n1 1 1 1\r\n", "output": "7\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
487/B
|
487
|
B
|
PyPy 3
|
TESTS
| 2 | 77 | 20,172,800 |
124967705
|
def adjust(A, l, s):
B = []
m, M = A[0], A[0]
for i in range(1, n):
m = min(A[i], m)
M = max(A[i], M)
if M - m > s:
B.append(i - 1)
m = M = A[i]
count = len(B) + 1
r = n - 1
m, M = A[-1], A[-1]
for i in reversed(range(n - 1)):
m = min(A[i], m)
M = max(A[i], M)
if len(B) != 0 and i == B[-1]:
size = r - B[-1]
if size < l:
m = min(A[i], m)
M = max(A[i], M)
if M - m <= s:
B[-1] -= 1
else:
return -1
else:
B.pop()
r = i
m, M = A[i], B[i]
else:
return count if r >= l else -1
n, s, l = map(int, input().split(' '))
A = list(map(int, input().split(' ')))
print(adjust(A, l, s))
| 35 | 280 | 30,208,000 |
134855235
|
from collections import deque
# f[i] = min(f[k]) + 1 for g[i] - 1 <= k <= i - l
# [g[i], i] is a valid seg
# sliding window of max/min elements
"""
debug qmin (increasing) qmax (decreasing), [-1] is max/min
debug deque([1]) deque([1])
debug deque([1, 2]) deque([2])
debug deque([3]) deque([2, 3])
debug deque([3, 4]) deque([2, 4])
debug deque([3, 4, 5]) deque([5])
debug deque([6]) deque([5, 6])
debug deque([6, 7]) deque([7])
g[i + 1] <= g[i] + 1 so it's also a sliding window minimum problem.
"""
def solve(N, S, L, A):
qmax, qmin = deque(), deque()
F = [N + 1] * (N + 1)
F[0] = 0
q = deque()
j = 0 # leftIdx
for i in range(1, N + 1):
x = A[i]
# maintain monotonic queue
while qmax and A[qmax[-1]] <= x: # increasing
qmax.pop()
qmax.append(i)
while qmin and A[qmin[-1]] >= x: # decreasing
qmin.pop()
qmin.append(i)
# increment j until max-min <= S
while j < i and qmax and qmin and A[qmax[0]] - A[qmin[0]] > S:
while qmax and qmax[0] <= j:
qmax.popleft()
while qmin and qmin[0] <= j:
qmin.popleft()
j += 1
# [j,i] is the longest segment now
if i >= L:
# insert i-L
while q and q[-1][1] >= F[i - L]:
q.pop()
q.append((i - L, F[i - L]))
# j - 1 <= k <= i - L
while q and q[0][0] < j - 1:
q.popleft()
if q:
F[i] = q[0][1] + 1
else:
F[i] = N + 1
return F[-1] if F[-1] <= N else -1
if __name__ == "__main__":
N, S, L = (int(_) for _ in input().split())
A = [0] + [int(_) for _ in input().split()]
print(solve(N, S, L, A))
|
Codeforces Round 278 (Div. 1)
|
CF
| 2,014 | 1 | 256 |
Strip
|
Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right.
Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:
- Each piece should contain at least l numbers.
- The difference between the maximal and the minimal number on the piece should be at most s.
Please help Alexandra to find the minimal number of pieces meeting the condition above.
|
The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105).
The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109).
|
Output the minimal number of strip pieces.
If there are no ways to split the strip, output -1.
| null |
For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2].
For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists.
|
[{"input": "7 2 2\n1 3 1 2 4 1 2", "output": "3"}, {"input": "7 2 2\n1 100 1 100 1 100 1", "output": "-1"}]
| 2,000 |
["binary search", "data structures", "dp", "two pointers"]
| 35 |
[{"input": "7 2 2\r\n1 3 1 2 4 1 2\r\n", "output": "3\r\n"}, {"input": "7 2 2\r\n1 100 1 100 1 100 1\r\n", "output": "-1\r\n"}, {"input": "1 0 1\r\n0\r\n", "output": "1\r\n"}, {"input": "6 565 2\r\n31 76 162 -182 -251 214\r\n", "output": "1\r\n"}, {"input": "1 0 1\r\n0\r\n", "output": "1\r\n"}, {"input": "1 0 1\r\n-1000000000\r\n", "output": "1\r\n"}, {"input": "1 100 2\r\n42\r\n", "output": "-1\r\n"}, {"input": "2 1000000000 1\r\n-1000000000 1000000000\r\n", "output": "2\r\n"}, {"input": "2 1000000000 2\r\n-1000000000 1000000000\r\n", "output": "-1\r\n"}, {"input": "10 3 3\r\n1 1 1 1 1 5 6 7 8 9\r\n", "output": "-1\r\n"}, {"input": "10 3 3\r\n1 1 1 2 2 5 6 7 8 9\r\n", "output": "3\r\n"}]
| false |
stdio
| null | true |
1000/D
|
1000
|
D
|
PyPy 3-64
|
TESTS
| 2 | 78 | 6,348,800 |
154497893
|
from math import sqrt,ceil
import os
import sys
from collections import defaultdict,deque
from io import BytesIO, IOBase
# prime = [True for i in range(5*10**5 + 1)]
# def SieveOfEratosthenes(n):
# p = 2
# while (p * p <= n):
# # If prime[p] is not changed, then it is a prime
# if (prime[p] == True):
# # Update all multiples of p
# for i in range(p ** 2, n + 1, p):
# prime[i] = False
# p += 1
# prime[0]= False
# prime[1]= False
# SieveOfEratosthenes(5*10**5)
MOD = 998244353
nmax = (2*(10**3))+2
fact = [1] * (nmax+1)
for i in range(2, nmax+1):
fact[i] = fact[i-1] * i % MOD
inv = [1] * (nmax+1)
for i in range(2, nmax+1):
inv[i] = pow(fact[i], MOD-2, MOD)
def C(n, m):
return fact[n] * inv[m] % MOD * inv[n-m] % MOD if 0 <= m <= n else 0
# import sys
# import math
# mod = 10**7+1
# LI=lambda:[int(k) for k in input().split()]
# input = lambda: sys.stdin.readline().rstrip()
# IN=lambda:int(input())
# S=lambda:input()
# r=range
# fact=[i for i in r(mod)]
# for i in reversed(r(2,int(mod**0.5))):
# i=i**2
# for j in range(i,mod,i):
# if fact[j]%i==0:
# fact[j]//=i
from collections import Counter
from functools import lru_cache
from collections import deque
def main():
# import bisect,math
# class SortedList():
# BUCKET_RATIO = 50
# REBUILD_RATIO = 170
# def __init__(self,buckets):
# buckets = list(buckets)
# buckets = sorted(buckets)
# self._build(buckets)
# def __iter__(self):
# for i in self.buckets:
# for j in i: yield j
# def __reversed__(self):
# for i in reversed(self.buckets):
# for j in reversed(i): yield j
# def __len__(self):
# return self.size
# def __contains__(self,x):
# if self.size == 0: return False
# bucket = self._find_bucket(x)
# i = bisect.bisect_left(bucket,x)
# return i != len(bucket) and bucket[i] == x
# def __getitem__(self,x):
# if x < 0: x += self.size
# if x < 0: raise IndexError
# for i in self.buckets:
# if x < len(i): return i[x]
# x -= len(i)
# raise IndexError
# def _build(self,buckets=None):
# if buckets is None: buckets = list(self)
# self.size = len(buckets)
# bucket_size = int(math.ceil(math.sqrt(self.size/self.BUCKET_RATIO)))
# tmp = []
# for i in range(bucket_size):
# t = buckets[(self.size*i)//bucket_size:(self.size*(i+1))//bucket_size]
# tmp.append(t)
# self.buckets = tmp
# def _find_bucket(self,x):
# for i in self.buckets:
# if x <= i[-1]:
# return i
# return i
# def add(self,x):
# # O(√N)
# if self.size == 0:
# self.buckets = [[x]]
# self.size = 1
# return True
# bucket = self._find_bucket(x)
# bisect.insort(bucket,x)
# self.size += 1
# if len(bucket) > len(self.buckets) * self.REBUILD_RATIO:
# self._build()
# return True
# def remove(self,x):
# # O(√N)
# if self.size == 0: return False
# bucket = self._find_bucket(x)
# i = bisect.bisect_left(bucket,x)
# if i == len(bucket) or bucket[i] != x: return False
# bucket.pop(i)
# self.size -= 1
# if len(bucket) == 0: self._build()
# return True
# def lt(self,x):
# # less than < x
# for i in reversed(self.buckets):
# if i[0] < x:
# return i[bisect.bisect_left(i,x) - 1]
# def le(self,x):
# # less than or equal to <= x
# for i in reversed(self.buckets):
# if i[0] <= x:
# return i[bisect.bisect_right(i,x) - 1]
# def gt(self,x):
# # greater than > x
# for i in self.buckets:
# if i[-1] > x:
# return i[bisect.bisect_right(i,x)]
# def ge(self,x):
# # greater than or equal to >= x
# for i in self.buckets:
# if i[-1] >= x:
# return i[bisect.bisect_left(i,x)]
# def index(self,x):
# # the number of elements < x
# ans = 0
# for i in self.buckets:
# if i[-1] >= x:
# return ans + bisect.bisect_left(i,x)
# ans += len(i)
# return ans
# def index_right(self,x):
# # the number of elements < x
# ans = 0
# for i in self.buckets:
# if i[-1] > x:
# return ans + bisect.bisect_right(i,x)
# ans += len(i)
# return ans
#----------------------------------------------------------------------------------------
n=int(input())
mod=998244353
arr=list(map(int,input().split()))
pos=[0]
for el in arr:
if el>0:
pos.append(pos[-1]+1)
else :
pos.append(pos[-1])
pos=pos[::-1]
dp=[0]*(n+1)
dp[n]=1
res=0
for i in range(n-2,-1,-1):
ans=0
for j in range(i+1,n+1):
posi=pos[i]-pos[j-1]
temp=0
if posi>=arr[i] and dp[j]:
temp=C(posi,arr[i])
temp*=dp[j]
temp%=mod
ans+=temp
ans%=mod
dp[i]=ans
res+=ans
res%=mod
print(res)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# endregion
if __name__ == '__main__':
main()
| 22 | 1,372 | 23,859,200 |
39716340
|
#!/usr/bin/python3
MOD = 998244353
def solve(N, A):
dp = [[0] * N for _ in range(N + 1)]
dp[0][0] = 1
for i in range(N):
for j in range(N):
c = dp[i][j]
dp[i + 1][j] += c
dp[i + 1][j] %= MOD
if j == 0:
if A[i] > 0 and A[i] < N:
dp[i + 1][A[i]] += c
dp[i + 1][A[i]] %= MOD
else:
dp[i + 1][j - 1] += c
dp[i + 1][j - 1] %= MOD
return (dp[N][0] + MOD - 1) % MOD
def main():
N = int(input())
A = [int(e) for e in input().split(' ')]
assert len(A) == N
print(solve(N, A))
if __name__ == '__main__':
main()
|
Educational Codeforces Round 46 (Rated for Div. 2)
|
ICPC
| 2,018 | 2 | 256 |
Yet Another Problem On a Subsequence
|
The sequence of integers $$$a_1, a_2, \dots, a_k$$$ is called a good array if $$$a_1 = k - 1$$$ and $$$a_1 > 0$$$. For example, the sequences $$$[3, -1, 44, 0], [1, -99]$$$ are good arrays, and the sequences $$$[3, 7, 8], [2, 5, 4, 1], [0]$$$ — are not.
A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences $$$[2, -3, 0, 1, 4]$$$, $$$[1, 2, 3, -3, -9, 4]$$$ are good, and the sequences $$$[2, -3, 0, 1]$$$, $$$[1, 2, 3, -3 -9, 4, 1]$$$ — are not.
For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.
|
The first line contains the number $$$n~(1 \le n \le 10^3)$$$ — the length of the initial sequence. The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n~(-10^9 \le a_i \le 10^9)$$$ — the sequence itself.
|
In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.
| null |
In the first test case, two good subsequences — $$$[a_1, a_2, a_3]$$$ and $$$[a_2, a_3]$$$.
In the second test case, seven good subsequences — $$$[a_1, a_2, a_3, a_4], [a_1, a_2], [a_1, a_3], [a_1, a_4], [a_2, a_3], [a_2, a_4]$$$ and $$$[a_3, a_4]$$$.
|
[{"input": "3\n2 1 1", "output": "2"}, {"input": "4\n1 1 1 1", "output": "7"}]
| 1,900 |
["combinatorics", "dp"]
| 22 |
[{"input": "3\r\n2 1 1\r\n", "output": "2\r\n"}, {"input": "4\r\n1 1 1 1\r\n", "output": "7\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
878/E
|
878
|
E
|
Python 3
|
TESTS
| 1 | 46 | 0 |
32123773
|
def find_ind(arr):
max_ind, max_el = 0, arr[0]
for i in range(1, len(arr)):
el = arr[i]
if el > max_el:
max_el = el
max_ind = i
return max_ind
def calc_max(arr):
l = len(arr)
if l == 0:
return 0
if l == 1:
return arr[0]
if l == 2:
return arr[0] + arr[1] * 2
i = find_ind(arr)
return calc_max(arr[:i - 1] + [arr[i - 1] + 2 * (arr[i] + 2 * calc_max(arr[i + 1:]))])
n, q = list(map(int, input().split()))
arr = list(map(int, input().split()))
rem = [list(map(int, input().split())) for i in range(q)]
for d in rem:
li, ri = d
print(calc_max(arr[li - 1:ri]))
| 55 | 999 | 80,281,600 |
303302080
|
import sys
MOD = 10**9 + 7
INF = 2 * 10**9 + 5
def upd(a):
a += (a >> 31) & MOD
return a
def init(n):
_2 = [1] * (n + 1)
for i in range(1, n + 1):
_2[i] = (_2[i-1] << 1) % MOD
return _2
def find(x, fa):
if fa[x] != x:
fa[x] = find(fa[x], fa)
return fa[x]
def merge(x, y, fa, L, s, f, _2):
fa[y] = x
if (L[x] > 30 and s[x] > 0) or (s[x] + (s[y] << L[x]) >= INF):
s[x] = INF
else:
s[x] += s[y] << L[x]
f[x] = (f[x] + f[y] * _2[L[x]]) % MOD
L[x] += L[y]
def calc(l, r, pr, _2):
return (pr[l] - pr[r+1] * _2[r-l+1] % MOD + MOD) % MOD
def main():
input = sys.stdin.read
data = input().split()
idx = 0
n = int(data[idx])
m = int(data[idx+1])
idx += 2
a = [0] * (n + 1)
for i in range(1, n+1):
a[i] = int(data[idx])
idx += 1
_2 = init(n)
pr = [0] * (n + 2)
for i in range(n, 0, -1):
pr[i] = ((pr[i+1] << 1) + a[i] + MOD) % MOD
fa = [i for i in range(n+1)]
L = [1] * (n+1)
f = [0] * (n+1)
s = [0] * (n+1)
for i in range(1, n+1):
f[i] = s[i] = a[i]
vec = [[] for _ in range(n+1)]
for i in range(1, m+1):
l = int(data[idx])
r = int(data[idx+1])
idx += 2
vec[r].append((l, i))
ans = [0] * (m+1)
ps = [0] * (n+1)
for i in range(1, n+1):
while find(i, fa) > 1 and s[find(i, fa)] >= 0:
merge(find(find(i, fa)-1, fa), find(i, fa), fa, L, s, f, _2)
ps[find(i, fa)] = (ps[find(find(i, fa)-1, fa)] + f[find(i, fa)]) % MOD
for q in vec[i]:
l, id = q
ans[id] = ((ps[find(i, fa)] - ps[find(l, fa)] + MOD) * 2 + calc(l, find(l, fa) + L[find(l, fa)] - 1, pr, _2) + MOD) % MOD
for i in range(1, m+1):
print(ans[i])
if __name__ == "__main__":
main()
|
Codeforces Round 443 (Div. 1)
|
CF
| 2,017 | 2 | 512 |
Numbers on the blackboard
|
A sequence of n integers is written on a blackboard. Soon Sasha will come to the blackboard and start the following actions: let x and y be two adjacent numbers (x before y), then he can remove them and write x + 2y instead of them. He will perform these operations until one number is left. Sasha likes big numbers and will get the biggest possible number.
Nikita wants to get to the blackboard before Sasha and erase some of the numbers. He has q options, in the option i he erases all numbers to the left of the li-th number and all numbers to the right of ri-th number, i. e. all numbers between the li-th and the ri-th, inclusive, remain on the blackboard. For each of the options he wants to know how big Sasha's final number is going to be. This number can be very big, so output it modulo 109 + 7.
|
The first line contains two integers n and q (1 ≤ n, q ≤ 105) — the number of integers on the blackboard and the number of Nikita's options.
The next line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the sequence on the blackboard.
Each of the next q lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n), describing Nikita's options.
|
For each option output Sasha's result modulo 109 + 7.
| null |
In the second sample Nikita doesn't erase anything. Sasha first erases the numbers 1 and 2 and writes 5. Then he erases 5 and -3 and gets -1. -1 modulo 109 + 7 is 109 + 6.
|
[{"input": "3 3\n1 2 3\n1 3\n1 2\n2 3", "output": "17\n5\n8"}, {"input": "3 1\n1 2 -3\n1 3", "output": "1000000006"}, {"input": "4 2\n1 1 1 -1\n1 4\n3 4", "output": "5\n1000000006"}]
| 3,300 |
["combinatorics", "dp"]
| 55 |
[{"input": "3 3\r\n1 2 3\r\n1 3\r\n1 2\r\n2 3\r\n", "output": "17\r\n5\r\n8\r\n"}, {"input": "3 1\r\n1 2 -3\r\n1 3\r\n", "output": "1000000006\r\n"}, {"input": "4 2\r\n1 1 1 -1\r\n1 4\r\n3 4\r\n", "output": "5\r\n1000000006\r\n"}, {"input": "4 1\r\n1 1 -3 1\r\n1 4\r\n", "output": "1\r\n"}, {"input": "1 1\r\n1000000000\r\n1 1\r\n", "output": "1000000000\r\n"}, {"input": "1 1\r\n-1000000000\r\n1 1\r\n", "output": "7\r\n"}, {"input": "2 3\r\n0 0\r\n1 2\r\n1 1\r\n2 2\r\n", "output": "0\r\n0\r\n0\r\n"}, {"input": "2 3\r\n1000000000 1000000000\r\n1 2\r\n1 1\r\n2 2\r\n", "output": "999999986\r\n1000000000\r\n1000000000\r\n"}, {"input": "10 10\r\n-7 4 4 -5 2 3 -9 7 -4 -2\r\n8 10\r\n8 9\r\n2 3\r\n2 9\r\n2 3\r\n7 8\r\n1 3\r\n2 6\r\n6 8\r\n3 3\r\n", "output": "1000000002\r\n1000000006\r\n12\r\n208\r\n12\r\n5\r\n17\r\n56\r\n13\r\n4\r\n"}, {"input": "10 10\r\n593536087 56559483 -439122178 -126803734 606390399 -809361217 444436245 71742850 -477364598 -818526589\r\n5 9\r\n5 7\r\n5 5\r\n1 9\r\n5 10\r\n2 2\r\n3 10\r\n4 5\r\n2 4\r\n4 4\r\n", "output": "384626549\r\n765412945\r\n606390399\r\n410699067\r\n747573385\r\n56559483\r\n72910946\r\n85977057\r\n924707673\r\n873196273\r\n"}, {"input": "10 10\r\n-616555628 133372392 -749502876 499498544 927177575 -838173566 -139786799 -676011158 155638259 102225904\r\n1 3\r\n3 3\r\n1 2\r\n2 10\r\n5 7\r\n4 5\r\n6 6\r\n8 9\r\n3 5\r\n7 7\r\n", "output": "151183418\r\n250497131\r\n650189163\r\n270536548\r\n971256859\r\n353853680\r\n161826441\r\n635265367\r\n958204491\r\n860213208\r\n"}, {"input": "10 1\r\n-11 81 -4 79 44 -11 -50 26 -38 13\r\n4 10\r\n", "output": "129\r\n"}, {"input": "10 1\r\n-858350203 62991893 757167826 -643742467 -122005341 210910071 973749788 -554405426 91646398 811009699\r\n1 6\r\n", "output": "208079143\r\n"}, {"input": "101 1\r\n-2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 -2 -1 -1 -1 1 1000000000\r\n1 101\r\n", "output": "165401054\r\n"}, {"input": "108 1\r\n-1 -1 -1 -2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1000000000\r\n1 108\r\n", "output": "171334758\r\n"}]
| false |
stdio
| null | true |
735/B
|
735
|
B
|
PyPy 3
|
TESTS
| 3 | 139 | 0 |
83912169
|
n, n1, n2 = [int(i) for i in input().split()]
arr = [int(i) for i in input().split()]
arr.sort(reverse=True)
print(sum(arr[:min(n1, n2)]) / min(n1, n2) + sum(arr[min(n1, n2):max(n1, n2)+1]) / max(n1, n2))
| 37 | 93 | 7,065,600 |
147723791
|
in1=input().split()
n=int(in1[0])
temp_n1=int(in1[1])
temp_n2=int(in1[2])
n1=min(temp_n1,temp_n2)
n2=max(temp_n1,temp_n2)
wealth=[int(i1) for i1 in input().split()]
wealth.sort(reverse=True)
avg1=sum(wealth[ :n1])/n1
avg2=sum(wealth[n1:n1+n2])/n2
print(avg1+avg2)
|
Codeforces Round 382 (Div. 2)
|
CF
| 2,016 | 2 | 256 |
Urbanization
|
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are n people who plan to move to the cities. The wealth of the i of them is equal to ai. Authorities plan to build two cities, first for n1 people and second for n2 people. Of course, each of n candidates can settle in only one of the cities. Thus, first some subset of candidates of size n1 settle in the first city and then some subset of size n2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth ai among all its residents divided by the number of them (n1 or n2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
|
The first line of the input contains three integers n, n1 and n2 (1 ≤ n, n1, n2 ≤ 100 000, n1 + n2 ≤ n) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000), the i-th of them is equal to the wealth of the i-th candidate.
|
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if $$\frac{|a-b|}{\max(1,b)} \leq 10^{-6}$$.
| null |
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (a3 + a4) / 2 + a2 = (3 + 2) / 2 + 4 = 6.5
|
[{"input": "2 1 1\n1 5", "output": "6.00000000"}, {"input": "4 2 1\n1 4 2 3", "output": "6.50000000"}]
| 1,100 |
["greedy", "number theory", "sortings"]
| 37 |
[{"input": "2 1 1\r\n1 5\r\n", "output": "6.00000000\r\n"}, {"input": "4 2 1\r\n1 4 2 3\r\n", "output": "6.50000000\r\n"}, {"input": "3 1 2\r\n1 2 3\r\n", "output": "4.50000000\r\n"}, {"input": "10 4 6\r\n3 5 7 9 12 25 67 69 83 96\r\n", "output": "88.91666667\r\n"}, {"input": "19 7 12\r\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 100000 100000\r\n", "output": "47052.10714286\r\n"}, {"input": "100 9 6\r\n109 711 40 95 935 48 228 253 308 726 816 534 252 8 966 363 162 508 84 83 807 506 748 178 45 30 106 108 764 698 825 198 336 353 158 790 64 262 403 334 577 571 742 541 946 602 279 621 910 776 421 886 29 133 114 394 762 965 339 263 750 530 49 80 124 31 322 292 27 590 960 278 111 932 849 491 561 744 469 511 106 271 156 160 836 363 149 473 457 543 976 809 490 29 85 626 265 88 995 946\r\n", "output": "1849.66666667\r\n"}, {"input": "69 6 63\r\n53475 22876 79144 6335 33763 79104 65441 45527 65847 94406 74670 43529 75330 19403 67629 56187 57949 23071 64910 54409 55348 18056 855 24961 50565 6622 26467 33989 22660 79469 41246 13965 79706 14422 16075 93378 81313 48173 13470 97348 2346 27452 59427 29925 29847 73823 32021 10988 24609 98855 90919 45939 17203 8439 43007 40138 55693 30314 71734 33458 66850 4011 20089 20546 92090 50842 78859 62756 40177\r\n", "output": "135712.88888889\r\n"}, {"input": "69 6 9\r\n2612 17461 69001 33130 10662 85485 88195 45974 16712 81365 67119 87797 15559 20197 74716 92979 97268 49466 68603 48351 99905 35606 54242 98603 68232 54398 82637 49647 38979 46171 54680 23334 15892 92186 69670 29711 67999 2220 32317 717 70667 68262 86760 55720 97158 61122 7251 138 21022 27197 12691 59331 13576 66999 38332 13574 83484 66646 17704 33065 98583 80259 64631 16745 69431 40747 82089 82788 32739\r\n", "output": "183129.44444444\r\n"}]
| false |
stdio
|
import sys
def main():
input_path, correct_output_path, submission_output_path = sys.argv[1:4]
with open(correct_output_path, 'r') as f:
correct_output = f.read().strip()
with open(submission_output_path, 'r') as f:
submission_output = f.read().strip()
try:
correct = float(correct_output)
submission = float(submission_output)
except:
print(0)
return
absolute_error = abs(submission - correct)
max_denominator = max(1.0, correct)
relative_error = absolute_error / max_denominator
if relative_error <= 1e-6:
print(100)
else:
print(0)
if __name__ == "__main__":
main()
| true |
463/D
|
463
|
D
|
PyPy 3
|
TESTS
| 1 | 93 | 1,536,000 |
204890117
|
def solve_1(s1, s2):
N = len(s1)
dp = [[0] * (N + 1) for _ in range(N + 1)]
for i in range(1, N + 1):
for j in range(1, N + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
return dp[-1][-1]
def solve():
n, k = map(int, input().split())
data = list()
for i in range(k):
seq = input().split()
data.append(seq)
result = n
for i in range(k - 1):
for j in range(i + 1, k):
res = solve_1(data[i], data[j])
result = min(result, res)
print(result)
solve()
| 40 | 124 | 29,388,800 |
115843377
|
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def main():
n,k = map(int,input().split())
arr = [list(map(int,input().split())) for _ in range(k)]
edg = [[1]*n for _ in range(n)]
for i in range(k):
for j in range(n):
for l in range(j+1):
edg[arr[i][j]-1][arr[i][l]-1] = 0
path = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if edg[i][j]:
path[i].append(j)
visi,poi,dp = [0]*n,[0]*n,[1]*n
for i in range(n):
if visi[i]:
continue
visi[i],st = 1,[i]
while len(st):
x,y = st[-1],path[st[-1]]
while poi[x] != len(y) and visi[y[poi[x]]]:
dp[x] = max(dp[x],dp[y[poi[x]]]+1)
poi[x] += 1
if poi[x] == len(y):
st.pop()
if len(st):
dp[st[-1]] = max(dp[st[-1]],dp[x]+1)
else:
st.append(y[poi[x]])
visi[st[-1]] = 1
poi[x] += 1
print(max(dp))
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self,file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd,self.buffer.getvalue())
self.buffer.truncate(0),self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self,file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s:self.buffer.write(s.encode("ascii"))
self.read = lambda:self.buffer.read().decode("ascii")
self.readline = lambda:self.buffer.readline().decode("ascii")
sys.stdin,sys.stdout = IOWrapper(sys.stdin),IOWrapper(sys.stdout)
input = lambda:sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
|
Codeforces Round 264 (Div. 2)
|
CF
| 2,014 | 2 | 256 |
Gargari and Permutations
|
Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have found k permutations. Each of them consists of numbers 1, 2, ..., n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?
You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
|
The first line contains two integers n and k (1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the next k lines contains integers 1, 2, ..., n in some order — description of the current permutation.
|
Print the length of the longest common subsequence.
| null |
The answer for the first test sample is subsequence [1, 2, 3].
|
[{"input": "4 3\n1 4 2 3\n4 1 2 3\n1 2 4 3", "output": "3"}]
| 1,900 |
["dfs and similar", "dp", "graphs", "implementation"]
| 40 |
[{"input": "4 3\r\n1 4 2 3\r\n4 1 2 3\r\n1 2 4 3\r\n", "output": "3\r\n"}, {"input": "6 3\r\n2 5 1 4 6 3\r\n5 1 4 3 2 6\r\n5 4 2 6 3 1\r\n", "output": "3\r\n"}, {"input": "41 4\r\n24 15 17 35 13 41 4 14 23 5 8 16 21 18 30 36 6 22 11 29 26 1 40 31 7 3 32 10 28 38 12 20 39 37 34 19 33 27 2 25 9\r\n22 13 25 24 38 35 29 12 15 8 11 37 3 19 4 23 18 32 30 40 36 21 16 34 27 9 5 41 39 2 14 17 31 33 26 7 1 10 20 6 28\r\n31 27 39 16 22 12 13 32 6 10 19 29 37 7 18 33 24 21 1 9 36 4 34 41 25 28 17 40 30 35 23 14 11 8 2 15 38 20 26 5 3\r\n8 18 39 38 7 34 16 31 15 1 40 20 37 4 25 11 17 19 33 26 6 14 13 41 12 32 2 21 10 35 27 9 28 5 30 24 22 23 29 3 36\r\n", "output": "4\r\n"}, {"input": "1 2\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "28 5\r\n3 14 12 16 13 27 20 8 1 10 24 11 5 9 7 18 17 23 22 25 28 19 4 21 26 6 15 2\r\n7 12 23 27 22 26 16 18 19 5 6 9 11 28 25 4 10 3 1 14 8 17 15 2 20 13 24 21\r\n21 20 2 5 19 15 12 4 18 9 23 16 11 14 8 6 25 27 13 17 10 26 7 24 28 1 3 22\r\n12 2 23 11 20 18 25 21 13 27 14 8 4 6 9 16 7 3 10 1 22 15 26 19 5 17 28 24\r\n13 2 6 19 22 23 4 1 28 10 18 17 21 8 9 3 26 11 12 27 14 20 24 25 15 5 16 7\r\n", "output": "3\r\n"}, {"input": "6 3\r\n2 5 1 4 6 3\r\n5 1 4 6 2 3\r\n5 4 2 6 3 1\r\n", "output": "4\r\n"}, {"input": "41 4\r\n24 15 17 35 13 41 4 14 23 5 8 16 21 18 30 36 6 22 11 29 26 1 40 31 7 3 32 10 28 38 12 20 39 37 34 19 33 27 2 25 9\r\n22 13 25 24 38 35 29 12 15 8 11 37 3 19 4 23 18 32 30 40 36 21 16 34 27 9 5 41 39 2 14 17 31 33 26 7 1 10 20 6 28\r\n31 27 39 16 22 12 13 32 6 10 19 29 37 7 18 33 24 21 1 9 36 4 34 41 25 28 17 40 30 35 23 14 11 8 2 15 38 20 26 5 3\r\n8 18 39 38 7 34 16 31 15 1 40 20 37 4 25 11 17 19 33 26 6 14 13 41 12 32 2 21 10 35 27 9 28 5 30 24 22 23 29 3 36\r\n", "output": "4\r\n"}, {"input": "37 3\r\n6 3 19 20 15 4 1 35 8 24 12 21 34 26 18 14 23 33 28 9 36 11 37 31 25 32 29 22 13 27 16 17 10 7 5 30 2\r\n10 3 35 17 34 21 14 8 26 28 11 19 27 7 4 23 24 22 12 13 16 1 25 29 5 31 30 20 32 18 15 9 2 36 37 33 6\r\n19 9 22 32 26 35 29 23 5 6 14 34 33 10 2 28 15 11 24 4 13 7 8 31 37 36 1 27 3 16 30 25 20 21 18 17 12\r\n", "output": "7\r\n"}]
| false |
stdio
| null | true |
594/A
|
594
|
A
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
181048619
|
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
min_res = 200_010
for i in range(n//2):
if a[i+n//2] - a[i] < min_res:
min_res = a[i+n//2] - a[i]
print(min_res)
| 50 | 217 | 16,179,200 |
189263054
|
# LUOGU_RID: 99814317
ans=0x3f3f3f3f
n=int(input())
a=list(map(int,input().split()))#输入n个点的位置坐标
b=n//2
a.sort()#给n个点的位置坐标从小到大排序
for i in range(b):#找出最后剩下的两个点的位置坐标的最小值
ans=min(ans,a[b+i]-a[i])#a[b+i]为剩下右侧的点的位置坐标，a[i]为剩下左侧的点的位置坐标
print(ans)
|
Codeforces Round 330 (Div. 1)
|
CF
| 2,015 | 2 | 256 |
Warrior and Archer
|
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
|
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
|
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
| null |
In the first sample one of the optimum behavior of the players looks like that:
1. Vova bans the position at coordinate 15;
2. Lesha bans the position at coordinate 3;
3. Vova bans the position at coordinate 31;
4. Lesha bans the position at coordinate 1.
After these actions only positions 0 and 7 will remain, and the distance between them is equal to 7.
In the second sample there are only two possible positions, so there will be no bans.
|
[{"input": "6\n0 1 3 7 15 31", "output": "7"}, {"input": "2\n73 37", "output": "36"}]
| 2,300 |
["games"]
| 50 |
[{"input": "6\r\n0 1 3 7 15 31\r\n", "output": "7\r\n"}, {"input": "2\r\n73 37\r\n", "output": "36\r\n"}, {"input": "2\r\n0 1000000000\r\n", "output": "1000000000\r\n"}, {"input": "8\r\n729541013 135019377 88372488 319157478 682081360 558614617 258129110 790518782\r\n", "output": "470242129\r\n"}, {"input": "2\r\n0 1\r\n", "output": "1\r\n"}, {"input": "8\r\n552283832 997699491 89302459 301640204 288141798 31112026 710831619 862166501\r\n", "output": "521171806\r\n"}, {"input": "4\r\n0 500000000 500000001 1000000000\r\n", "output": "500000000\r\n"}, {"input": "18\r\n515925896 832652240 279975694 570998878 28122427 209724246 898414431 709461320 358922485 439508829 403574907 358500312 596248410 968234748 187793884 728450713 30350176 528924900\r\n", "output": "369950401\r\n"}, {"input": "20\r\n713900269 192811911 592111899 609607891 585084800 601258511 223103775 876894656 751583891 230837577 971499807 312977833 344314550 397998873 558637732 216574673 913028292 762852863 464376621 61315042\r\n", "output": "384683838\r\n"}, {"input": "10\r\n805513144 38998401 16228409 266085559 293487744 471510400 138613792 649258082 904651590 244678415\r\n", "output": "277259335\r\n"}, {"input": "6\r\n0 166666666 333333333 499999998 666666665 833333330\r\n", "output": "499999997\r\n"}, {"input": "16\r\n1 62500001 125000001 187500000 250000000 312500000 375000000 437500001 500000000 562500000 625000000 687500001 750000001 812500002 875000002 937500000\r\n", "output": "499999999\r\n"}, {"input": "12\r\n5 83333336 166666669 250000001 333333336 416666670 500000004 583333336 666666667 750000001 833333334 916666671\r\n", "output": "499999998\r\n"}, {"input": "20\r\n54 50000046 100000041 150000049 200000061 250000039 300000043 350000054 400000042 450000045 500000076 550000052 600000064 650000065 700000055 750000046 800000044 850000042 900000052 950000054\r\n", "output": "499999988\r\n"}]
| false |
stdio
| null | true |
376/B
|
376
|
B
|
Python 3
|
TESTS
| 3 | 46 | 307,200 |
5692747
|
n,m=map(int,input().split())
f=[[0 for _ in range(n)] for _ in range(n)]
for i in range(m):
a,b,c=map(int,input().split())
f[a-1][b-1]=c
for i in range(n):
for j in range(n):
if f[i][j]>0:
for k in range(n):
if f[j][k]>0:
if f[i][j]>=f[j][k]:
f[i][j]-=f[j][k]
f[i][k]=f[j][k]
f[j][k]=0
else:
f[j][k]-=f[i][j]
f[j][k]=f[i][k]
f[i][j]=0
for i in range(n):
f[i][i]=0
print(sum(map(sum,f)))
| 29 | 46 | 0 |
189698875
|
n, m = map(int, input().split())
own = dict()
for _ in range(m):
a, b, c = map(int, input().split())
if a in own:
own[a] += c
else:
own[a] = c
if b in own:
own[b] -= c
else:
own[b] = -c
ans = 0
for i in own.values():
ans += i if i > 0 else 0
print(ans)
|
Codeforces Round 221 (Div. 2)
|
CF
| 2,013 | 1 | 256 |
I.O.U.
|
Imagine that there is a group of three friends: A, B and С. A owes B 20 rubles and B owes C 20 rubles. The total sum of the debts is 40 rubles. You can see that the debts are not organized in a very optimal manner. Let's rearrange them like that: assume that A owes C 20 rubles and B doesn't owe anything to anybody. The debts still mean the same but the total sum of the debts now equals 20 rubles.
This task is a generalisation of a described example. Imagine that your group of friends has n people and you know the debts between the people. Optimize the given debts without changing their meaning. In other words, finally for each friend the difference between the total money he should give and the total money he should take must be the same. Print the minimum sum of all debts in the optimal rearrangement of the debts. See the notes to the test samples to better understand the problem.
|
The first line contains two integers n and m (1 ≤ n ≤ 100; 0 ≤ m ≤ 104). The next m lines contain the debts. The i-th line contains three integers ai, bi, ci (1 ≤ ai, bi ≤ n; ai ≠ bi; 1 ≤ ci ≤ 100), which mean that person ai owes person bi ci rubles.
Assume that the people are numbered by integers from 1 to n.
It is guaranteed that the same pair of people occurs at most once in the input. The input doesn't simultaneously contain pair of people (x, y) and pair of people (y, x).
|
Print a single integer — the minimum sum of debts in the optimal rearrangement.
| null |
In the first sample, you can assume that person number 1 owes 8 rubles to person number 2, 1 ruble to person number 3 and 1 ruble to person number 4. He doesn't owe anybody else anything. In the end, the total debt equals 10.
In the second sample, there are no debts.
In the third sample, you can annul all the debts.
|
[{"input": "5 3\n1 2 10\n2 3 1\n2 4 1", "output": "10"}, {"input": "3 0", "output": "0"}, {"input": "4 3\n1 2 1\n2 3 1\n3 1 1", "output": "0"}]
| 1,300 |
["implementation"]
| 29 |
[{"input": "5 3\r\n1 2 10\r\n2 3 1\r\n2 4 1\r\n", "output": "10\r\n"}, {"input": "3 0\r\n", "output": "0\r\n"}, {"input": "4 3\r\n1 2 1\r\n2 3 1\r\n3 1 1\r\n", "output": "0\r\n"}, {"input": "20 28\r\n1 5 6\r\n1 12 7\r\n1 13 4\r\n1 15 7\r\n1 20 3\r\n2 4 1\r\n2 15 6\r\n3 5 3\r\n3 8 10\r\n3 13 8\r\n3 20 6\r\n4 6 10\r\n4 12 8\r\n4 19 5\r\n5 17 8\r\n6 9 9\r\n6 16 2\r\n6 19 9\r\n7 14 6\r\n8 9 3\r\n8 16 10\r\n9 11 7\r\n9 17 8\r\n11 13 8\r\n11 17 17\r\n11 19 1\r\n15 20 2\r\n17 20 1\r\n", "output": "124\r\n"}, {"input": "20 36\r\n1 2 13\r\n1 3 1\r\n1 6 4\r\n1 12 8\r\n1 13 9\r\n1 15 3\r\n1 18 4\r\n2 10 2\r\n2 15 2\r\n2 18 6\r\n3 7 8\r\n3 16 19\r\n4 7 1\r\n4 18 4\r\n5 9 2\r\n5 15 9\r\n5 17 4\r\n5 18 5\r\n6 11 7\r\n6 13 1\r\n6 14 9\r\n7 10 4\r\n7 12 10\r\n7 15 9\r\n7 17 8\r\n8 14 4\r\n10 13 8\r\n10 19 9\r\n11 12 5\r\n12 17 6\r\n13 15 8\r\n13 19 4\r\n14 15 9\r\n14 16 8\r\n17 19 8\r\n17 20 7\r\n", "output": "147\r\n"}, {"input": "20 40\r\n1 13 4\r\n2 3 3\r\n2 4 5\r\n2 7 7\r\n2 17 10\r\n3 5 3\r\n3 6 9\r\n3 10 4\r\n3 12 2\r\n3 13 2\r\n3 14 3\r\n4 5 4\r\n4 8 7\r\n4 13 9\r\n5 6 14\r\n5 14 5\r\n7 11 5\r\n7 12 13\r\n7 15 7\r\n8 14 5\r\n8 16 7\r\n8 18 17\r\n9 11 8\r\n9 19 19\r\n10 12 4\r\n10 16 3\r\n10 18 10\r\n10 20 9\r\n11 13 9\r\n11 20 2\r\n12 13 8\r\n12 18 2\r\n12 20 3\r\n13 17 1\r\n13 20 4\r\n14 16 8\r\n16 19 3\r\n18 19 3\r\n18 20 7\r\n19 20 10\r\n", "output": "165\r\n"}, {"input": "50 10\r\n1 5 1\r\n2 34 2\r\n3 8 10\r\n5 28 4\r\n7 28 6\r\n13 49 9\r\n15 42 7\r\n16 26 7\r\n18 47 5\r\n20 41 10\r\n", "output": "60\r\n"}, {"input": "50 46\r\n1 6 10\r\n1 18 1\r\n1 24 10\r\n1 33 2\r\n1 40 8\r\n3 16 7\r\n4 26 8\r\n4 32 2\r\n4 34 6\r\n5 29 8\r\n6 44 3\r\n8 20 5\r\n8 42 13\r\n10 13 5\r\n10 25 7\r\n10 27 9\r\n10 29 10\r\n11 23 4\r\n12 28 7\r\n12 30 10\r\n12 40 10\r\n13 18 2\r\n13 33 2\r\n14 15 7\r\n14 43 10\r\n14 47 3\r\n16 27 10\r\n17 21 6\r\n17 30 9\r\n19 40 4\r\n22 24 8\r\n22 25 7\r\n22 38 18\r\n25 38 1\r\n27 31 7\r\n27 40 8\r\n30 36 8\r\n31 34 1\r\n32 49 6\r\n33 35 4\r\n33 50 7\r\n38 47 1\r\n42 47 2\r\n42 50 5\r\n43 44 9\r\n47 50 5\r\n", "output": "228\r\n"}, {"input": "100 48\r\n1 56 6\r\n2 42 3\r\n3 52 1\r\n9 50 8\r\n10 96 8\r\n11 39 2\r\n12 51 6\r\n12 68 7\r\n13 40 5\r\n14 18 10\r\n14 70 6\r\n15 37 4\r\n15 38 8\r\n15 82 6\r\n15 85 5\r\n16 48 4\r\n16 50 9\r\n16 71 9\r\n17 18 3\r\n17 100 10\r\n20 73 3\r\n22 32 9\r\n22 89 9\r\n23 53 3\r\n24 53 1\r\n27 78 10\r\n30 50 5\r\n33 94 8\r\n34 87 9\r\n35 73 3\r\n36 51 8\r\n37 88 10\r\n37 97 2\r\n40 47 8\r\n40 90 6\r\n44 53 3\r\n44 65 3\r\n47 48 8\r\n48 72 10\r\n49 98 2\r\n53 68 10\r\n53 71 9\r\n57 62 2\r\n63 76 10\r\n66 90 9\r\n71 76 8\r\n72 80 5\r\n75 77 7\r\n", "output": "253\r\n"}, {"input": "4 3\r\n1 4 1\r\n2 3 1\r\n4 2 2\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
888/G
|
888
|
G
|
PyPy 3
|
TESTS
| 2 | 124 | 5,120,000 |
217245536
|
V = 2
MX = 3;
class Node:
def __init__(self) -> None:
self.childs = [None]*V
self.ended = False
self.val = -1
self.cnt = 0
root = Node()
def add(x):
cur = root
for i in range(MX,-1,-1): # store from msb...lsb
ind = (x>>i)&1
if not cur.childs[ind]: cur.childs[ind] = Node()
cur = cur.childs[ind]
cur.ended = 1;
cur.cnt+=1
cur.val = x
def reset():
global root
root = Node()
def remove(x):
cur = root
path = []
for i in range(MX,-1,-1): # store from msb...lsb
ind = (x>>i)&1
path.append((cur,ind))
cur = cur.childs[ind]
cur.cnt -=1
if cur.cnt==0:
while path:
p,i = path.pop()
p.childs[i] = None
if p.childs[i^1]: break
"""
Find max xor of x in trie
"""
def min_xor(x):
cur = root
for i in range(MX,-1,-1):
if not cur: return float('inf')
ind = (x>>i)&1
if cur.childs[ind]: cur = cur.childs[ind]
else: cur = cur.childs[ind^1]
return cur.val^x
n = int(input())
A = list(map(int,input().split()))
A.sort() # order doesnt matter
ans = 0
def dnq(l,r,b): #
global ans
if l>=r: return
reset()
m = l
while m<=r and (A[m]&(1<<b))==0:
add(A[m])
m +=1
if m==l or m>r:
dnq(l,r,b-1)
return
xr = float('inf')
for j in range(m,r+1):
xr = min(xr,min_xor(A[j]))
ans += xr if xr<float('inf') else 0
dnq(l,m-1,b-1)
dnq(m,r,b-1)
dnq(0,n-1,29)
print(ans)
| 79 | 997 | 149,504,000 |
225470974
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def insert(x, la):
j = 0
for i in p:
if x & i:
if not G[j] >> 25:
la += 1
G[j] ^= la << 25
j = la
else:
j = G[j] >> 25
else:
if not G[j] & 33554431:
la += 1
G[j] ^= la
j = la
else:
j = G[j] & 33554431
cnt[j] += 1
return la
def xor_min(x, s):
j, ans = s, 0
for i in range(30 - len(st), -1, -1):
if x & pow2[i]:
if G[j] >> 25:
j = G[j] >> 25
else:
ans ^= pow2[i]
j = G[j] & 33554431
else:
if G[j] & 33554431:
j = G[j] & 33554431
else:
ans ^= pow2[i]
j = G[j] >> 25
return ans
n = int(input())
a = list(map(int, input().split()))
a = list(set(a))
a.sort()
pow2 = [1]
l = 30
for _ in range(l):
pow2.append(2 * pow2[-1])
p = pow2[::-1]
n = len(a)
G, cnt, la = [0] * 5000000, [0] * 5000000, 0
for i in a:
la = insert(i, la)
h = l
l0 = [-1] * (la + 1)
l0[0] = 0
st = [0]
ans = 0
while st:
i = st[-1]
j, k = G[i] & 33554431, G[i] >> 25
if l0[j] == -1 and l0[k] == -1 and min(cnt[j], cnt[k]):
l = l0[i]
m = l + cnt[j]
r = m + cnt[k]
ans0 = pow2[30]
if (m - l) * (r - m) <= 5000:
for u in range(l, m):
au = a[u]
for v in range(m, r):
ans0 = min(ans0, au ^ a[v])
elif m - l <= r - m:
for u in range(l, m):
ans0 = min(ans0, xor_min(a[u], k))
else:
for u in range(m, r):
ans0 = min(ans0, xor_min(a[u], j))
ans0 |= pow2[h]
ans += ans0
if l0[j] == -1:
if cnt[j] > 1:
st.append(j)
h -= 1
l0[j] = l0[i]
elif l0[k] == -1:
if cnt[k] > 1:
st.append(k)
h -= 1
l0[k] = l0[i] + cnt[j]
else:
st.pop()
h += 1
print(ans)
|
Educational Codeforces Round 32
|
ICPC
| 2,017 | 2 | 256 |
Xor-MST
|
You are given a complete undirected graph with n vertices. A number ai is assigned to each vertex, and the weight of an edge between vertices i and j is equal to ai xor aj.
Calculate the weight of the minimum spanning tree in this graph.
|
The first line contains n (1 ≤ n ≤ 200000) — the number of vertices in the graph.
The second line contains n integers a1, a2, ..., an (0 ≤ ai < 230) — the numbers assigned to the vertices.
|
Print one number — the weight of the minimum spanning tree in the graph.
| null | null |
[{"input": "5\n1 2 3 4 5", "output": "8"}, {"input": "4\n1 2 3 4", "output": "8"}]
| 2,300 |
["bitmasks", "constructive algorithms", "data structures"]
| 79 |
[{"input": "5\r\n1 2 3 4 5\r\n", "output": "8\r\n"}, {"input": "4\r\n1 2 3 4\r\n", "output": "8\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
888/G
|
888
|
G
|
PyPy 3-64
|
TESTS
| 2 | 109 | 9,113,600 |
184858836
|
def calc(a,k):
if len(a)<=1:
return 0
if k<0:
return 0
c0=[]
c1=[]
for i in a:
if (i>>k)&1:
c1.append(i^(1<<k))
else:
c0.append(i)
res=1<<30
if len(c0)>=2:
res=min(res,calc(c0,k-1))
if len(c1)>=2:
res=min(res,calc(c1,k-1))
if len(c0)==len(c1)==1:
res=min(res,(1<<k)^c0[0]^c1[0])
return res
def solve(a,k):
if len(a)==0 or k<0:
return 0
c0=[]
c1=[]
for i in a:
if (i>>k)&1:
c1.append(i^(1<<k))
else:
c0.append(i)
res=solve(c0,k-1)+solve(c1,k-1)
if c0 and c1:
res+=(1<<k)+calc(c0+c1,k-1)
return res
n=int(input())
a=list(map(int,input().split()))
print(solve(a,30))
| 79 | 997 | 149,504,000 |
225470974
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def insert(x, la):
j = 0
for i in p:
if x & i:
if not G[j] >> 25:
la += 1
G[j] ^= la << 25
j = la
else:
j = G[j] >> 25
else:
if not G[j] & 33554431:
la += 1
G[j] ^= la
j = la
else:
j = G[j] & 33554431
cnt[j] += 1
return la
def xor_min(x, s):
j, ans = s, 0
for i in range(30 - len(st), -1, -1):
if x & pow2[i]:
if G[j] >> 25:
j = G[j] >> 25
else:
ans ^= pow2[i]
j = G[j] & 33554431
else:
if G[j] & 33554431:
j = G[j] & 33554431
else:
ans ^= pow2[i]
j = G[j] >> 25
return ans
n = int(input())
a = list(map(int, input().split()))
a = list(set(a))
a.sort()
pow2 = [1]
l = 30
for _ in range(l):
pow2.append(2 * pow2[-1])
p = pow2[::-1]
n = len(a)
G, cnt, la = [0] * 5000000, [0] * 5000000, 0
for i in a:
la = insert(i, la)
h = l
l0 = [-1] * (la + 1)
l0[0] = 0
st = [0]
ans = 0
while st:
i = st[-1]
j, k = G[i] & 33554431, G[i] >> 25
if l0[j] == -1 and l0[k] == -1 and min(cnt[j], cnt[k]):
l = l0[i]
m = l + cnt[j]
r = m + cnt[k]
ans0 = pow2[30]
if (m - l) * (r - m) <= 5000:
for u in range(l, m):
au = a[u]
for v in range(m, r):
ans0 = min(ans0, au ^ a[v])
elif m - l <= r - m:
for u in range(l, m):
ans0 = min(ans0, xor_min(a[u], k))
else:
for u in range(m, r):
ans0 = min(ans0, xor_min(a[u], j))
ans0 |= pow2[h]
ans += ans0
if l0[j] == -1:
if cnt[j] > 1:
st.append(j)
h -= 1
l0[j] = l0[i]
elif l0[k] == -1:
if cnt[k] > 1:
st.append(k)
h -= 1
l0[k] = l0[i] + cnt[j]
else:
st.pop()
h += 1
print(ans)
|
Educational Codeforces Round 32
|
ICPC
| 2,017 | 2 | 256 |
Xor-MST
|
You are given a complete undirected graph with n vertices. A number ai is assigned to each vertex, and the weight of an edge between vertices i and j is equal to ai xor aj.
Calculate the weight of the minimum spanning tree in this graph.
|
The first line contains n (1 ≤ n ≤ 200000) — the number of vertices in the graph.
The second line contains n integers a1, a2, ..., an (0 ≤ ai < 230) — the numbers assigned to the vertices.
|
Print one number — the weight of the minimum spanning tree in the graph.
| null | null |
[{"input": "5\n1 2 3 4 5", "output": "8"}, {"input": "4\n1 2 3 4", "output": "8"}]
| 2,300 |
["bitmasks", "constructive algorithms", "data structures"]
| 79 |
[{"input": "5\r\n1 2 3 4 5\r\n", "output": "8\r\n"}, {"input": "4\r\n1 2 3 4\r\n", "output": "8\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
177/D1
|
177
|
D2
|
Python 3
|
TESTS2
| 3 | 124 | 0 |
117258709
|
def range_sum(parr, l, r):
if l==r and l==0:
return parr[l]
if l==r:
return parr[l]-parr[l-1]
if l==0:
return parr[r]
return parr[r] - parr[l-1]
n,m,c = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
pb = [0]*m
pb[0] = b[0]
for i in range(1,m):
pb[i] =(b[i]+ pb[i-1])%c
aux = [0]*n
#region 1
i= 0
while i<min(n-m+1,m):
aux[i]= i+1
i+=1
#region 3
k = 0
j=n-1
while k<min(n-m+1,m):
aux[j]= k+1
k+=1
j-= 1
#region 2
k = i
while k<=j:
aux[k]=m
k += 1
op = [0]*n
k=0
while k<n:
if k<=j:
op[k] = (a[k] + range_sum(pb, 0, aux[k]-1))%c
else:
op[k] = (a[k] + range_sum(pb, m-aux[k],m-1))%c
print(op[k], end=' ')
k+=1
| 12 | 92 | 0 |
176975249
|
'''
Online Python Compiler.
Code, Compile, Run and Debug python program online.
Write your code in this editor and press "Run" button to execute it.
'''
n,m,c=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
prefsum=[0]
for j in range(m): prefsum.append(prefsum[-1]+b[j])
ans=[]
for i in range(n):
ans.append(prefsum[min(m,i+1)]-prefsum[max(0,i-(n-m))]+a[i])
for ele in ans:
print(ele%c,end=" ")
|
ABBYY Cup 2.0 - Easy
|
ICPC
| 2,012 | 2 | 256 |
Encrypting Messages
|
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
|
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
- 1 ≤ m ≤ n ≤ 103
- 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
- 1 ≤ m ≤ n ≤ 105
- 1 ≤ c ≤ 103
|
Print n space-separated integers — the result of encrypting the original message.
| null |
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
|
[{"input": "4 3 2\n1 1 1 1\n1 1 1", "output": "0 1 1 0"}, {"input": "3 1 5\n1 2 3\n4", "output": "0 1 2"}]
| 1,200 |
["brute force"]
| 12 |
[{"input": "4 3 2\r\n1 1 1 1\r\n1 1 1\r\n", "output": "0 1 1 0\r\n"}, {"input": "3 1 5\r\n1 2 3\r\n4\r\n", "output": "0 1 2\r\n"}, {"input": "5 2 7\r\n0 0 1 2 4\r\n3 5\r\n", "output": "3 1 2 3 2\r\n"}, {"input": "20 15 17\r\n4 9 14 11 15 16 15 4 0 10 7 12 10 1 8 6 7 14 1 13\r\n6 3 14 8 8 11 16 4 5 9 2 13 6 14 15\r\n", "output": "10 1 3 8 3 15 7 14 1 12 3 10 15 16 16 5 4 15 13 11\r\n"}]
| false |
stdio
| null | true |
247/D
|
250
|
D
|
Python 3
|
TESTS
| 3 | 92 | 0 |
13550585
|
# http://codeforces.com/problemset/problem/250/D
# inputFile.readline
from math import sqrt
def calcPath(x1, y1, x2, y2):
return sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2))
# with open('h.in', 'r') as inputFile, open('h.out', 'w') as outputFile:
(n, m, a, b) = [int(x) for x in input().strip().split(' ')]
A = [int(x) for x in input().strip().split(' ')]
B = [int(x) for x in input().strip().split(' ')]
L = [int(x) for x in input().strip().split(' ')]
# print(n,m,a,b)
# print(A,B,L)
i = 0
j = 0
min_path = 0
min_i = 0
min_j = 0
# curr_path = 0
flagDec = False
if n == 1 and m == 1:
print("1 1")
else:
while i < m:
c_min_path = 0
curr_path = 0
while j < n:
curr_path = calcPath(0, 0, a, A[j]) + calcPath(a, A[j], b, B[i]) + L[i]
if c_min_path < curr_path and c_min_path != 0:
# res[min_path] = [j, i + 1]
min_path = c_min_path
min_i = i + 1
min_j = j
i += 1
if flagDec:
j -= 1
flagDec = False
# print(j,i,curr_path)
break
c_min_path = curr_path
# print(res)
# min_i = i+1
# min_j = j+1
# print(min_j,min_i,min_path)
j += 1
flagDec = True
if j == n:
break
# print(str(min_j)+" "+str(min_i))
# print(res)
print(str(min_j) + " " + str(min_i))
| 33 | 794 | 14,745,600 |
106349764
|
import sys
def pro():
return sys.stdin.readline().strip()
def rop():
return map(int, pro().split())
def main():
s = list(rop())
a = list(rop())
q = list(rop())
o = list(rop())
p = -1
t = (1e100, -1, -1)
for i in range(s[1]):
while not((p == - 1 or s[2] * q[i] - s[3] * a[p] >= 0)
and (p + 1 == s[0] or s[2] * q[i] - s[3] * a[p+1] < 0)):
p += 1
if p != -1:
t = min(t, (o[i] + (s[2] ** 2 + a[p] ** 2) ** 0.5 + ((s[3] - s[2]) ** 2 + (q[i] - a[p]) ** 2) ** 0.5, p, i))
if p + 1 != s[0]:
t = min(t, (o[i] + (s[2] ** 2 + a[p + 1] ** 2) ** 0.5 + ((s[3] - s[2]) ** 2 + (q[i] - a[p + 1]) ** 2) ** 0.5, p + 1, i))
print(t[1] + 1, t[2] + 1)
main()
|
CROC-MBTU 2012, Final Round
|
CF
| 2,012 | 1 | 256 |
Building Bridge
|
Two villages are separated by a river that flows from the north to the south. The villagers want to build a bridge across the river to make it easier to move across the villages.
The river banks can be assumed to be vertical straight lines x = a and x = b (0 < a < b).
The west village lies in a steppe at point O = (0, 0). There are n pathways leading from the village to the river, they end at points Ai = (a, yi). The villagers there are plain and simple, so their pathways are straight segments as well.
The east village has reserved and cunning people. Their village is in the forest on the east bank of the river, but its exact position is not clear. There are m twisted paths leading from this village to the river and ending at points Bi = (b, y'i). The lengths of all these paths are known, the length of the path that leads from the eastern village to point Bi, equals li.
The villagers want to choose exactly one point on the left bank of river Ai, exactly one point on the right bank Bj and connect them by a straight-line bridge so as to make the total distance between the villages (the sum of |OAi| + |AiBj| + lj, where |XY| is the Euclidean distance between points X and Y) were minimum. The Euclidean distance between points (x1, y1) and (x2, y2) equals $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$.
Help them and find the required pair of points.
|
The first line contains integers n, m, a, b (1 ≤ n, m ≤ 105, 0 < a < b < 106).
The second line contains n integers in the ascending order: the i-th integer determines the coordinate of point Ai and equals yi (|yi| ≤ 106).
The third line contains m integers in the ascending order: the i-th integer determines the coordinate of point Bi and equals y'i (|y'i| ≤ 106).
The fourth line contains m more integers: the i-th of them determines the length of the path that connects the eastern village and point Bi, and equals li (1 ≤ li ≤ 106).
It is guaranteed, that there is such a point C with abscissa at least b, that |BiC| ≤ li for all i (1 ≤ i ≤ m). It is guaranteed that no two points Ai coincide. It is guaranteed that no two points Bi coincide.
|
Print two integers — the numbers of points on the left (west) and right (east) banks, respectively, between which you need to build a bridge. You can assume that the points on the west bank are numbered from 1 to n, in the order in which they are given in the input. Similarly, the points on the east bank are numbered from 1 to m in the order in which they are given in the input.
If there are multiple solutions, print any of them. The solution will be accepted if the final length of the path will differ from the answer of the jury by no more than 10 - 6 in absolute or relative value.
| null | null |
[{"input": "3 2 3 5\n-2 -1 4\n-1 2\n7 3", "output": "2 2"}]
| 1,900 |
[]
| 33 |
[{"input": "3 2 3 5\r\n-2 -1 4\r\n-1 2\r\n7 3\r\n", "output": "2 2"}, {"input": "1 1 10 20\r\n5\r\n-5\r\n1\r\n", "output": "1 1"}, {"input": "2 2 1 2\r\n-1 10\r\n8 9\r\n3 7\r\n", "output": "1 1"}, {"input": "10 20 50 60\r\n-96 -75 32 37 42 43 44 57 61 65\r\n-95 -90 -86 -79 -65 -62 -47 -11 -8 -6 1 8 23 25 32 51 73 88 94 100\r\n138 75 132 116 49 43 96 166 96 161 146 112 195 192 201 186 251 254 220 227\r\n", "output": "2 6"}]
| false |
stdio
| null | true |
888/G
|
888
|
G
|
Python 3
|
TESTS
| 2 | 108 | 1,740,800 |
189721742
|
# LUOGU_RID: 100250825
class trie:
def __init__(self):
self.ch=[None,None]
def add(self,s):
cur=self
for i in s:
if not cur.ch[int(i)]:
cur.ch[int(i)]=trie()
cur=cur.ch[int(i)]
def MST(self):
tot=0
if self.ch[0]:
tot=self.ch[0].MST()
if self.ch[1]:
tot+=self.ch[1].MST()
if self.ch[0] and self.ch[1]:
tot+=int("1"+bestedge(self.ch[0],self.ch[1]),base=2)
return tot
def bestedge(p,q):
res="2"
if p.ch[0] and q.ch[0] or p.ch[1] and q.ch[1]:
if p.ch[0] and q.ch[0]:
res='0'+bestedge(p.ch[0],q.ch[0])
if p.ch[1] and q.ch[1]:
res=min(res,'0'+bestedge(p.ch[1],q.ch[1]))
return res
if p.ch[0] and q.ch[1]:
res='1'+bestedge(p.ch[0],q.ch[1])
if p.ch[1] and q.ch[0]:
res=min(res,'1'+bestedge(p.ch[1],q.ch[0]))
if res=="2":
return ""
return res
T=trie()
n=int(input())
for i in map(int,input().split()):
T.add(bin(i+(2**5))[2:])
print(T.MST())
| 79 | 997 | 149,504,000 |
225470974
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def insert(x, la):
j = 0
for i in p:
if x & i:
if not G[j] >> 25:
la += 1
G[j] ^= la << 25
j = la
else:
j = G[j] >> 25
else:
if not G[j] & 33554431:
la += 1
G[j] ^= la
j = la
else:
j = G[j] & 33554431
cnt[j] += 1
return la
def xor_min(x, s):
j, ans = s, 0
for i in range(30 - len(st), -1, -1):
if x & pow2[i]:
if G[j] >> 25:
j = G[j] >> 25
else:
ans ^= pow2[i]
j = G[j] & 33554431
else:
if G[j] & 33554431:
j = G[j] & 33554431
else:
ans ^= pow2[i]
j = G[j] >> 25
return ans
n = int(input())
a = list(map(int, input().split()))
a = list(set(a))
a.sort()
pow2 = [1]
l = 30
for _ in range(l):
pow2.append(2 * pow2[-1])
p = pow2[::-1]
n = len(a)
G, cnt, la = [0] * 5000000, [0] * 5000000, 0
for i in a:
la = insert(i, la)
h = l
l0 = [-1] * (la + 1)
l0[0] = 0
st = [0]
ans = 0
while st:
i = st[-1]
j, k = G[i] & 33554431, G[i] >> 25
if l0[j] == -1 and l0[k] == -1 and min(cnt[j], cnt[k]):
l = l0[i]
m = l + cnt[j]
r = m + cnt[k]
ans0 = pow2[30]
if (m - l) * (r - m) <= 5000:
for u in range(l, m):
au = a[u]
for v in range(m, r):
ans0 = min(ans0, au ^ a[v])
elif m - l <= r - m:
for u in range(l, m):
ans0 = min(ans0, xor_min(a[u], k))
else:
for u in range(m, r):
ans0 = min(ans0, xor_min(a[u], j))
ans0 |= pow2[h]
ans += ans0
if l0[j] == -1:
if cnt[j] > 1:
st.append(j)
h -= 1
l0[j] = l0[i]
elif l0[k] == -1:
if cnt[k] > 1:
st.append(k)
h -= 1
l0[k] = l0[i] + cnt[j]
else:
st.pop()
h += 1
print(ans)
|
Educational Codeforces Round 32
|
ICPC
| 2,017 | 2 | 256 |
Xor-MST
|
You are given a complete undirected graph with n vertices. A number ai is assigned to each vertex, and the weight of an edge between vertices i and j is equal to ai xor aj.
Calculate the weight of the minimum spanning tree in this graph.
|
The first line contains n (1 ≤ n ≤ 200000) — the number of vertices in the graph.
The second line contains n integers a1, a2, ..., an (0 ≤ ai < 230) — the numbers assigned to the vertices.
|
Print one number — the weight of the minimum spanning tree in the graph.
| null | null |
[{"input": "5\n1 2 3 4 5", "output": "8"}, {"input": "4\n1 2 3 4", "output": "8"}]
| 2,300 |
["bitmasks", "constructive algorithms", "data structures"]
| 79 |
[{"input": "5\r\n1 2 3 4 5\r\n", "output": "8\r\n"}, {"input": "4\r\n1 2 3 4\r\n", "output": "8\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
888/G
|
888
|
G
|
PyPy 3-64
|
TESTS
| 2 | 155 | 103,833,600 |
224998215
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(l, u, v):
return (l << 40) ^ (u << 20) ^ v
def insert(x, c, la):
j = 0
for i in p:
if x & i:
if not G[j] >> 25:
la += 1
G[j] ^= la << 25
j = la
else:
j = G[j] >> 25
else:
if not G[j] & 33554431:
la += 1
G[j] ^= la
j = la
else:
j = G[j] & 33554431
color[j] = c
return la
def xor_min(x, c):
j, ans = 0, 0
for i in p:
if not x & i:
if color[G[j] >> 25] == c:
j = G[j] >> 25
else:
ans += i
j = G[j] & 33554431
else:
if color[G[j] & 33554431] == c:
j = G[j] & 33554431
else:
ans += i
j = G[j] >> 25
return ans
n = int(input())
a = list(map(int, input().split()))
a = list(set(a))
a.sort()
pow2 = [1]
l = 30
for _ in range(l):
pow2.append(2 * pow2[-1])
p = pow2[::-1]
n = len(a)
st = [f(l, 0, n)]
ans = 0
G, color, c, la = [0] * 6000000, [0] * 6000000, 0, 0
while st:
x = st.pop()
l0, u, v = x >> 40, (x >> 20) & 1048575, x & 1048575
c += 1
if u == v or l0 < 0:
continue
p2 = pow2[l0]
if p2 & a[u] == p2 & a[v - 1]:
st.append(f(l0 - 1, u, v))
continue
m = u
while not a[m] & p2:
m += 1
x, y = [u, m], [m, v]
if m - u > v - m:
x, y = y, x
ans0 = pow2[l0 + 1]
if (m - u) * (v - m) < 500:
for i in range(x[0], x[1]):
ai = a[i]
for j in range(y[0], y[1]):
ans0 = min(ans0, ai ^ a[j])
else:
for i in range(x[0], x[1]):
la = insert(a[i], c, la)
for i in range(y[0], y[1]):
ans0 = min(ans0, xor_min(a[i], c))
ans += ans0
st.append(f(l0 - 1, u, m))
st.append(f(l0 - 1, m, v))
print(ans)
| 79 | 997 | 149,504,000 |
225470974
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def insert(x, la):
j = 0
for i in p:
if x & i:
if not G[j] >> 25:
la += 1
G[j] ^= la << 25
j = la
else:
j = G[j] >> 25
else:
if not G[j] & 33554431:
la += 1
G[j] ^= la
j = la
else:
j = G[j] & 33554431
cnt[j] += 1
return la
def xor_min(x, s):
j, ans = s, 0
for i in range(30 - len(st), -1, -1):
if x & pow2[i]:
if G[j] >> 25:
j = G[j] >> 25
else:
ans ^= pow2[i]
j = G[j] & 33554431
else:
if G[j] & 33554431:
j = G[j] & 33554431
else:
ans ^= pow2[i]
j = G[j] >> 25
return ans
n = int(input())
a = list(map(int, input().split()))
a = list(set(a))
a.sort()
pow2 = [1]
l = 30
for _ in range(l):
pow2.append(2 * pow2[-1])
p = pow2[::-1]
n = len(a)
G, cnt, la = [0] * 5000000, [0] * 5000000, 0
for i in a:
la = insert(i, la)
h = l
l0 = [-1] * (la + 1)
l0[0] = 0
st = [0]
ans = 0
while st:
i = st[-1]
j, k = G[i] & 33554431, G[i] >> 25
if l0[j] == -1 and l0[k] == -1 and min(cnt[j], cnt[k]):
l = l0[i]
m = l + cnt[j]
r = m + cnt[k]
ans0 = pow2[30]
if (m - l) * (r - m) <= 5000:
for u in range(l, m):
au = a[u]
for v in range(m, r):
ans0 = min(ans0, au ^ a[v])
elif m - l <= r - m:
for u in range(l, m):
ans0 = min(ans0, xor_min(a[u], k))
else:
for u in range(m, r):
ans0 = min(ans0, xor_min(a[u], j))
ans0 |= pow2[h]
ans += ans0
if l0[j] == -1:
if cnt[j] > 1:
st.append(j)
h -= 1
l0[j] = l0[i]
elif l0[k] == -1:
if cnt[k] > 1:
st.append(k)
h -= 1
l0[k] = l0[i] + cnt[j]
else:
st.pop()
h += 1
print(ans)
|
Educational Codeforces Round 32
|
ICPC
| 2,017 | 2 | 256 |
Xor-MST
|
You are given a complete undirected graph with n vertices. A number ai is assigned to each vertex, and the weight of an edge between vertices i and j is equal to ai xor aj.
Calculate the weight of the minimum spanning tree in this graph.
|
The first line contains n (1 ≤ n ≤ 200000) — the number of vertices in the graph.
The second line contains n integers a1, a2, ..., an (0 ≤ ai < 230) — the numbers assigned to the vertices.
|
Print one number — the weight of the minimum spanning tree in the graph.
| null | null |
[{"input": "5\n1 2 3 4 5", "output": "8"}, {"input": "4\n1 2 3 4", "output": "8"}]
| 2,300 |
["bitmasks", "constructive algorithms", "data structures"]
| 79 |
[{"input": "5\r\n1 2 3 4 5\r\n", "output": "8\r\n"}, {"input": "4\r\n1 2 3 4\r\n", "output": "8\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
177/D1
|
177
|
D1
|
Python 3
|
TESTS1
| 3 | 62 | 0 |
176972603
|
'''
Online Python Compiler.
Code, Compile, Run and Debug python program online.
Write your code in this editor and press "Run" button to execute it.
'''
n,m,c=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
prefsum=[b[0]]
for j in range(1,m): prefsum.append(prefsum[-1]+b[j])
prev=[b[-1]]
for j in range(-2,-(m+1),-1):prev.append(prev[-1]+b[j])
ans=[]
i,j=0,m-1
diff=[0 for times in range(n+1)]
while j<n:
diff[i]+=1
diff[j+1]-=1
i+=1
j+=1
for u in range(1,n+1):
diff[u]=diff[u]+diff[u-1]
flag=0
for y in range(n):
if flag==0:
ans.append(prefsum[diff[y]-1]+a[y])
if y+1<n and diff[y+1]<diff[y]:
flag=1
else:
ans.append(prev[diff[y]-1]+a[y])
for ele in ans:
print(ele%c,end=" ")
| 12 | 124 | 204,800 |
14750244
|
len_message, len_key, mod = map(int, input().split())
message = list(map(int, input().split()))
key = list(map(int, input().split()))
sum, low, high = 0, 0, 0
result = message[:]
for i in range(len_message):
if high < len_key:
sum += key[high]
high += 1
if i + len_key > len_message:
sum -= key[low]
low += 1
result[i] = (result[i] + sum) % mod
print(' '.join(map(str, result)))
|
ABBYY Cup 2.0 - Easy
|
ICPC
| 2,012 | 2 | 256 |
Encrypting Messages
|
The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.
A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.
Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.
Help the Beaver to write a program that will encrypt messages in the described manner.
|
The first input line contains three integers n, m and c, separated by single spaces.
The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message.
The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.
The input limitations for getting 30 points are:
- 1 ≤ m ≤ n ≤ 103
- 1 ≤ c ≤ 103
The input limitations for getting 100 points are:
- 1 ≤ m ≤ n ≤ 105
- 1 ≤ c ≤ 103
|
Print n space-separated integers — the result of encrypting the original message.
| null |
In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer.
|
[{"input": "4 3 2\n1 1 1 1\n1 1 1", "output": "0 1 1 0"}, {"input": "3 1 5\n1 2 3\n4", "output": "0 1 2"}]
| 1,200 |
["brute force"]
| 12 |
[{"input": "4 3 2\r\n1 1 1 1\r\n1 1 1\r\n", "output": "0 1 1 0\r\n"}, {"input": "3 1 5\r\n1 2 3\r\n4\r\n", "output": "0 1 2\r\n"}, {"input": "5 2 7\r\n0 0 1 2 4\r\n3 5\r\n", "output": "3 1 2 3 2\r\n"}, {"input": "20 15 17\r\n4 9 14 11 15 16 15 4 0 10 7 12 10 1 8 6 7 14 1 13\r\n6 3 14 8 8 11 16 4 5 9 2 13 6 14 15\r\n", "output": "10 1 3 8 3 15 7 14 1 12 3 10 15 16 16 5 4 15 13 11\r\n"}]
| false |
stdio
| null | true |
151/B
|
151
|
B
|
Python 3
|
TESTS
| 0 | 15 | 0 |
206245008
|
n = int(input())
x={}
for i in range(n) :
nb , name = map(str , input().split() )
mx_tascore , max_pizzsco , max_girsco = 0 ,0 , 0
for j in range(int(nb)) :
number = input()
if (int(number[0]))>(int(number[1]))>(int(number[3]))>(int(number[4]))>(int(number[6])>(int(number[7])) ) :
max_pizzsco += 1
elif number[0 : 2] == number[3 : 5] == number[6 :] :
mx_tascore += 1
else:
max_girsco += 1
x[name] = [mx_tascore ,max_pizzsco , max_girsco]
mxxx = [[] ,[] ,[]]
for i in range(3) :
mx = None
mx_name =None
for key ,value in x.items() :
if mx is None or mx <= value[i]:
mx = value[i]
for key , value in x.items():
if mx == value[i] :
mxxx[i].append(key)
print("If you want to call a taxi, you should call:" + " " + ",".join(mxxx[0]) +".")
print("If you want to order a pizza, you should call:" + " " +",".join(mxxx[1]) +"." )
print("If you want to go to a cafe with a wonderful girl, you should call:"+" "+ ",".join(mxxx[2])+".")
| 38 | 77 | 0 |
187718251
|
n = int(input())
#taxi, pizza, girl
all = [[], [], []]
names = []
for i in range(n):
s, name = input().split(" ")
s = int(s)
names.append(name)
taxi = 0
pizza = 0
girl = 0
for ii in range(s):
phone = [i for i in input().split("-")]
phone = list(phone[0]) + list(phone[1]) + list(phone[2])
nums = [int(i) for i in phone]
if(len(list(set(nums))) == 1):
taxi += 1
else:
a = [nums[i+1] - nums[i] for i in range(5)]
is_pizza = 1
for ii in a:
if(ii > -1):
is_pizza = 0
if(is_pizza):
pizza += 1
else:
girl += 1
all[0].append(taxi)
all[1].append(pizza)
all[2].append(girl)
print("If you want to call a taxi, you should call: ", end = "")
count = all[0].count(max(all[0]))
for i in range(n):
if(all[0][i] == max(all[0])):
print(names[i], end = "")
count -= 1
if(count):
print(", ", end = "")
else:
print(".")
print("If you want to order a pizza, you should call: ", end = "")
count = all[1].count(max(all[1]))
for i in range(n):
if(all[1][i] == max(all[1])):
print(names[i], end = "")
count -= 1
if(count):
print(", ", end = "")
else:
print(".")
print("If you want to go to a cafe with a wonderful girl, you should call: ", end = "")
count = all[2].count(max(all[2]))
for i in range(n):
if(all[2][i] == max(all[2])):
print(names[i], end = "")
count -= 1
if(count):
print(", ", end = "")
else:
print(".")
|
Codeforces Round 107 (Div. 2)
|
CF
| 2,012 | 2 | 256 |
Phone Numbers
|
Winters are just damn freezing cold in Nvodsk! That's why a group of n friends prefers to take a taxi, order a pizza and call girls. The phone numbers in the city consist of three pairs of digits (for example, 12-34-56). Each friend has a phonebook of size si (that's the number of phone numbers). We know that taxi numbers consist of six identical digits (for example, 22-22-22), the numbers of pizza deliveries should necessarily be decreasing sequences of six different digits (for example, 98-73-21), all other numbers are the girls' numbers.
You are given your friends' phone books. Calculate which friend is best to go to when you are interested in each of those things (who has maximal number of phone numbers of each type).
If the phone book of one person contains some number two times, you should count it twice. That is, each number should be taken into consideration the number of times it occurs in the phone book.
|
The first line contains an integer n (1 ≤ n ≤ 100) — the number of friends.
Then follow n data blocks that describe each friend's phone books. Each block is presented in the following form: first goes the line that contains integer si and string namei (0 ≤ si ≤ 100) — the number of phone numbers in the phone book of the i-th friend and the name of the i-th friend. The name is a non-empty sequence of uppercase and lowercase Latin letters, containing no more than 20 characters. Next si lines contain numbers as "XX-XX-XX", where X is arbitrary digits from 0 to 9.
|
In the first line print the phrase "If you want to call a taxi, you should call: ". Then print names of all friends whose phone books contain maximal number of taxi phone numbers.
In the second line print the phrase "If you want to order a pizza, you should call: ". Then print names of all friends who have maximal number of pizza phone numbers.
In the third line print the phrase "If you want to go to a cafe with a wonderful girl, you should call: ". Then print names of all friends who have maximal number of girls' phone numbers.
Print the names in the order in which they are given in the input data. Separate two consecutive names with a comma and a space. Each line should end with exactly one point. For clarifications concerning the output form, see sample tests. It is necessary that you follow the output form strictly. Extra spaces are not allowed.
| null |
In the first sample you are given four friends. Fedorov's phone book contains one taxi number and one pizza delivery number, Melnikov's phone book only has 3 numbers of girls, Rogulenko's one has 6 taxi numbers and one pizza delivery number, Kaluzhin's one contains 2 taxi numbers and one pizza delivery number.
Thus, if you need to order a taxi, you should obviously call Rogulenko, if you need to order a pizza you should call anybody of the following: Rogulenko, Fedorov, Kaluzhin (each of them has one number). Melnikov has maximal number of phone numbers of girls.
|
[{"input": "4\n2 Fedorov\n22-22-22\n98-76-54\n3 Melnikov\n75-19-09\n23-45-67\n99-99-98\n7 Rogulenko\n22-22-22\n11-11-11\n33-33-33\n44-44-44\n55-55-55\n66-66-66\n95-43-21\n3 Kaluzhin\n11-11-11\n99-99-99\n98-65-32", "output": "If you want to call a taxi, you should call: Rogulenko.\nIf you want to order a pizza, you should call: Fedorov, Rogulenko, Kaluzhin.\nIf you want to go to a cafe with a wonderful girl, you should call: Melnikov."}, {"input": "3\n5 Gleb\n66-66-66\n55-55-55\n01-01-01\n65-43-21\n12-34-56\n3 Serega\n55-55-55\n87-65-43\n65-55-21\n5 Melnik\n12-42-12\n87-73-01\n36-04-12\n88-12-22\n82-11-43", "output": "If you want to call a taxi, you should call: Gleb.\nIf you want to order a pizza, you should call: Gleb, Serega.\nIf you want to go to a cafe with a wonderful girl, you should call: Melnik."}, {"input": "3\n3 Kulczynski\n22-22-22\n65-43-21\n98-12-00\n4 Pachocki\n11-11-11\n11-11-11\n11-11-11\n98-76-54\n0 Smietanka", "output": "If you want to call a taxi, you should call: Pachocki.\nIf you want to order a pizza, you should call: Kulczynski, Pachocki.\nIf you want to go to a cafe with a wonderful girl, you should call: Kulczynski."}]
| 1,200 |
["implementation", "strings"]
| 38 |
[{"input": "4\r\n2 Fedorov\r\n22-22-22\r\n98-76-54\r\n3 Melnikov\r\n75-19-09\r\n23-45-67\r\n99-99-98\r\n7 Rogulenko\r\n22-22-22\r\n11-11-11\r\n33-33-33\r\n44-44-44\r\n55-55-55\r\n66-66-66\r\n95-43-21\r\n3 Kaluzhin\r\n11-11-11\r\n99-99-99\r\n98-65-32\r\n", "output": "If you want to call a taxi, you should call: Rogulenko.\r\nIf you want to order a pizza, you should call: Fedorov, Rogulenko, Kaluzhin.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Melnikov.\r\n"}, {"input": "3\r\n5 Gleb\r\n66-66-66\r\n55-55-55\r\n01-01-01\r\n65-43-21\r\n12-34-56\r\n3 Serega\r\n55-55-55\r\n87-65-43\r\n65-55-21\r\n5 Melnik\r\n12-42-12\r\n87-73-01\r\n36-04-12\r\n88-12-22\r\n82-11-43\r\n", "output": "If you want to call a taxi, you should call: Gleb.\r\nIf you want to order a pizza, you should call: Gleb, Serega.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Melnik.\r\n"}, {"input": "3\r\n3 Kulczynski\r\n22-22-22\r\n65-43-21\r\n98-12-00\r\n4 Pachocki\r\n11-11-11\r\n11-11-11\r\n11-11-11\r\n98-76-54\r\n0 Smietanka\r\n", "output": "If you want to call a taxi, you should call: Pachocki.\r\nIf you want to order a pizza, you should call: Kulczynski, Pachocki.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Kulczynski.\r\n"}, {"input": "4\r\n0 Gleb\r\n0 Sergey\r\n0 Sasha\r\n0 HrenSGori\r\n", "output": "If you want to call a taxi, you should call: Gleb, Sergey, Sasha, HrenSGori.\r\nIf you want to order a pizza, you should call: Gleb, Sergey, Sasha, HrenSGori.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Gleb, Sergey, Sasha, HrenSGori.\r\n"}, {"input": "5\r\n0 PmfItzXdroG\r\n0 HRykTUCkxgOaD\r\n0 fcHIUkrn\r\n2 eUvyUuXFvvuYobrFBxe\r\n98-76-32\r\n02-21-39\r\n8 VUMvHy\r\n97-65-41\r\n65-70-26\r\n54-49-11\r\n33-33-33\r\n76-54-31\r\n11-11-11\r\n82-95-22\r\n98-75-31\r\n", "output": "If you want to call a taxi, you should call: VUMvHy.\r\nIf you want to order a pizza, you should call: VUMvHy.\r\nIf you want to go to a cafe with a wonderful girl, you should call: VUMvHy.\r\n"}, {"input": "5\r\n2 ZaxsHjkGMPxZgwzpya\r\n94-20-75\r\n96-54-32\r\n2 gAiJXEYwXU\r\n11-11-11\r\n77-77-77\r\n1 j\r\n86-43-10\r\n1 dRJrc\r\n98-76-21\r\n2 UAiXZTnBKDoKb\r\n35-19-89\r\n98-65-40\r\n", "output": "If you want to call a taxi, you should call: gAiJXEYwXU.\r\nIf you want to order a pizza, you should call: ZaxsHjkGMPxZgwzpya, j, dRJrc, UAiXZTnBKDoKb.\r\nIf you want to go to a cafe with a wonderful girl, you should call: ZaxsHjkGMPxZgwzpya, UAiXZTnBKDoKb.\r\n"}, {"input": "5\r\n4 vKHeRjJubHZ\r\n11-11-11\r\n99-99-99\r\n00-00-00\r\n52-73-46\r\n6 hckQfheNMOgZVsa\r\n96-53-20\r\n50-69-33\r\n64-78-80\r\n77-77-77\r\n06-10-48\r\n33-39-96\r\n1 RykElQYdYbQfqlrk\r\n97-43-21\r\n4 GDptSUmbYqkjW\r\n87-42-10\r\n56-87-67\r\n86-54-20\r\n65-43-10\r\n6 jUEgOK\r\n87-65-40\r\n05-90-59\r\n06-32-30\r\n44-57-02\r\n48-78-94\r\n55-55-55\r\n", "output": "If you want to call a taxi, you should call: vKHeRjJubHZ.\r\nIf you want to order a pizza, you should call: GDptSUmbYqkjW.\r\nIf you want to go to a cafe with a wonderful girl, you should call: hckQfheNMOgZVsa, jUEgOK.\r\n"}, {"input": "10\r\n5 eeleGlOFWbcnIPPtnll\r\n55-55-55\r\n00-00-00\r\n98-65-32\r\n76-43-10\r\n98-76-54\r\n2 DMBiqRyQJkFvHPJNJp\r\n28-97-50\r\n87-64-10\r\n4 bAfmtnKHohIX\r\n61-58-93\r\n77-77-77\r\n53-17-51\r\n96-43-10\r\n3 cDX\r\n22-22-22\r\n77-77-77\r\n63-30-64\r\n1 HCeHJ\r\n44-44-44\r\n6 HgSpfAolwoaBQ\r\n96-93-53\r\n98-53-10\r\n33-33-33\r\n66-66-66\r\n87-54-32\r\n11-11-11\r\n0 hn\r\n7 qGRocddf\r\n74-34-87\r\n97-53-20\r\n76-32-10\r\n54-32-10\r\n98-74-21\r\n33-33-33\r\n00-00-00\r\n5 XrdtbTC\r\n99-99-99\r\n86-53-20\r\n96-34-97\r\n75-43-20\r\n85-32-10\r\n0 gDLEXYNyoDSgSLJSec\r\n", "output": "If you want to call a taxi, you should call: HgSpfAolwoaBQ.\r\nIf you want to order a pizza, you should call: qGRocddf.\r\nIf you want to go to a cafe with a wonderful girl, you should call: bAfmtnKHohIX.\r\n"}, {"input": "3\r\n5 hieu\r\n11-22-33\r\n22-33-55\r\n33-66-22\r\n99-00-22\r\n55-33-11\r\n4 duong\r\n11-11-11\r\n22-22-22\r\n33-33-33\r\n44-44-44\r\n3 quan\r\n98-76-54\r\n76-54-32\r\n65-43-21\r\n", "output": "If you want to call a taxi, you should call: duong.\r\nIf you want to order a pizza, you should call: quan.\r\nIf you want to go to a cafe with a wonderful girl, you should call: hieu.\r\n"}, {"input": "2\r\n3 ha\r\n11-11-11\r\n98-76-54\r\n12-34-56\r\n1 haha\r\n98-76-55\r\n", "output": "If you want to call a taxi, you should call: ha.\r\nIf you want to order a pizza, you should call: ha.\r\nIf you want to go to a cafe with a wonderful girl, you should call: ha, haha.\r\n"}, {"input": "2\r\n2 Alex\r\n12-12-12\r\n99-87-76\r\n2 Mula\r\n22-22-22\r\n99-87-76\r\n", "output": "If you want to call a taxi, you should call: Mula.\r\nIf you want to order a pizza, you should call: Alex, Mula.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Alex.\r\n"}, {"input": "2\r\n2 Alex\r\n12-12-12\r\n99-98-76\r\n2 Mula\r\n22-22-22\r\n99-98-76\r\n", "output": "If you want to call a taxi, you should call: Mula.\r\nIf you want to order a pizza, you should call: Alex, Mula.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Alex.\r\n"}, {"input": "3\r\n5 Gleb\r\n66-66-66\r\n55-55-55\r\n01-01-01\r\n65-43-21\r\n12-34-56\r\n8 Serega\r\n55-55-55\r\n87-65-43\r\n65-55-21\r\n11-22-33\r\n11-22-33\r\n11-22-33\r\n11-22-33\r\n11-22-33\r\n5 Melnik\r\n12-42-12\r\n87-73-01\r\n36-04-12\r\n88-12-22\r\n82-11-43\r\n", "output": "If you want to call a taxi, you should call: Gleb.\r\nIf you want to order a pizza, you should call: Gleb, Serega.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Serega.\r\n"}, {"input": "2\r\n2 sur\r\n32-32-32\r\n43-43-43\r\n2 sun\r\n22-22-22\r\n23-41-31\r\n", "output": "If you want to call a taxi, you should call: sun.\r\nIf you want to order a pizza, you should call: sur, sun.\r\nIf you want to go to a cafe with a wonderful girl, you should call: sur.\r\n"}, {"input": "3\r\n1 Fedorov\r\n21-21-21\r\n1 Melnikov\r\n99-99-99\r\n1 Rogulenko\r\n95-43-21\r\n", "output": "If you want to call a taxi, you should call: Melnikov.\r\nIf you want to order a pizza, you should call: Rogulenko.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Fedorov.\r\n"}, {"input": "3\r\n5 Gleb\r\n66-66-66\r\n55-55-55\r\n01-01-01\r\n65-43-21\r\n12-34-56\r\n8 Serega\r\n55-55-55\r\n87-65-43\r\n65-55-21\r\n56-34-12\r\n56-34-12\r\n56-34-12\r\n56-34-12\r\n56-34-12\r\n5 Melnik\r\n12-42-12\r\n87-73-01\r\n36-04-12\r\n88-12-22\r\n82-11-43\r\n", "output": "If you want to call a taxi, you should call: Gleb.\r\nIf you want to order a pizza, you should call: Gleb, Serega.\r\nIf you want to go to a cafe with a wonderful girl, you should call: Serega.\r\n"}]
| false |
stdio
| null | true |
412/D
|
412
|
D
|
Python 3
|
PRETESTS
| 0 | 46 | 0 |
6411160
|
import sys
line = str.split(input()," ")
n = int(line[0])
m = int(line[1])
ab=[]
for i in range(0,m):
line = str.split(input()," ")
ab.append([int(line[0]),int(line[1])])
#print("n:"+str(n))
#print("m:"+str(m))
#print("ab:"+str(ab))
def contains(a,b,ab):
f = [a,b]
for i in ab:
if (f==i):
return True
return False
persons=[]
for i in range(0,n):
persons.append(i+1)
donepersons=[]
success = False
def next(personlist,donepersons,ab,last):
#print(str(personlist)+" - "+str(donepersons) + " - "+str(last))
global success
if (len(personlist)==0):
success = True
sys.stdout.write(str(donepersons[0]))
for i in range(1,n):
sys.stdout.write(" "+str(donepersons[i]))
return
if (last==None):
for p in personlist:
if (success == True):
return
newpersonlist = personlist
newdonepersons = donepersons
newlast = p
newpersonlist.remove(p)
newdonepersons.append(p)
next(newpersonlist,newdonepersons,ab,newlast)
for p in personlist:
if (success == True):
return
if (contains(last,p,ab)):
continue
else:
newlast = p
newpersonlist = personlist
newdonepersons = donepersons
newpersonlist.remove(p)
newdonepersons.append(p)
next(newpersonlist,newdonepersons,ab,newlast)
next(persons,donepersons,ab,None)
if (success == False):
sys.stdout.write(str(-1))
| 33 | 343 | 9,830,400 |
159254152
|
n,m=map(int,input().split())
g=[set()for i in range(n)]
for i in range(m):
a,b=map(int,input().split())
g[a-1].add(b-1)
c=[0]*n
for i in range(n): c[i]=i
for i in range(n):
j=i
while j>0 and c[j] in g[c[j-1]]:
c[j],c[j-1]=c[j-1],c[j]
j-=1
for i in c: print(i+1,end=' ')
|
Coder-Strike 2014 - Round 1
|
CF
| 2,014 | 1 | 256 |
Giving Awards
|
The employees of the R1 company often spend time together: they watch football, they go camping, they solve contests. So, it's no big deal that sometimes someone pays for someone else.
Today is the day of giving out money rewards. The R1 company CEO will invite employees into his office one by one, rewarding each one for the hard work this month. The CEO knows who owes money to whom. And he also understands that if he invites person x to his office for a reward, and then immediately invite person y, who has lent some money to person x, then they can meet. Of course, in such a situation, the joy of person x from his brand new money reward will be much less. Therefore, the R1 CEO decided to invite the staff in such an order that the described situation will not happen for any pair of employees invited one after another.
However, there are a lot of employees in the company, and the CEO doesn't have a lot of time. Therefore, the task has been assigned to you. Given the debt relationships between all the employees, determine in which order they should be invited to the office of the R1 company CEO, or determine that the described order does not exist.
|
The first line contains space-separated integers n and m $$(2\leq n\leq3\cdot10^{4};1\leq m\leq min(10^{5},\frac{n(n-1)}{2}))$$ — the number of employees in R1 and the number of debt relations. Each of the following m lines contains two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi), these integers indicate that the person number ai owes money to a person a number bi. Assume that all the employees are numbered from 1 to n.
It is guaranteed that each pair of people p, q is mentioned in the input data at most once. In particular, the input data will not contain pairs p, q and q, p simultaneously.
|
Print -1 if the described order does not exist. Otherwise, print the permutation of n distinct integers. The first number should denote the number of the person who goes to the CEO office first, the second number denote the person who goes second and so on.
If there are multiple correct orders, you are allowed to print any of them.
| null | null |
[{"input": "2 1\n1 2", "output": "2 1"}, {"input": "3 3\n1 2\n2 3\n3 1", "output": "2 1 3"}]
| 2,000 |
["dfs and similar"]
| 33 |
[{"input": "2 1\r\n1 2\r\n", "output": "2 1 \r\n"}, {"input": "3 3\r\n1 2\r\n2 3\r\n3 1\r\n", "output": "2 1 3 \r\n"}, {"input": "10 45\r\n10 5\r\n10 7\r\n6 1\r\n5 8\r\n3 5\r\n6 5\r\n1 2\r\n6 10\r\n2 9\r\n9 5\r\n4 1\r\n7 5\r\n1 8\r\n6 8\r\n10 9\r\n7 2\r\n7 9\r\n4 10\r\n7 3\r\n4 8\r\n10 3\r\n10 8\r\n2 10\r\n8 2\r\n4 2\r\n5 2\r\n9 1\r\n4 5\r\n1 3\r\n9 6\r\n3 8\r\n5 1\r\n6 4\r\n4 7\r\n8 7\r\n7 1\r\n9 3\r\n3 4\r\n6 2\r\n3 2\r\n6 7\r\n6 3\r\n4 9\r\n8 9\r\n1 10\r\n", "output": "2 3 1 5 7 8 4 6 9 10 \r\n"}, {"input": "4 6\r\n3 4\r\n4 1\r\n4 2\r\n2 1\r\n3 2\r\n3 1\r\n", "output": "1 2 4 3 \r\n"}, {"input": "5 10\r\n4 2\r\n4 5\r\n3 1\r\n2 1\r\n3 4\r\n3 2\r\n5 1\r\n5 2\r\n4 1\r\n5 3\r\n", "output": "1 2 4 3 5 \r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m = map(int, f.readline().split())
edges = set()
for _ in range(m):
a, b = map(int, f.readline().split())
edges.add((a, b))
# Read submission output
with open(submission_path) as f:
line = f.readline().strip()
if line == '-1':
# Check if it's actually impossible
# To determine this, we'd need to know if any valid order exists, which is complex.
# However, the problem requires the checker to verify the submission's output.
# Since the submission claims it's impossible, but how can the checker know?
# This suggests that the checker cannot handle the -1 case properly.
# Wait, the problem says if the order does not exist, output -1.
# But how can the checker verify if the submission's -1 is correct?
# This is a problem because the checker would need to solve the problem itself.
# However, the checker can't do that. So for this problem, the checker can't validate a submission that outputs -1.
# Therefore, in the checker code, if the submission outputs -1, we have to check if a valid order exists.
# But that's equivalent to solving the problem. Which is impossible for a checker to do without the correct algorithm.
# Hence, this problem's checker is incomplete. However, given the problem statement, perhaps the checker is expected to handle cases where the submission outputs -1 correctly.
# So, the approach here is that the checker cannot fully validate submissions that output -1. Thus, in practice, such cases would have to be handled by the contest system's test cases, which would have the correct answer (either a permutation or -1).
# However, in the problem statement, the output is either a permutation or -1. So the checker must check two possibilities:
# 1. The submission outputs a valid permutation (so -1 is wrong)
# 2. The submission outputs -1, and no valid permutation exists.
# But how can the checker distinguish between the two? It can't unless it solves the problem.
# Therefore, the checker code here can't handle the -1 case. Thus, the problem may have test cases where the correct answer is -1, and the checker would need to check that the submission's output is -1 and that no valid permutation exists. But how?
# This is a problem. Therefore, perhaps the correct approach is that in this problem, the checker can only check the permutation case. If the submission outputs -1, the checker can't validate it unless it knows whether such a permutation exists (which requires solving the problem).
# Given that the checker's code is supposed to be written without solving the problem, this is impossible. Therefore, the problem is not a 'checker' type but 'diff'? But the problem allows multiple correct outputs, so it's a checker. However, the -1 case complicates things.
# However, according to the problem statement, the correct answer is either -1 or a permutation. So if the submission outputs -1, the checker must check whether the correct answer is -1. But how?
# This is a contradiction. Therefore, in practice, perhaps the test cases where the answer is -1 would be evaluated as 'diff' (exact match), and the rest as 'checker'.
# However, the problem statement's output allows any permutation. So if the reference output is a permutation, then the submission must also be a permutation. If the reference output is -1, then the submission must be -1.
# Therefore, the checker must compare the submission's output with the reference output in the case where the reference output is -1. Otherwise, check the permutation.
# So, the checker must read the reference output. If the reference output is -1, then the submission must also output -1. Else, the submission must output a valid permutation.
# Therefore, the checker logic is:
# Read the reference output. If it's -1, then submission must be -1.
# Otherwise, submission must be a valid permutation.
# So, the code must compare the submission against the reference output in the case where the reference is -1.
# So, the code should proceed as follows:
# 1. Read the reference output (output_path).
with open(output_path) as ref_f:
ref_line = ref_f.readline().strip()
if ref_line == '-1':
# The correct answer is -1; submission must be -1
if line == '-1':
print(1)
return
else:
print(0)
return
else:
# The correct answer is a permutation. The submission must be a valid permutation.
# So proceed with permutation checks.
pass
# If submission is -1 but reference is not, invalid.
if line == '-1':
print(0)
return
# Parse submission's permutation
perm = list(map(int, line.split()))
if len(perm) != n:
print(0)
return
if set(perm) != set(range(1, n+1)):
print(0)
return
# Check consecutive pairs
for i in range(n-1):
x = perm[i]
y = perm[i+1]
if (x, y) in edges:
print(0)
return
print(1)
if __name__ == '__main__':
main()
| true |
436/C
|
436
|
C
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
176955206
|
import sys
from array import array
from bisect import *
input = lambda: sys.stdin.buffer.readline().decode().strip()
inp = lambda dtype: [dtype(x) for x in input().split()]
inp_2d = lambda dtype, n: [dtype(input()) for _ in range(n)]
inp_2ds = lambda dtype, n: [inp(dtype) for _ in range(n)]
ceil1 = lambda a, b: (a + b - 1) // b
debug = lambda *x: print(*x, file=sys.stderr)
out = []
class disjointset:
def __init__(self, n):
self.rank, self.parent = array('i', [0] * (n + 1)), array('i', [i for i in range(n + 1)])
self.n = n
def find(self, x):
xcopy = x
while x != self.parent[x]:
x = self.parent[x]
while xcopy != x:
self.parent[xcopy], xcopy = x, self.parent[xcopy]
return x
def union(self, x, y):
xpar, ypar = self.find(x), self.find(y)
if xpar == ypar:
return
par, child = xpar, ypar
if self.rank[xpar] < self.rank[ypar]:
par, child = ypar, xpar
elif self.rank[xpar] == self.rank[ypar]:
self.rank[xpar] += 1
self.parent[child] = par
self.n -= 1
def kruskal(self, edges):
ixs = sorted(range(len(edges)), key=edges.__getitem__)
ret = 0
# loop over v-1
for i in ixs:
w, u, v = edges[i]
upar, vpar = self.find(u), self.find(v)
if upar != vpar:
self.union(upar, vpar)
ret += w
g.add_edge(u, v)
return ret + self.n * n * m
class graph:
def __init__(self, n):
self.n = n
self.gdict = [array('i') for _ in range(n + 1)]
self.vis = array('b', [0] * (self.n + 1))
def add_edge(self, node1, node2):
self.gdict[node1].append(node2)
self.gdict[node2].append(node1)
def dfs(self, root):
stk = array('i', [root])
self.vis[root] = True
while stk:
node = stk.pop()
for ch in self.gdict[node]:
if not self.vis[ch]:
self.vis[ch] = True
print(ch, root)
stk.append(ch)
n, m, k, w = inp(int)
a = [inp_2d(str, n) for _ in range(k)]
edges = []
dis, g = disjointset(k), graph(k)
for i in range(k):
for j in range(i + 1, k):
cost = w * sum(array('b', [a[i][r][c] != a[j][r][c] for r in range(n) for c in range(m)]))
if cost < n * m:
edges.append((cost, i + 1, j + 1))
print(dis.kruskal(edges))
for i in range(1, k + 1):
if not g.vis[i]:
print(i, 0)
g.dfs(i)
| 30 | 1,964 | 62,873,600 |
232798881
|
import heapq
total_bytes = 0
messages_sent = []
def send_message(messages, weight, start_ind):
global total_bytes
global messages_sent
num_messages = len(messages)
num_rows = len(messages[1])
num_cols = len(messages[1][0])
sent = [False] * num_messages
# Send the first message
total_bytes += num_rows * num_cols
mst = ["1 0"]
sent[start_ind] = True
queue = []
start_message = messages[start_ind]
max_bytes = num_rows * num_cols
for curr_ind in range(1, num_messages):
curr_message = messages[curr_ind]
diff = 0
if curr_ind != start_ind:
for i in range(num_rows):
for j in range(num_cols):
if curr_message[i][j] != start_message[i][j]:
diff += 1
if max_bytes > diff * weight:
heapq.heappush(queue, (diff * weight, (start_ind, curr_ind)))
else:
heapq.heappush(queue, (max_bytes, (0, curr_ind)))
while queue:
curr_bytes, (prev_ind, curr_ind) = heapq.heappop(queue)
if sent[curr_ind]:
continue
mst.append(f"{curr_ind} {prev_ind}")
sent[curr_ind] = True
total_bytes += curr_bytes
curr_message = messages[curr_ind]
for next_ind in range(1, num_messages):
if next_ind != curr_ind and not sent[next_ind]:
next_message = messages[next_ind]
diff = 0
for i in range(num_rows):
for j in range(num_cols):
if curr_message[i][j] != next_message[i][j]:
diff += 1
if max_bytes > diff * weight:
heapq.heappush(queue, (diff * weight, (curr_ind, next_ind)))
else:
heapq.heappush(queue, (max_bytes, (0, next_ind)))
return mst
def main():
global total_bytes
global messages_sent
num_rows, num_cols, num_messages, weight = map(int, input().split())
messages = [[]]
for _ in range(num_messages):
message = [input() for _ in range(num_rows)]
messages.append(message)
result = send_message(messages, weight, 1)
print(total_bytes)
for line in result:
print(line)
if __name__ == "__main__":
main()
|
Zepto Code Rush 2014
|
CF
| 2,014 | 2 | 256 |
Dungeons and Candies
|
During the loading of the game "Dungeons and Candies" you are required to get descriptions of k levels from the server. Each description is a map of an n × m checkered rectangular field. Some cells of the field contain candies (each cell has at most one candy). An empty cell is denoted as "." on the map, but if a cell has a candy, it is denoted as a letter of the English alphabet. A level may contain identical candies, in this case the letters in the corresponding cells of the map will be the same.
When you transmit information via a network, you want to minimize traffic — the total size of the transferred data. The levels can be transmitted in any order. There are two ways to transmit the current level A:
1. You can transmit the whole level A. Then you need to transmit n·m bytes via the network.
2. You can transmit the difference between level A and some previously transmitted level B (if it exists); this operation requires to transmit dA, B·w bytes, where dA, B is the number of cells of the field that are different for A and B, and w is a constant. Note, that you should compare only the corresponding cells of levels A and B to calculate dA, B. You cannot transform the maps of levels, i.e. rotate or shift them relatively to each other.
Your task is to find a way to transfer all the k levels and minimize the traffic.
|
The first line contains four integers n, m, k, w (1 ≤ n, m ≤ 10; 1 ≤ k, w ≤ 1000). Then follows the description of k levels. Each level is described by n lines, each line contains m characters. Each character is either a letter of the English alphabet or a dot ("."). Please note that the case of the letters matters.
|
In the first line print the required minimum number of transferred bytes.
Then print k pairs of integers x1, y1, x2, y2, ..., xk, yk, describing the way to transfer levels. Pair xi, yi means that level xi needs to be transferred by way yi. If yi equals 0, that means that the level must be transferred using the first way, otherwise yi must be equal to the number of a previously transferred level. It means that you will transfer the difference between levels yi and xi to transfer level xi. Print the pairs in the order of transferring levels. The levels are numbered 1 through k in the order they follow in the input.
If there are multiple optimal solutions, you can print any of them.
| null | null |
[{"input": "2 3 3 2\nA.A\n...\nA.a\n..C\nX.Y\n...", "output": "14\n1 0\n2 1\n3 1"}, {"input": "1 1 4 1\nA\n.\nB\n.", "output": "3\n1 0\n2 0\n4 2\n3 0"}, {"input": "1 3 5 2\nABA\nBBB\nBBA\nBAB\nABB", "output": "11\n1 0\n3 1\n2 3\n4 2\n5 1"}]
| 1,800 |
["dsu", "graphs", "greedy", "trees"]
| 30 |
[{"input": "2 3 3 2\r\nA.A\r\n...\r\nA.a\r\n..C\r\nX.Y\r\n...\r\n", "output": "14\r\n1 0\r\n2 1\r\n3 1\r\n"}, {"input": "1 1 4 1\r\nA\r\n.\r\nB\r\n.\r\n", "output": "3\r\n1 0\r\n2 0\r\n4 2\r\n3 0\r\n"}, {"input": "1 3 5 2\r\nABA\r\nBBB\r\nBBA\r\nBAB\r\nABB\r\n", "output": "11\r\n1 0\r\n3 1\r\n2 3\r\n4 2\r\n5 1\r\n"}, {"input": "2 2 5 1\r\n..\r\nBA\r\n.A\r\nB.\r\n..\r\nA.\r\nAB\r\n.B\r\n..\r\n..\r\n", "output": "12\r\n1 0\r\n2 1\r\n3 1\r\n5 3\r\n4 5\r\n"}, {"input": "3 3 10 2\r\nBA.\r\n..A\r\n.BB\r\nB..\r\n..B\r\n.AA\r\nB..\r\nAB.\r\n..A\r\nBAB\r\n.A.\r\n.B.\r\n..B\r\nA..\r\n...\r\n...\r\n.B.\r\nBA.\r\n..B\r\n.AB\r\n.B.\r\nB.A\r\n.A.\r\n.BA\r\n..B\r\n...\r\n.A.\r\n.AA\r\n..A\r\n.B.\r\n", "output": "67\r\n1 0\r\n10 1\r\n2 1\r\n3 2\r\n4 1\r\n7 4\r\n9 7\r\n5 9\r\n6 9\r\n8 4\r\n"}, {"input": "3 1 5 1\r\nB\r\nA\r\nB\r\nA\r\nA\r\nB\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\n", "output": "5\r\n1 0\r\n2 1\r\n3 2\r\n4 3\r\n5 3\r\n"}, {"input": "3 2 10 1\r\nAB\r\nBA\r\nAB\r\nAA\r\nAA\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBA\r\nAB\r\nAA\r\nBB\r\nAB\r\nBA\r\nBB\r\nBB\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBB\r\nAB\r\nAA\r\n", "output": "16\r\n1 0\r\n3 1\r\n8 3\r\n2 3\r\n4 2\r\n9 4\r\n6 4\r\n7 6\r\n10 6\r\n5 10\r\n"}, {"input": "2 3 10 2\r\nABB\r\nABA\r\nAAB\r\nBAB\r\nAAA\r\nBBA\r\nBBB\r\nBAA\r\nBBB\r\nABB\r\nABA\r\nBBA\r\nBBB\r\nAAB\r\nABA\r\nABB\r\nBBA\r\nBAB\r\nBBB\r\nBBB\r\n", "output": "38\r\n1 0\r\n5 1\r\n7 5\r\n4 7\r\n9 4\r\n10 5\r\n6 1\r\n3 6\r\n8 1\r\n2 0\r\n"}, {"input": "1 1 1 1\r\n.\r\n", "output": "1\r\n1 0\r\n"}]
| false |
stdio
|
import sys
def read_levels(input_path):
with open(input_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip()]
first_line = lines[0].split()
n, m, k, w = map(int, first_line)
levels = [None] # 1-based indexing
idx = 1
for _ in range(k):
level = []
for _ in range(n):
if idx >= len(lines):
return None, None, None, None, None
level.append(lines[idx].strip())
idx += 1
levels.append(level)
return n, m, k, w, levels
def main(input_path, output_path, submission_path):
n, m, k, w, levels = read_levels(input_path)
if levels is None:
return 0
# Read reference output's first line
with open(output_path) as f_ref:
ref_lines = f_ref.read().splitlines()
if not ref_lines:
return 0
ref_total = int(ref_lines[0])
# Read submission output
with open(submission_path) as f_sub:
sub_lines = f_sub.read().splitlines()
if not sub_lines:
return 0
try:
sub_total = int(sub_lines[0])
except:
return 0
# Check sub_total matches reference
if sub_total != ref_total:
return 0
# Parse pairs
tokens = []
for line in sub_lines[1:]:
tokens.extend(line.strip().split())
if len(tokens) != 2 * k:
return 0
pairs = []
try:
for i in range(0, len(tokens), 2):
xi = int(tokens[i])
yi = int(tokens[i+1])
pairs.append((xi, yi))
except:
return 0
# Check all xi are unique and 1..k
seen_xi = set()
for xi, _ in pairs:
if xi < 1 or xi > k or xi in seen_xi:
return 0
seen_xi.add(xi)
if len(seen_xi) != k:
return 0
# Check pairs order and calculate total
transmitted = set()
total = 0
for xi, yi in pairs:
if yi != 0 and yi not in transmitted:
return 0
if yi == 0:
cost = n * m
else:
# Compute d between xi and yi
d = 0
for row in range(n):
for col in range(m):
if levels[xi][row][col] != levels[yi][row][col]:
d += 1
cost = d * w
total += cost
transmitted.add(xi)
if total != sub_total:
return 0
return 1
if __name__ == "__main__":
input_path, output_path, submission_path = sys.argv[1:]
score = main(input_path, output_path, submission_path)
print(score)
| true |
436/C
|
436
|
C
|
Python 3
|
TESTS
| 2 | 61 | 614,400 |
10516965
|
import random
#dsu
p = []
def finds(v):
if p[v] != v:
p[v] = finds(p[v])
return p[v]
def union(v1,v2):
r1 = finds(v1)
r2 = finds(v2)
if r1 != r2:
if random.choice([0,1]) == 0:
p[r1] = r2
else:
p[r2] = r1
#input
I = lambda:map(int,input().split())
n,m,k,w = I()
#solution
t = [[None for _ in range(n)] for _ in range(k)]
for i in range(k):
for j in range(n):
t[i][j] = list(input())
def dif(a,b):
ans = 0
for i in range(n):
for j in range(m):
ans += int(a[i][j] != b[i][j])
return ans * w
#G = [[] for _ in range(k)]
edge = []
for i in range(k):
for j in range(i + 1,k):
wig = dif (t[i],t[j])
if wig < n*m:
edge.append((i,j,wig))
#G[i].append((j,wig))
#G[j].append((i,wig))
edge.sort(key=lambda x: x[2]) # sort by the wig parameter
#Kruskal's algorithm
p = list(range(0,k)) # parent array
path = [-1] * k
ans = 0
for i in edge:
a,b,w = i
if finds(a) != finds(b):
path[a] = b
path[b] = a
union(a,b)
ans += w
mark = [False] * k
st = set()
for x in range(k):
d = finds(x)
if d not in st:
mark[x] = True
st.add(d)
ans += n * m * len(st)
print (ans)
for i in range(k):
if mark[i] == True:
print (i + 1,0)
else:
print (i + 1,path[i] + 1)
| 30 | 1,466 | 70,041,600 |
87968659
|
def put():
return map(int, input().split())
def diff(x,y):
ans = 0
for i in range(n*m):
if s[x][i]!= s[y][i]:
ans+=1
return ans
def find(i):
if i==p[i]:
return i
p[i] = find(p[i])
return p[i]
def union(i,j):
if rank[i]>rank[j]:
i,j = j,i
elif rank[i]==rank[j]:
rank[j]+=1
p[i]= j
def dfs(i,p):
if i!=0:
print(i,p)
for j in tree[i]:
if j!=p:
dfs(j,i)
n,m,k,w = put()
s = ['']*k
for i in range(k):
for j in range(n):
s[i]+=input()
edge = []
k+=1
rank = [0]*(k)
p = list(range(k))
cost = 0
tree = [[] for i in range(k)]
for i in range(k):
for j in range(i+1,k):
if i==0:
z=n*m
else:
z = diff(i-1,j-1)*w
edge.append((z,i,j))
edge.sort()
for z,i,j in edge:
u = find(i)
v = find(j)
if u!=v:
union(u,v)
cost+= z
tree[i].append(j)
tree[j].append(i)
print(cost)
dfs(0,-1)
|
Zepto Code Rush 2014
|
CF
| 2,014 | 2 | 256 |
Dungeons and Candies
|
During the loading of the game "Dungeons and Candies" you are required to get descriptions of k levels from the server. Each description is a map of an n × m checkered rectangular field. Some cells of the field contain candies (each cell has at most one candy). An empty cell is denoted as "." on the map, but if a cell has a candy, it is denoted as a letter of the English alphabet. A level may contain identical candies, in this case the letters in the corresponding cells of the map will be the same.
When you transmit information via a network, you want to minimize traffic — the total size of the transferred data. The levels can be transmitted in any order. There are two ways to transmit the current level A:
1. You can transmit the whole level A. Then you need to transmit n·m bytes via the network.
2. You can transmit the difference between level A and some previously transmitted level B (if it exists); this operation requires to transmit dA, B·w bytes, where dA, B is the number of cells of the field that are different for A and B, and w is a constant. Note, that you should compare only the corresponding cells of levels A and B to calculate dA, B. You cannot transform the maps of levels, i.e. rotate or shift them relatively to each other.
Your task is to find a way to transfer all the k levels and minimize the traffic.
|
The first line contains four integers n, m, k, w (1 ≤ n, m ≤ 10; 1 ≤ k, w ≤ 1000). Then follows the description of k levels. Each level is described by n lines, each line contains m characters. Each character is either a letter of the English alphabet or a dot ("."). Please note that the case of the letters matters.
|
In the first line print the required minimum number of transferred bytes.
Then print k pairs of integers x1, y1, x2, y2, ..., xk, yk, describing the way to transfer levels. Pair xi, yi means that level xi needs to be transferred by way yi. If yi equals 0, that means that the level must be transferred using the first way, otherwise yi must be equal to the number of a previously transferred level. It means that you will transfer the difference between levels yi and xi to transfer level xi. Print the pairs in the order of transferring levels. The levels are numbered 1 through k in the order they follow in the input.
If there are multiple optimal solutions, you can print any of them.
| null | null |
[{"input": "2 3 3 2\nA.A\n...\nA.a\n..C\nX.Y\n...", "output": "14\n1 0\n2 1\n3 1"}, {"input": "1 1 4 1\nA\n.\nB\n.", "output": "3\n1 0\n2 0\n4 2\n3 0"}, {"input": "1 3 5 2\nABA\nBBB\nBBA\nBAB\nABB", "output": "11\n1 0\n3 1\n2 3\n4 2\n5 1"}]
| 1,800 |
["dsu", "graphs", "greedy", "trees"]
| 30 |
[{"input": "2 3 3 2\r\nA.A\r\n...\r\nA.a\r\n..C\r\nX.Y\r\n...\r\n", "output": "14\r\n1 0\r\n2 1\r\n3 1\r\n"}, {"input": "1 1 4 1\r\nA\r\n.\r\nB\r\n.\r\n", "output": "3\r\n1 0\r\n2 0\r\n4 2\r\n3 0\r\n"}, {"input": "1 3 5 2\r\nABA\r\nBBB\r\nBBA\r\nBAB\r\nABB\r\n", "output": "11\r\n1 0\r\n3 1\r\n2 3\r\n4 2\r\n5 1\r\n"}, {"input": "2 2 5 1\r\n..\r\nBA\r\n.A\r\nB.\r\n..\r\nA.\r\nAB\r\n.B\r\n..\r\n..\r\n", "output": "12\r\n1 0\r\n2 1\r\n3 1\r\n5 3\r\n4 5\r\n"}, {"input": "3 3 10 2\r\nBA.\r\n..A\r\n.BB\r\nB..\r\n..B\r\n.AA\r\nB..\r\nAB.\r\n..A\r\nBAB\r\n.A.\r\n.B.\r\n..B\r\nA..\r\n...\r\n...\r\n.B.\r\nBA.\r\n..B\r\n.AB\r\n.B.\r\nB.A\r\n.A.\r\n.BA\r\n..B\r\n...\r\n.A.\r\n.AA\r\n..A\r\n.B.\r\n", "output": "67\r\n1 0\r\n10 1\r\n2 1\r\n3 2\r\n4 1\r\n7 4\r\n9 7\r\n5 9\r\n6 9\r\n8 4\r\n"}, {"input": "3 1 5 1\r\nB\r\nA\r\nB\r\nA\r\nA\r\nB\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\n", "output": "5\r\n1 0\r\n2 1\r\n3 2\r\n4 3\r\n5 3\r\n"}, {"input": "3 2 10 1\r\nAB\r\nBA\r\nAB\r\nAA\r\nAA\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBA\r\nAB\r\nAA\r\nBB\r\nAB\r\nBA\r\nBB\r\nBB\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBB\r\nAB\r\nAA\r\n", "output": "16\r\n1 0\r\n3 1\r\n8 3\r\n2 3\r\n4 2\r\n9 4\r\n6 4\r\n7 6\r\n10 6\r\n5 10\r\n"}, {"input": "2 3 10 2\r\nABB\r\nABA\r\nAAB\r\nBAB\r\nAAA\r\nBBA\r\nBBB\r\nBAA\r\nBBB\r\nABB\r\nABA\r\nBBA\r\nBBB\r\nAAB\r\nABA\r\nABB\r\nBBA\r\nBAB\r\nBBB\r\nBBB\r\n", "output": "38\r\n1 0\r\n5 1\r\n7 5\r\n4 7\r\n9 4\r\n10 5\r\n6 1\r\n3 6\r\n8 1\r\n2 0\r\n"}, {"input": "1 1 1 1\r\n.\r\n", "output": "1\r\n1 0\r\n"}]
| false |
stdio
|
import sys
def read_levels(input_path):
with open(input_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip()]
first_line = lines[0].split()
n, m, k, w = map(int, first_line)
levels = [None] # 1-based indexing
idx = 1
for _ in range(k):
level = []
for _ in range(n):
if idx >= len(lines):
return None, None, None, None, None
level.append(lines[idx].strip())
idx += 1
levels.append(level)
return n, m, k, w, levels
def main(input_path, output_path, submission_path):
n, m, k, w, levels = read_levels(input_path)
if levels is None:
return 0
# Read reference output's first line
with open(output_path) as f_ref:
ref_lines = f_ref.read().splitlines()
if not ref_lines:
return 0
ref_total = int(ref_lines[0])
# Read submission output
with open(submission_path) as f_sub:
sub_lines = f_sub.read().splitlines()
if not sub_lines:
return 0
try:
sub_total = int(sub_lines[0])
except:
return 0
# Check sub_total matches reference
if sub_total != ref_total:
return 0
# Parse pairs
tokens = []
for line in sub_lines[1:]:
tokens.extend(line.strip().split())
if len(tokens) != 2 * k:
return 0
pairs = []
try:
for i in range(0, len(tokens), 2):
xi = int(tokens[i])
yi = int(tokens[i+1])
pairs.append((xi, yi))
except:
return 0
# Check all xi are unique and 1..k
seen_xi = set()
for xi, _ in pairs:
if xi < 1 or xi > k or xi in seen_xi:
return 0
seen_xi.add(xi)
if len(seen_xi) != k:
return 0
# Check pairs order and calculate total
transmitted = set()
total = 0
for xi, yi in pairs:
if yi != 0 and yi not in transmitted:
return 0
if yi == 0:
cost = n * m
else:
# Compute d between xi and yi
d = 0
for row in range(n):
for col in range(m):
if levels[xi][row][col] != levels[yi][row][col]:
d += 1
cost = d * w
total += cost
transmitted.add(xi)
if total != sub_total:
return 0
return 1
if __name__ == "__main__":
input_path, output_path, submission_path = sys.argv[1:]
score = main(input_path, output_path, submission_path)
print(score)
| true |
412/D
|
412
|
D
|
Python 3
|
PRETESTS
| 3 | 62 | 0 |
6409547
|
from sys import setrecursionlimit
setrecursionlimit(100500)
def dfs(u):
print(u + 1, end = ' ')
visited[u] = True
for x in edges[u]:
if not visited[x]:
dfs(x)
n, m = map(int, input().split())
visited, drains, edges = [False] * n, [0] * n, [[] for x in range(n)]
for x in range(m):
a, b = map(int, input().split())
edges[b - 1].append(a - 1)
drains[a - 1] += 1
for x in sorted(range(n), key = lambda a: drains[a]):
if not visited[x]:
dfs(x)
| 33 | 374 | 11,673,600 |
6413301
|
import sys
n,m=map(int,sys.stdin.readline().split())
P={}
for i in range(m):
a,b=map(int,sys.stdin.readline().split())
P[(a-1,b-1)]=1
A=[-1]*n
A[0]=0
for i in range(1,n):
j=1
A[i]=i
x=i
while(x>0 and (A[x-1],A[x]) in P):
A[x-1],A[x]=A[x],A[x-1]
x-=1
Anss=""
for i in range(n):
Anss+=str(A[i]+1)+" "
sys.stdout.write(Anss)
|
Coder-Strike 2014 - Round 1
|
CF
| 2,014 | 1 | 256 |
Giving Awards
|
The employees of the R1 company often spend time together: they watch football, they go camping, they solve contests. So, it's no big deal that sometimes someone pays for someone else.
Today is the day of giving out money rewards. The R1 company CEO will invite employees into his office one by one, rewarding each one for the hard work this month. The CEO knows who owes money to whom. And he also understands that if he invites person x to his office for a reward, and then immediately invite person y, who has lent some money to person x, then they can meet. Of course, in such a situation, the joy of person x from his brand new money reward will be much less. Therefore, the R1 CEO decided to invite the staff in such an order that the described situation will not happen for any pair of employees invited one after another.
However, there are a lot of employees in the company, and the CEO doesn't have a lot of time. Therefore, the task has been assigned to you. Given the debt relationships between all the employees, determine in which order they should be invited to the office of the R1 company CEO, or determine that the described order does not exist.
|
The first line contains space-separated integers n and m $$(2\leq n\leq3\cdot10^{4};1\leq m\leq min(10^{5},\frac{n(n-1)}{2}))$$ — the number of employees in R1 and the number of debt relations. Each of the following m lines contains two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi), these integers indicate that the person number ai owes money to a person a number bi. Assume that all the employees are numbered from 1 to n.
It is guaranteed that each pair of people p, q is mentioned in the input data at most once. In particular, the input data will not contain pairs p, q and q, p simultaneously.
|
Print -1 if the described order does not exist. Otherwise, print the permutation of n distinct integers. The first number should denote the number of the person who goes to the CEO office first, the second number denote the person who goes second and so on.
If there are multiple correct orders, you are allowed to print any of them.
| null | null |
[{"input": "2 1\n1 2", "output": "2 1"}, {"input": "3 3\n1 2\n2 3\n3 1", "output": "2 1 3"}]
| 2,000 |
["dfs and similar"]
| 33 |
[{"input": "2 1\r\n1 2\r\n", "output": "2 1 \r\n"}, {"input": "3 3\r\n1 2\r\n2 3\r\n3 1\r\n", "output": "2 1 3 \r\n"}, {"input": "10 45\r\n10 5\r\n10 7\r\n6 1\r\n5 8\r\n3 5\r\n6 5\r\n1 2\r\n6 10\r\n2 9\r\n9 5\r\n4 1\r\n7 5\r\n1 8\r\n6 8\r\n10 9\r\n7 2\r\n7 9\r\n4 10\r\n7 3\r\n4 8\r\n10 3\r\n10 8\r\n2 10\r\n8 2\r\n4 2\r\n5 2\r\n9 1\r\n4 5\r\n1 3\r\n9 6\r\n3 8\r\n5 1\r\n6 4\r\n4 7\r\n8 7\r\n7 1\r\n9 3\r\n3 4\r\n6 2\r\n3 2\r\n6 7\r\n6 3\r\n4 9\r\n8 9\r\n1 10\r\n", "output": "2 3 1 5 7 8 4 6 9 10 \r\n"}, {"input": "4 6\r\n3 4\r\n4 1\r\n4 2\r\n2 1\r\n3 2\r\n3 1\r\n", "output": "1 2 4 3 \r\n"}, {"input": "5 10\r\n4 2\r\n4 5\r\n3 1\r\n2 1\r\n3 4\r\n3 2\r\n5 1\r\n5 2\r\n4 1\r\n5 3\r\n", "output": "1 2 4 3 5 \r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m = map(int, f.readline().split())
edges = set()
for _ in range(m):
a, b = map(int, f.readline().split())
edges.add((a, b))
# Read submission output
with open(submission_path) as f:
line = f.readline().strip()
if line == '-1':
# Check if it's actually impossible
# To determine this, we'd need to know if any valid order exists, which is complex.
# However, the problem requires the checker to verify the submission's output.
# Since the submission claims it's impossible, but how can the checker know?
# This suggests that the checker cannot handle the -1 case properly.
# Wait, the problem says if the order does not exist, output -1.
# But how can the checker verify if the submission's -1 is correct?
# This is a problem because the checker would need to solve the problem itself.
# However, the checker can't do that. So for this problem, the checker can't validate a submission that outputs -1.
# Therefore, in the checker code, if the submission outputs -1, we have to check if a valid order exists.
# But that's equivalent to solving the problem. Which is impossible for a checker to do without the correct algorithm.
# Hence, this problem's checker is incomplete. However, given the problem statement, perhaps the checker is expected to handle cases where the submission outputs -1 correctly.
# So, the approach here is that the checker cannot fully validate submissions that output -1. Thus, in practice, such cases would have to be handled by the contest system's test cases, which would have the correct answer (either a permutation or -1).
# However, in the problem statement, the output is either a permutation or -1. So the checker must check two possibilities:
# 1. The submission outputs a valid permutation (so -1 is wrong)
# 2. The submission outputs -1, and no valid permutation exists.
# But how can the checker distinguish between the two? It can't unless it solves the problem.
# Therefore, the checker code here can't handle the -1 case. Thus, the problem may have test cases where the correct answer is -1, and the checker would need to check that the submission's output is -1 and that no valid permutation exists. But how?
# This is a problem. Therefore, perhaps the correct approach is that in this problem, the checker can only check the permutation case. If the submission outputs -1, the checker can't validate it unless it knows whether such a permutation exists (which requires solving the problem).
# Given that the checker's code is supposed to be written without solving the problem, this is impossible. Therefore, the problem is not a 'checker' type but 'diff'? But the problem allows multiple correct outputs, so it's a checker. However, the -1 case complicates things.
# However, according to the problem statement, the correct answer is either -1 or a permutation. So if the submission outputs -1, the checker must check whether the correct answer is -1. But how?
# This is a contradiction. Therefore, in practice, perhaps the test cases where the answer is -1 would be evaluated as 'diff' (exact match), and the rest as 'checker'.
# However, the problem statement's output allows any permutation. So if the reference output is a permutation, then the submission must also be a permutation. If the reference output is -1, then the submission must be -1.
# Therefore, the checker must compare the submission's output with the reference output in the case where the reference output is -1. Otherwise, check the permutation.
# So, the checker must read the reference output. If the reference output is -1, then the submission must also output -1. Else, the submission must output a valid permutation.
# Therefore, the checker logic is:
# Read the reference output. If it's -1, then submission must be -1.
# Otherwise, submission must be a valid permutation.
# So, the code must compare the submission against the reference output in the case where the reference is -1.
# So, the code should proceed as follows:
# 1. Read the reference output (output_path).
with open(output_path) as ref_f:
ref_line = ref_f.readline().strip()
if ref_line == '-1':
# The correct answer is -1; submission must be -1
if line == '-1':
print(1)
return
else:
print(0)
return
else:
# The correct answer is a permutation. The submission must be a valid permutation.
# So proceed with permutation checks.
pass
# If submission is -1 but reference is not, invalid.
if line == '-1':
print(0)
return
# Parse submission's permutation
perm = list(map(int, line.split()))
if len(perm) != n:
print(0)
return
if set(perm) != set(range(1, n+1)):
print(0)
return
# Check consecutive pairs
for i in range(n-1):
x = perm[i]
y = perm[i+1]
if (x, y) in edges:
print(0)
return
print(1)
if __name__ == '__main__':
main()
| true |
846/F
|
846
|
F
|
PyPy 3-64
|
TESTS
| 2 | 171 | 40,960,000 |
182256696
|
from sys import stdin
input=lambda :stdin.readline()[:-1]
n=int(input())
a=list(map(lambda x:int(x)-1,input().split()))
c=[[] for i in range(10**6)]
for i in range(n):
c[a[i]].append(i)
ans=1/n
for i in range(10**6):
if len(c[i])==0:
continue
x=[0]+c[i]+[n-1]
res=n*(n-1)
m=len(x)
for i in range(1,m):
d=x[i]-x[i-1]
res-=d*(d-1)
ans+=res/(n*n)
print(ans)
| 31 | 1,091 | 85,811,200 |
197617831
|
n=int(input())
a=list(map(int,input().split()))
lastocc=[0]*1000006
ans=[0]*n
ans[0]=1
lastocc[a[0]]=1
for i in range(1,n):
ans[i]=ans[i-1]+(i+1-lastocc[a[i]])
lastocc[a[i]]=i+1
print((2*sum(ans)-n)/(n*n))
|
Educational Codeforces Round 28
|
ICPC
| 2,017 | 2 | 256 |
Random Query
|
You are given an array a consisting of n positive integers. You pick two integer numbers l and r from 1 to n, inclusive (numbers are picked randomly, equiprobably and independently). If l > r, then you swap values of l and r. You have to calculate the expected value of the number of unique elements in segment of the array from index l to index r, inclusive (1-indexed).
|
The first line contains one integer number n (1 ≤ n ≤ 106). The second line contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 106) — elements of the array.
|
Print one number — the expected number of unique elements in chosen segment.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if $${ \operatorname* { m i n } } ( | x - y |, { \frac { | x - y | } { x } } ) \leq 1 0 ^ { - 4 }$$, where x is jury's answer, and y is your answer.
| null | null |
[{"input": "2\n1 2", "output": "1.500000"}, {"input": "2\n2 2", "output": "1.000000"}]
| 1,800 |
["data structures", "math", "probabilities", "two pointers"]
| 31 |
[{"input": "2\r\n1 2\r\n", "output": "1.500000\r\n"}, {"input": "2\r\n2 2\r\n", "output": "1.000000\r\n"}, {"input": "10\r\n9 6 8 5 5 2 8 9 2 2\r\n", "output": "3.100000\r\n"}, {"input": "20\r\n49 33 9 8 50 21 12 44 23 39 24 10 17 4 17 40 24 19 27 21\r\n", "output": "7.010000\r\n"}, {"input": "1\r\n1000000\r\n", "output": "1.000000\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
correct_output_path = sys.argv[2]
submission_output_path = sys.argv[3]
# Read correct output
with open(correct_output_path, 'r') as f:
correct_line = f.readline().strip()
x = float(correct_line)
# Read submission output
with open(submission_output_path, 'r') as f:
submission_line = f.readline().strip()
try:
y = float(submission_line)
except:
# Invalid output format, score 0
print(0)
return
# Compute errors
absolute_error = abs(x - y)
if absolute_error <= 1e-4:
print(1)
return
# Compute relative error
relative_error = absolute_error / x
if relative_error <= 1e-4:
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
474/D
|
474
|
D
|
PyPy 3-64
|
TESTS
| 2 | 77 | 5,734,400 |
190973874
|
import sys
input=sys.stdin.readline
t,k=map(int,input().split())
dp=[0]*(100001)
for i in range(1,k+1):dp[i]+=1
dp[k]+=1
for i in range(k+1,len(dp)):
dp[i]=(dp[i-1]+dp[i-k])%(10**9+7)
for i in range(len(dp)):
dp[i]=(dp[i]+dp[i-1])%(10**9+7)
for _ in range(t):
l,r=map(int,input().split())
print(dp[r]-dp[l-1])
| 43 | 156 | 10,444,800 |
209800957
|
from sys import stdin
input = lambda: stdin.readline().rstrip('\r\n')
M = int(1e9) + 7
N = int(1e5) + 5
t, k = map(int, input().split())
dp = [0]*N
dp[0] = 1
for i in range(1, N):
dp[i] = dp[i-1] + (i >= k) * dp[i-k]
dp[i] %= M
for i in range(1, N):
dp[i] += dp[i-1]
dp[i] %= M
for _ in range(t):
a, b = map(int, input().split())
print((dp[b] - dp[a-1]) % M)
|
Codeforces Round 271 (Div. 2)
|
CF
| 2,014 | 1.5 | 256 |
Flowers
|
We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.
But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.
Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (109 + 7).
|
Input contains several test cases.
The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.
The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.
|
Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).
| null |
- For K = 2 and length 1 Marmot can eat (R).
- For K = 2 and length 2 Marmot can eat (RR) and (WW).
- For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).
- For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).
|
[{"input": "3 2\n1 3\n2 3\n4 4", "output": "6\n5\n5"}]
| 1,700 |
["dp"]
| 43 |
[{"input": "3 2\r\n1 3\r\n2 3\r\n4 4\r\n", "output": "6\r\n5\r\n5\r\n"}, {"input": "1 1\r\n1 3\r\n", "output": "14\r\n"}, {"input": "1 2\r\n64329 79425\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
351/E
|
351
|
E
|
Python 3
|
TESTS
| 1 | 92 | 0 |
228759640
|
def min_inversions(n, p):
inv_count = 0
for i in range(n):
# Count of numbers greater than |p[i]| that come after it
greater_count = sum(1 for j in range(i+1, n) if abs(p[j]) > abs(p[i]))
# Count of numbers smaller than |p[i]| that come after it
smaller_count = sum(1 for j in range(i+1, n) if abs(p[j]) < abs(p[i]))
# If p[i] is negative
if p[i] < 0:
inv_count += min(greater_count, smaller_count)
# If p[i] is positive
else:
inv_count += smaller_count if smaller_count <= greater_count else greater_count
return inv_count
# Input and Output handling
n = int(input().strip())
p = list(map(int, input().split()))
print(min_inversions(n, p))
| 36 | 280 | 3,686,400 |
150804902
|
import sys
input = sys.stdin.buffer.readline
def process(A):
n = len(A)
S = [1 for i in range(n)]
d = {}
for i in range(n):
ai = abs(A[i])
if ai not in d:
d[ai] = []
d[ai].append(i)
L = sorted(d)
answer = 0
while len(L) > 0:
ai = L.pop()
for i in d[ai]:
S[i] = 0
for i in d[ai]:
answer+=min(sum(S[:i]), sum(S[i+1:]))
if sum(S[:i]) < sum(S[i+1:]):
A[i] = -1*ai
else:
A[i] = ai
return answer
n = int(input())
A = [int(x) for x in input().split()]
print(process(A))
|
Codeforces Round 204 (Div. 1)
|
CF
| 2,013 | 2 | 256 |
Jeff and Permutation
|
Jeff's friends know full well that the boy likes to get sequences and arrays for his birthday. Thus, Jeff got sequence p1, p2, ..., pn for his birthday.
Jeff hates inversions in sequences. An inversion in sequence a1, a2, ..., an is a pair of indexes i, j (1 ≤ i < j ≤ n), such that an inequality ai > aj holds.
Jeff can multiply some numbers of the sequence p by -1. At that, he wants the number of inversions in the sequence to be minimum. Help Jeff and find the minimum number of inversions he manages to get.
|
The first line contains integer n (1 ≤ n ≤ 2000). The next line contains n integers — sequence p1, p2, ..., pn (|pi| ≤ 105). The numbers are separated by spaces.
|
In a single line print the answer to the problem — the minimum number of inversions Jeff can get.
| null | null |
[{"input": "2\n2 1", "output": "0"}, {"input": "9\n-2 0 -1 0 -1 2 1 0 -1", "output": "6"}]
| 2,200 |
["greedy"]
| 36 |
[{"input": "2\r\n2 1\r\n", "output": "0\r\n"}, {"input": "9\r\n-2 0 -1 0 -1 2 1 0 -1\r\n", "output": "6\r\n"}, {"input": "9\r\n0 0 1 1 0 0 1 0 1\r\n", "output": "5\r\n"}, {"input": "8\r\n0 1 2 -1 -2 1 -2 2\r\n", "output": "3\r\n"}, {"input": "24\r\n-1 -1 2 2 0 -2 2 -1 0 0 2 -2 3 0 2 -3 0 -3 -1 1 0 0 -1 -2\r\n", "output": "55\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "31\r\n-2 2 -2 -1 0 0 1 2 1 1 -1 -2 1 -1 -2 2 0 1 -1 -2 -1 -2 -1 2 2 2 2 1 1 0 1\r\n", "output": "74\r\n"}, {"input": "9\r\n1 -1 -1 0 -1 0 1 1 1\r\n", "output": "1\r\n"}, {"input": "5\r\n1 0 1 -2 1\r\n", "output": "1\r\n"}, {"input": "31\r\n-5 -5 5 3 -1 3 1 -3 -3 -1 -5 -3 -2 -4 -3 3 5 -2 1 0 -1 1 -3 1 -1 1 3 3 2 1 0\r\n", "output": "70\r\n"}, {"input": "53\r\n-3 2 -3 -5 -2 7 0 -2 1 6 -1 2 5 -3 3 -6 -2 -5 -3 -6 4 -4 -2 6 1 -7 -6 -4 0 2 -5 -1 -2 -6 2 2 7 -2 -3 1 0 -4 3 4 -2 7 -3 7 7 3 -5 -5 3\r\n", "output": "289\r\n"}, {"input": "24\r\n-3 -4 3 -3 3 2 -1 -3 -4 0 -4 0 2 3 3 -1 2 1 2 -2 3 -2 1 0\r\n", "output": "46\r\n"}, {"input": "50\r\n-6 1 -3 7 -5 -5 4 0 3 -5 1 2 -1 0 7 0 6 3 -5 4 4 3 -7 -1 4 4 -5 3 7 1 4 2 6 -4 0 3 -3 -2 -3 1 -5 3 -4 2 -2 7 -1 3 -7 4\r\n", "output": "260\r\n"}, {"input": "17\r\n-56007 -97423 -66458 -17041 49374 60662 42188 56222 28689 -4117 -1712 11034 17161 43908 -65064 -76642 -73934\r\n", "output": "13\r\n"}, {"input": "12\r\n0 1 0 1 1 -1 1 -1 0 1 0 -1\r\n", "output": "12\r\n"}]
| false |
stdio
| null | true |
247/D
|
250
|
D
|
Python 3
|
TESTS
| 3 | 92 | 0 |
13547255
|
# http://codeforces.com/problemset/problem/250/D
from math import sqrt
def calcPath(x1, y1, x2, y2):
return sqrt(pow(x1-x2,2)+pow(y1-y2,2))
# with open('h.in', 'r') as inputFile, open('h.out', 'w') as outputFile:
(n,m,a,b) = [int(x) for x in input().strip().split(' ')]
A = [int(x) for x in input().strip().split(' ')]
B = [int(x) for x in input().strip().split(' ')]
L = [int(x) for x in input().strip().split(' ')]
# print(n,m,a,b)
# print(A,B,L)
i = 0
res = {}
while i<m:
min_path = 0
curr_path = 0
j = 0
while j<n:
curr_path = calcPath(0,0,a,A[j])+ calcPath(a,A[j],b,B[i])+L[i]
if min_path < curr_path and min_path != 0:
break
min_path = curr_path
res[min_path] = [i+1,j+1];
j += 1
i += 1
min_k = min(res.keys())
# print(res)
print(str(res[min_k][0])+" "+str(res[min_k][1]))
| 33 | 1,122 | 11,878,400 |
13858446
|
from math import sqrt,fabs
def dist(x1, y1, x2, y2):
return sqrt(pow(abs(x1 - x2), 2) + pow(abs(y1 - y2), 2))
def calcOptimumRightPoint(startX, startY):
l = float("inf")
idx = -1
for i in range(len(B)):
d = dist(startX, startY, b, B[i]) + L[i]
if d <= l:
l = d
idx = i
return idx
n,m,a,b = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]
L = [int(x) for x in input().split()]
optimumRightPoint = calcOptimumRightPoint(0,0)
intersectLeft = (a * B[optimumRightPoint]) / b
l = float("inf")
optimumLeftPoint = -1
for i in range(len(A)):
if fabs(intersectLeft-A[i]) < l:
l = fabs(intersectLeft-A[i])
optimumLeftPoint = i
optimumRightPoint = calcOptimumRightPoint(a, A[optimumLeftPoint])
print(optimumLeftPoint + 1, optimumRightPoint + 1)
|
CROC-MBTU 2012, Final Round
|
CF
| 2,012 | 1 | 256 |
Building Bridge
|
Two villages are separated by a river that flows from the north to the south. The villagers want to build a bridge across the river to make it easier to move across the villages.
The river banks can be assumed to be vertical straight lines x = a and x = b (0 < a < b).
The west village lies in a steppe at point O = (0, 0). There are n pathways leading from the village to the river, they end at points Ai = (a, yi). The villagers there are plain and simple, so their pathways are straight segments as well.
The east village has reserved and cunning people. Their village is in the forest on the east bank of the river, but its exact position is not clear. There are m twisted paths leading from this village to the river and ending at points Bi = (b, y'i). The lengths of all these paths are known, the length of the path that leads from the eastern village to point Bi, equals li.
The villagers want to choose exactly one point on the left bank of river Ai, exactly one point on the right bank Bj and connect them by a straight-line bridge so as to make the total distance between the villages (the sum of |OAi| + |AiBj| + lj, where |XY| is the Euclidean distance between points X and Y) were minimum. The Euclidean distance between points (x1, y1) and (x2, y2) equals $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$.
Help them and find the required pair of points.
|
The first line contains integers n, m, a, b (1 ≤ n, m ≤ 105, 0 < a < b < 106).
The second line contains n integers in the ascending order: the i-th integer determines the coordinate of point Ai and equals yi (|yi| ≤ 106).
The third line contains m integers in the ascending order: the i-th integer determines the coordinate of point Bi and equals y'i (|y'i| ≤ 106).
The fourth line contains m more integers: the i-th of them determines the length of the path that connects the eastern village and point Bi, and equals li (1 ≤ li ≤ 106).
It is guaranteed, that there is such a point C with abscissa at least b, that |BiC| ≤ li for all i (1 ≤ i ≤ m). It is guaranteed that no two points Ai coincide. It is guaranteed that no two points Bi coincide.
|
Print two integers — the numbers of points on the left (west) and right (east) banks, respectively, between which you need to build a bridge. You can assume that the points on the west bank are numbered from 1 to n, in the order in which they are given in the input. Similarly, the points on the east bank are numbered from 1 to m in the order in which they are given in the input.
If there are multiple solutions, print any of them. The solution will be accepted if the final length of the path will differ from the answer of the jury by no more than 10 - 6 in absolute or relative value.
| null | null |
[{"input": "3 2 3 5\n-2 -1 4\n-1 2\n7 3", "output": "2 2"}]
| 1,900 |
[]
| 33 |
[{"input": "3 2 3 5\r\n-2 -1 4\r\n-1 2\r\n7 3\r\n", "output": "2 2"}, {"input": "1 1 10 20\r\n5\r\n-5\r\n1\r\n", "output": "1 1"}, {"input": "2 2 1 2\r\n-1 10\r\n8 9\r\n3 7\r\n", "output": "1 1"}, {"input": "10 20 50 60\r\n-96 -75 32 37 42 43 44 57 61 65\r\n-95 -90 -86 -79 -65 -62 -47 -11 -8 -6 1 8 23 25 32 51 73 88 94 100\r\n138 75 132 116 49 43 96 166 96 161 146 112 195 192 201 186 251 254 220 227\r\n", "output": "2 6"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 122 | 7,065,600 |
129628906
|
#265B
t = int(input())
tree = []
for i in range(t):
tree.append(int(input()))
m = min(tree)
total = 0
for i in range(t):
if (i == t - 1):
total+= (tree[i] - m)
else:
total+= (tree[i] - m) * 2
ans = total + (m * t) + t
print(ans)
| 15 | 124 | 2,662,400 |
177435498
|
import sys
input = lambda: sys.stdin.buffer.readline().decode().strip()
inl = lambda: list(map(int, input().split()))
n = int(input())
a = [int(input()) for _ in range(n)]
# 1. intial value
ans = n + a[0]
# 2. diff between each two tress
for i in range(len(a) - 1):
ans += abs(a[i] - a[i + 1]) + 1
# 3. print ans
print(ans)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 122 | 0 |
105161618
|
n=int(input())
count=0
h=int(input())
count=h+1
while(n>1):
n-=1
h1=int(input())
if(h>h1):
count+=h-h1
h=h-h1
count+=1
count+=h1-h
count+=1
h=h1
print(count)
| 15 | 186 | 0 |
155483755
|
import sys
input = sys.stdin.readline
c, h = 0, 0
k = int(input())
for _ in range(k):
n = int(input())
c += abs(n - h) + 1
h += n - h
print(c+k-1)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 122 | 0 |
114982362
|
n = int(input())
p=int(input())
t=p+1
for i in range(1,n):
h=int(input())
if(h<p):
t+=(p-h)+2
else:
t+=2+(h-p)
print(t)
| 15 | 186 | 10,547,200 |
172097524
|
import math
import copy
import itertools
import bisect
import sys
input = sys.stdin.readline
def ilst():
return list(map(int,input().split()))
def inum():
return map(int,input().split())
def islst():
return list(map(str,input().split()))
n = int(input())
l = []
for _ in range(n):
l.append(int(input()))
ans = l[0]+1
for i in range(1,n):
if l[i] <= l[i-1]:
ans += l[i-1]-l[i]+2
else:
ans += l[i]-l[i-1]+2
print(ans)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 122 | 0 |
178157411
|
"""
https://codeforces.com/problemset/problem/265/B
"""
x = int(input())
trees = [int(input()) for _ in range(x)]
hauteur = trees[0]
total = trees[0]
indice = 1
while indice < x:
if hauteur <= trees[indice]:
total += trees[indice] - hauteur
else:
total += trees[indice]
hauteur = trees[indice]
indice += 1
total = total + 2 * x - 1
print(total)
| 15 | 218 | 3,174,400 |
179967935
|
import sys
input = sys.stdin.readline
n = int (input())
prev = 0
ans = n + (n-1)
while n > 0:
x = int (input())
ans += abs(prev-x)
prev = x
n-=1
print (ans)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
32/A
|
32
|
A
|
Python 3
|
TESTS
| 3 | 218 | 0 |
46680942
|
n, d = map(int, input().split())
a = [int(i) for i in input().split()]
m = []
for k in range(n):
l = [a[i]-a[k] for i in range(n) if abs(a[i]-a[k]) <= 10]
m.append(len(l))
print(max(m)*(max(m)-1))
| 32 | 122 | 307,200 |
27425665
|
n, d = (int(x) for x in input().split())
a = sorted([int(x) for x in input().split()])
s = 0
l = 0
for r in range(1, n):
while a[r] - a[l] > d:
l += 1
s += (r - l) * 2
print(s)
|
Codeforces Beta Round 32 (Div. 2, Codeforces format)
|
CF
| 2,010 | 2 | 256 |
Reconnaissance
|
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most d centimeters. Captain Bob has n soldiers in his detachment. Their heights are a1, a2, ..., an centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1, 2) and (2, 1) should be regarded as different.
|
The first line contains two integers n and d (1 ≤ n ≤ 1000, 1 ≤ d ≤ 109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains n space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
|
Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed d.
| null | null |
[{"input": "5 10\n10 20 50 60 65", "output": "6"}, {"input": "5 1\n55 30 29 31 55", "output": "6"}]
| 800 |
["brute force"]
| 32 |
[{"input": "5 10\r\n10 20 50 60 65\r\n", "output": "6\r\n"}, {"input": "5 1\r\n55 30 29 31 55\r\n", "output": "6\r\n"}, {"input": "6 10\r\n4 6 4 1 9 3\r\n", "output": "30\r\n"}, {"input": "7 100\r\n19 1694 261 162 1 234 513\r\n", "output": "8\r\n"}, {"input": "8 42\r\n37 53 74 187 568 22 5 65\r\n", "output": "20\r\n"}, {"input": "10 4\r\n11 6 76 49 28 20 57 152 5 32\r\n", "output": "4\r\n"}, {"input": "100 100\r\n51 93 101 960 2 477 213 129 663 925 254 78 1486 274 160 481 132 156 412 372 5 57 152 298 1771 7 359 468 254 406 202 929 221 366 552 97 555 29 822 118 539 140 992 854 7 163 134 103 940 30 409 1003 398 43 555 79 107 40 23 103 643 171 310 382 770 337 18 189 570 177 29 54 855 171 205 291 299 935 620 180 114 358 88 292 118 400 218 537 369 60 683 192 13 537 59 824 264 191 3 300\r\n", "output": "2404\r\n"}]
| false |
stdio
| null | true |
813/F
|
813
|
F
|
PyPy 3-64
|
TESTS
| 2 | 62 | 0 |
173855331
|
from collections import defaultdict, deque
from math import inf
def bfsBipartite(graph,s=1):
colors = dict()
cola = deque()
cola.append(s)
colors[s] = 1
while len(cola)> 0:
u = cola.popleft()
ady = filter(lambda x: graph[u][x]!= inf,graph[u].keys())
for v in ady:
if colors.get(v,False):
if colors[v] == colors[u]:
return "NO"
if not colors.get(v,False):
colors[v] = 2 if colors[u] == 1 else 1
cola.append(v)
return "YES"
def main():
n,q = map(int,input().split())
graph = defaultdict(dict)
for i in range(q):
u,v = map(int,input().split())
if graph[u].get(v,inf) != inf:
graph[u][v] = inf
graph[v][u] = inf
else:
graph[u][v] = 1
graph[v][u] = 1
print(bfsBipartite(graph))
main()
| 25 | 1,185 | 83,660,800 |
215838651
|
from collections import defaultdict
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v, w):
return (u * n2 + v) * n2 + w
def g(u, v):
return u * n2 + v
def get_root(s):
while s ^ root[s]:
s = root[s]
return s
def unite(s, t, i):
s, t = get_root(s), get_root(t)
if s == t:
return
if rank[s] < rank[t]:
s, t = t, s
c1, c2, c3, c4 = s, t, 0, root[t]
if rank[s] == rank[t]:
rank[s] += 1
c3 = 1
root[t] = s
st1.append(f(c1, c2, c3))
st2.append(g(i, c4))
return
def same(s, t):
return True if get_root(s) == get_root(t) else False
def undo(x, y):
s, z = divmod(x, n2 * n2)
t, c = divmod(z, n2)
rt = y % n2
rank[s] -= c
root[t] = rt
return
n, q = map(int, input().split())
d = defaultdict(lambda : -1)
u, v = [], []
x, y = [0] * q, [0] * q
for i in range(q):
x0, y0 = map(int, input().split())
if x0 > y0:
x0, y0 = y0, x0
x[i], y[i] = x0, y0
if d[(x0, y0)] == -1:
d[(x0, y0)] = i
else:
u.append(d[(x0, y0)])
v.append(i - 1)
d[(x0, y0)] = -1
for i in d.values():
if i ^ -1:
u.append(i)
v.append(q - 1)
l1 = pow(2, (q + 1).bit_length())
l2 = 2 * l1
s0 = [0] * l2
for u0, v0 in zip(u, v):
l0 = u0 + l1
r0 = v0 + l1
while l0 <= r0:
if l0 & 1:
s0[l0] += 1
l0 += 1
l0 >>= 1
if not r0 & 1 and l0 <= r0:
s0[r0] += 1
r0 -= 1
r0 >>= 1
for i in range(1, l2):
s0[i] += s0[i - 1]
now = [0] + list(s0)
tree = [-1] * now[l2]
m = len(u)
for i in range(m):
l0 = u[i] + l1
r0 = v[i] + l1
while l0 <= r0:
if l0 & 1:
tree[now[l0]] = u[i]
now[l0] += 1
l0 += 1
l0 >>= 1
if not r0 & 1 and l0 <= r0:
tree[now[r0]] = u[i]
now[r0] += 1
r0 -= 1
r0 >>= 1
n += 5
n2 = 2 * n
root = [i for i in range(n2)]
rank = [1 for _ in range(n2)]
s0 = [0] + s0
st1, st2 = [], []
now = 1
ng = 0
for i in range(s0[1], s0[2]):
j = tree[i]
u0, v0 = get_root(x[j]), get_root(y[j])
u1, v1 = get_root(x[j] + n), get_root(y[j] + n)
if u0 == v0 and u0 == v1:
continue
if u0 == u1:
ng -= 1
if v0 == v1:
ng -= 1
unite(u0, v1, 1)
unite(u1, v0, 1)
if same(u0, u1):
ng += 1
visit = [0] * l2
cnt = [0] * l2
cnt[1] = ng
ans = []
while len(ans) ^ q:
if now >= l1:
ans.append("YES" if not ng else "NO")
if now >= l1 or visit[now << 1 ^ 1]:
visit[now] = 1
while st1 and st2[-1] // n2 == now:
undo(st1.pop(), st2.pop())
now >>= 1
ng = cnt[now]
continue
now = now << 1 if not visit[now << 1] else now << 1 ^ 1
for i in range(s0[now], s0[now + 1]):
j = tree[i]
u0, v0 = get_root(x[j]), get_root(y[j])
u1, v1 = get_root(x[j] + n), get_root(y[j] + n)
if u0 == v0 and u0 == v1:
continue
if u0 == u1:
ng -= 1
if v0 == v1:
ng -= 1
unite(u0, v1, now)
unite(u1, v0, now)
if same(u0, u1):
ng += 1
cnt[now] = ng
sys.stdout.write("\n".join(map(str, ans)))
|
Educational Codeforces Round 22
|
ICPC
| 2,017 | 6 | 256 |
Bipartite Checking
|
You are given an undirected graph consisting of n vertices. Initially there are no edges in the graph. Also you are given q queries, each query either adds one undirected edge to the graph or removes it. After each query you have to check if the resulting graph is bipartite (that is, you can paint all vertices of the graph into two colors so that there is no edge connecting two vertices of the same color).
|
The first line contains two integers n and q (2 ≤ n, q ≤ 100000).
Then q lines follow. ith line contains two numbers xi and yi (1 ≤ xi < yi ≤ n). These numbers describe ith query: if there is an edge between vertices xi and yi, then remove it, otherwise add it.
|
Print q lines. ith line must contain YES if the graph is bipartite after ith query, and NO otherwise.
| null | null |
[{"input": "3 5\n2 3\n1 3\n1 2\n1 2\n1 2", "output": "YES\nYES\nNO\nYES\nNO"}]
| 2,500 |
["data structures", "dsu", "graphs"]
| 25 |
[{"input": "3 5\r\n2 3\r\n1 3\r\n1 2\r\n1 2\r\n1 2\r\n", "output": "YES\r\nYES\r\nNO\r\nYES\r\nNO\r\n"}, {"input": "5 10\r\n1 5\r\n2 5\r\n2 4\r\n1 4\r\n4 5\r\n2 4\r\n2 5\r\n1 4\r\n2 3\r\n1 2\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nYES\r\nYES\r\nYES\r\n"}, {"input": "10 20\r\n1 10\r\n5 7\r\n1 2\r\n3 5\r\n3 6\r\n4 9\r\n3 4\r\n6 9\r\n4 8\r\n6 9\r\n7 8\r\n3 8\r\n7 10\r\n2 7\r\n3 7\r\n5 9\r\n6 7\r\n4 6\r\n2 10\r\n8 10\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "10 30\r\n5 6\r\n5 9\r\n4 9\r\n6 7\r\n7 9\r\n3 10\r\n5 6\r\n5 7\r\n6 10\r\n2 4\r\n2 6\r\n2 5\r\n3 7\r\n1 8\r\n8 9\r\n3 4\r\n3 5\r\n1 9\r\n6 7\r\n4 8\r\n4 5\r\n1 5\r\n2 3\r\n4 10\r\n1 7\r\n2 8\r\n3 10\r\n1 7\r\n1 7\r\n3 8\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "10 40\r\n6 9\r\n1 5\r\n2 6\r\n2 5\r\n7 9\r\n7 9\r\n5 6\r\n5 8\r\n6 9\r\n1 7\r\n5 6\r\n1 7\r\n1 9\r\n4 5\r\n4 6\r\n6 8\r\n7 8\r\n1 8\r\n5 7\r\n1 7\r\n8 9\r\n5 6\r\n6 7\r\n1 4\r\n3 7\r\n9 10\r\n1 7\r\n4 7\r\n4 10\r\n3 8\r\n7 10\r\n3 6\r\n1 10\r\n6 10\r\n8 9\r\n8 10\r\n7 10\r\n2 5\r\n1 9\r\n3 6\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nNO\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "30 40\r\n5 15\r\n13 16\r\n12 17\r\n19 23\r\n1 27\r\n16 25\r\n20 21\r\n6 18\r\n10 17\r\n7 13\r\n20 24\r\n4 17\r\n8 12\r\n12 25\r\n25 29\r\n4 7\r\n1 14\r\n2 21\r\n4 26\r\n2 13\r\n20 24\r\n23 24\r\n8 16\r\n16 18\r\n8 10\r\n25 28\r\n4 22\r\n11 25\r\n13 24\r\n19 22\r\n18 20\r\n22 30\r\n4 13\r\n28 29\r\n6 13\r\n18 22\r\n18 28\r\n4 20\r\n14 21\r\n5 6\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "50 60\r\n7 36\r\n43 45\r\n12 17\r\n10 40\r\n30 47\r\n18 30\r\n3 9\r\n5 6\r\n13 49\r\n5 26\r\n4 20\r\n5 50\r\n27 41\r\n3 21\r\n15 43\r\n24 41\r\n6 30\r\n40 50\r\n8 13\r\n9 21\r\n2 47\r\n23 26\r\n21 22\r\n15 31\r\n28 38\r\n1 50\r\n24 35\r\n2 13\r\n4 33\r\n14 42\r\n10 28\r\n3 5\r\n18 19\r\n9 40\r\n11 21\r\n22 36\r\n6 11\r\n36 44\r\n20 35\r\n7 38\r\n9 33\r\n29 31\r\n6 14\r\n22 32\r\n27 48\r\n19 31\r\n39 47\r\n12 50\r\n8 38\r\n35 36\r\n1 43\r\n7 49\r\n10 25\r\n10 21\r\n14 15\r\n1 44\r\n8 32\r\n17 50\r\n42 45\r\n13 44\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\n"}]
| false |
stdio
| null | true |
247/D
|
250
|
D
|
Python 3
|
TESTS
| 1 | 78 | 307,200 |
51285153
|
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
n, m, a, b = mints()
A = list(mints())
B = list(mints())
l = list(mints())
j = -1
r = (1e100,-1,-1)
for i in range(m):
while not((j == -1 or a*B[i]-b*A[j] >= 0) \
and (j+1 == n or a*B[i]-b*A[j+1] < 0)):
j += 1
if j != -1:
r = min(r,(l[i]+sqrt(a*a+A[j]*A[j])+sqrt((b-a)**2+(B[i]-A[j])**2),j,i))
if j+1 != n:
r = min(r,(l[i]+sqrt(a*a+A[j+1]*A[j+1])+sqrt((b-a)**2+(B[i]-A[j+1])**2),j,i))
print(r[1]+1,r[2]+1)
| 33 | 794 | 14,745,600 |
106349764
|
import sys
def pro():
return sys.stdin.readline().strip()
def rop():
return map(int, pro().split())
def main():
s = list(rop())
a = list(rop())
q = list(rop())
o = list(rop())
p = -1
t = (1e100, -1, -1)
for i in range(s[1]):
while not((p == - 1 or s[2] * q[i] - s[3] * a[p] >= 0)
and (p + 1 == s[0] or s[2] * q[i] - s[3] * a[p+1] < 0)):
p += 1
if p != -1:
t = min(t, (o[i] + (s[2] ** 2 + a[p] ** 2) ** 0.5 + ((s[3] - s[2]) ** 2 + (q[i] - a[p]) ** 2) ** 0.5, p, i))
if p + 1 != s[0]:
t = min(t, (o[i] + (s[2] ** 2 + a[p + 1] ** 2) ** 0.5 + ((s[3] - s[2]) ** 2 + (q[i] - a[p + 1]) ** 2) ** 0.5, p + 1, i))
print(t[1] + 1, t[2] + 1)
main()
|
CROC-MBTU 2012, Final Round
|
CF
| 2,012 | 1 | 256 |
Building Bridge
|
Two villages are separated by a river that flows from the north to the south. The villagers want to build a bridge across the river to make it easier to move across the villages.
The river banks can be assumed to be vertical straight lines x = a and x = b (0 < a < b).
The west village lies in a steppe at point O = (0, 0). There are n pathways leading from the village to the river, they end at points Ai = (a, yi). The villagers there are plain and simple, so their pathways are straight segments as well.
The east village has reserved and cunning people. Their village is in the forest on the east bank of the river, but its exact position is not clear. There are m twisted paths leading from this village to the river and ending at points Bi = (b, y'i). The lengths of all these paths are known, the length of the path that leads from the eastern village to point Bi, equals li.
The villagers want to choose exactly one point on the left bank of river Ai, exactly one point on the right bank Bj and connect them by a straight-line bridge so as to make the total distance between the villages (the sum of |OAi| + |AiBj| + lj, where |XY| is the Euclidean distance between points X and Y) were minimum. The Euclidean distance between points (x1, y1) and (x2, y2) equals $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$.
Help them and find the required pair of points.
|
The first line contains integers n, m, a, b (1 ≤ n, m ≤ 105, 0 < a < b < 106).
The second line contains n integers in the ascending order: the i-th integer determines the coordinate of point Ai and equals yi (|yi| ≤ 106).
The third line contains m integers in the ascending order: the i-th integer determines the coordinate of point Bi and equals y'i (|y'i| ≤ 106).
The fourth line contains m more integers: the i-th of them determines the length of the path that connects the eastern village and point Bi, and equals li (1 ≤ li ≤ 106).
It is guaranteed, that there is such a point C with abscissa at least b, that |BiC| ≤ li for all i (1 ≤ i ≤ m). It is guaranteed that no two points Ai coincide. It is guaranteed that no two points Bi coincide.
|
Print two integers — the numbers of points on the left (west) and right (east) banks, respectively, between which you need to build a bridge. You can assume that the points on the west bank are numbered from 1 to n, in the order in which they are given in the input. Similarly, the points on the east bank are numbered from 1 to m in the order in which they are given in the input.
If there are multiple solutions, print any of them. The solution will be accepted if the final length of the path will differ from the answer of the jury by no more than 10 - 6 in absolute or relative value.
| null | null |
[{"input": "3 2 3 5\n-2 -1 4\n-1 2\n7 3", "output": "2 2"}]
| 1,900 |
[]
| 33 |
[{"input": "3 2 3 5\r\n-2 -1 4\r\n-1 2\r\n7 3\r\n", "output": "2 2"}, {"input": "1 1 10 20\r\n5\r\n-5\r\n1\r\n", "output": "1 1"}, {"input": "2 2 1 2\r\n-1 10\r\n8 9\r\n3 7\r\n", "output": "1 1"}, {"input": "10 20 50 60\r\n-96 -75 32 37 42 43 44 57 61 65\r\n-95 -90 -86 -79 -65 -62 -47 -11 -8 -6 1 8 23 25 32 51 73 88 94 100\r\n138 75 132 116 49 43 96 166 96 161 146 112 195 192 201 186 251 254 220 227\r\n", "output": "2 6"}]
| false |
stdio
| null | true |
32/A
|
32
|
A
|
Python 3
|
TESTS
| 3 | 92 | 0 |
177074613
|
"""According to the regulations of Berland's army,
a reconnaissance unit should consist of exactly two soldiers.
Since these two soldiers shouldn't differ much, their heights can differ by at most d centimeters.
Captain Bob has n soldiers in his detachment.
Their heights are a1, a2, ..., an centimeters.
Some soldiers are of the same height.
Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1, 2) and (2, 1) should be regarded as different.
Input
The first line contains two integers n and d (1 <= n <= 1000, 1 <= d <= 1e9)
- amount of soldiers in Bob's detachment and the maximum allowed height difference respectively.
The second line contains n space-separated integers - heights of all the soldiers in Bob's detachment.
These numbers don't exceed 1e9.
Output
Output one number - amount of ways to form a reconnaissance unit of two soldiers,
whose height difference doesn't exceed d."""
n, d = map(int, input().split())
sHeight = input().split()
sHeight.sort()
result = 0
for i in range(0, n - 1):
for j in range(i + 1, n):
if (int(sHeight[j]) - int(sHeight[i])) <= d:
result += 2
else:
break
print(str(result))
| 32 | 122 | 1,945,600 |
202802705
|
_, b = map(int, input().split())
a = list(map(int, input().split()))
a.sort();k = 0
for i in range(len(a)):
for j in range(i+1, len(a)):
if abs(a[j] - a[i]) <= b:
k += 1;
print(k *2)
|
Codeforces Beta Round 32 (Div. 2, Codeforces format)
|
CF
| 2,010 | 2 | 256 |
Reconnaissance
|
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most d centimeters. Captain Bob has n soldiers in his detachment. Their heights are a1, a2, ..., an centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1, 2) and (2, 1) should be regarded as different.
|
The first line contains two integers n and d (1 ≤ n ≤ 1000, 1 ≤ d ≤ 109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains n space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
|
Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed d.
| null | null |
[{"input": "5 10\n10 20 50 60 65", "output": "6"}, {"input": "5 1\n55 30 29 31 55", "output": "6"}]
| 800 |
["brute force"]
| 32 |
[{"input": "5 10\r\n10 20 50 60 65\r\n", "output": "6\r\n"}, {"input": "5 1\r\n55 30 29 31 55\r\n", "output": "6\r\n"}, {"input": "6 10\r\n4 6 4 1 9 3\r\n", "output": "30\r\n"}, {"input": "7 100\r\n19 1694 261 162 1 234 513\r\n", "output": "8\r\n"}, {"input": "8 42\r\n37 53 74 187 568 22 5 65\r\n", "output": "20\r\n"}, {"input": "10 4\r\n11 6 76 49 28 20 57 152 5 32\r\n", "output": "4\r\n"}, {"input": "100 100\r\n51 93 101 960 2 477 213 129 663 925 254 78 1486 274 160 481 132 156 412 372 5 57 152 298 1771 7 359 468 254 406 202 929 221 366 552 97 555 29 822 118 539 140 992 854 7 163 134 103 940 30 409 1003 398 43 555 79 107 40 23 103 643 171 310 382 770 337 18 189 570 177 29 54 855 171 205 291 299 935 620 180 114 358 88 292 118 400 218 537 369 60 683 192 13 537 59 824 264 191 3 300\r\n", "output": "2404\r\n"}]
| false |
stdio
| null | true |
846/F
|
846
|
F
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
229354204
|
n = int(input())
a = list(map(int, input().split()))
e = [.0] * n
ds, st = 1, 1
ans = 0.0
sll = len(set(a))
while st < n:
ans += ds * 1 / n
if a[st] != a[st - 1]:
ds += 1
st += 1
ans += ds / n
e[0] = ans
i, ds = 1, 0
while i < n:
e[i] = e[i - 1]
if a[i] != a[i - 1]:
ds += 1
if a[i] == a[i - 1]:
i += 1
continue
e[i] += ds * 1 / n
e[i] -= (sll - ds) * 1 / n
i += 1
print(sum(e) / n)
| 31 | 1,091 | 122,163,200 |
226537020
|
import math
import sys
def solve():
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
last = {}
all_possible = (n * n)
not_pos = {}
for i in range(n):
if a[i] not in last:
last[a[i]] = i
not_pos[a[i]] = i * i
else:
not_pos[a[i]] += (i - last[a[i]] - 1) * (i - last[a[i]] - 1)
last[a[i]] = i
#print(not_pos)
for i in last:
not_pos[i] += (n - last[i] - 1) * (n - last[i] - 1)
res = 0
for i in not_pos:
res += (all_possible - not_pos[i]) / all_possible
#print(not_pos)
print("%.9f" % (res))
return
if __name__ == "__main__":
mt = 0
if mt:
t = int(sys.stdin.readline())
for i in range(t):
solve()
else:
solve()
|
Educational Codeforces Round 28
|
ICPC
| 2,017 | 2 | 256 |
Random Query
|
You are given an array a consisting of n positive integers. You pick two integer numbers l and r from 1 to n, inclusive (numbers are picked randomly, equiprobably and independently). If l > r, then you swap values of l and r. You have to calculate the expected value of the number of unique elements in segment of the array from index l to index r, inclusive (1-indexed).
|
The first line contains one integer number n (1 ≤ n ≤ 106). The second line contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 106) — elements of the array.
|
Print one number — the expected number of unique elements in chosen segment.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if $${ \operatorname* { m i n } } ( | x - y |, { \frac { | x - y | } { x } } ) \leq 1 0 ^ { - 4 }$$, where x is jury's answer, and y is your answer.
| null | null |
[{"input": "2\n1 2", "output": "1.500000"}, {"input": "2\n2 2", "output": "1.000000"}]
| 1,800 |
["data structures", "math", "probabilities", "two pointers"]
| 31 |
[{"input": "2\r\n1 2\r\n", "output": "1.500000\r\n"}, {"input": "2\r\n2 2\r\n", "output": "1.000000\r\n"}, {"input": "10\r\n9 6 8 5 5 2 8 9 2 2\r\n", "output": "3.100000\r\n"}, {"input": "20\r\n49 33 9 8 50 21 12 44 23 39 24 10 17 4 17 40 24 19 27 21\r\n", "output": "7.010000\r\n"}, {"input": "1\r\n1000000\r\n", "output": "1.000000\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
correct_output_path = sys.argv[2]
submission_output_path = sys.argv[3]
# Read correct output
with open(correct_output_path, 'r') as f:
correct_line = f.readline().strip()
x = float(correct_line)
# Read submission output
with open(submission_output_path, 'r') as f:
submission_line = f.readline().strip()
try:
y = float(submission_line)
except:
# Invalid output format, score 0
print(0)
return
# Compute errors
absolute_error = abs(x - y)
if absolute_error <= 1e-4:
print(1)
return
# Compute relative error
relative_error = absolute_error / x
if relative_error <= 1e-4:
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 124 | 0 |
104007793
|
n = int(input())
curr = 0
time = 0
prev = 0
for i in range(n):
h = int(input())
if prev<=h:
curr = prev
else:
time+=(prev-h)
curr = prev-h
time+= (h-curr)+1
prev = h
print(time+(n-1))# for jumps
| 15 | 248 | 0 |
214675105
|
n = int(input())
o = 0
t = 0
for i in range(n):
x = int(input())
if x < o:
t += o - x + 1
o = x
else:
t += x - o + 1
o = x
t += 1
print(t - 1)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 124 | 0 |
112649662
|
n = int(input())
count = 0
g = int(input())
count+=g+1
for i in range(n-1):
a = int(input())
if(a>g):
count+=a-g+1+1
g = a
else:
count+=g-a+1+1
print(count)
| 15 | 248 | 0 |
220679465
|
n=int(input())
ans=2*n-1
pre=0
for i in range(n):
cur=int(input())
ans+=abs(cur-pre)
pre=cur
print(ans)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
PyPy 3-64
|
TESTS
| 2 | 92 | 1,331,200 |
204337506
|
n, x = int(input()), 0
s = n * 2
for i in range(n):
s += abs(x - i)
x = i
print(s)
| 15 | 248 | 0 |
230813787
|
n=int(input())
h=0
t=0
for j in range(n):
hi = int(input())
if(j==0):
h=hi
t+=(hi+1)
else:
t+=(abs(hi-h)+2)
h=hi
# print(t)
print(t)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
803/G
|
803
|
G
|
Python 3
|
TESTS
| 2 | 46 | 204,800 |
26910361
|
from sys import stdin, stdout
n, k = map(int, stdin.readline().split())
values = [0] + list(map(int, stdin.readline().split()))
m = int(stdin.readline())
questions = []
ar = set()
for i in range(m):
questions.append(tuple(map(int, stdin.readline().strip().split())))
ar.add(int(questions[-1][1]))
ar.add(int(questions[-1][2]))
compress = [0 for i in range(len(ar) + 1)]
ar = sorted(list(ar))
d = {}
for i in range(len(ar)):
compress[i + 1] = ar[i]
d[ar[i]] = i + 1
def update(l, r, ind, lb, rb, value):
if (l == lb and r == rb):
tree[ind] = value
else:
m = (lb + rb) // 2
if (l <= m):
update(l, min(m, r), ind * 2, l, m, value)
if (r > m):
update(max(l, m + 1), r, ind * 2 + 1, m + 1, rb, value)
def get(l, r, ind, lb, rb):
if (l == lb and r == rb):
return tree[ind]
else:
m = (lb + rb) // 2
ans = float('inf')
if (l <= m):
ans = get(l, min(m, r), ind * 2, l, m)
if (r > m):
ans = min(ans, get(max(l, m + 1), r, ind * 2 + 1, m + 1, rb))
return ans
def getvalue(ind):
if (ind < len(compress)):
return (values[compress[ind] % (n + 1) + (compress[ind] == n + 1)])
else:
return 0
size = 1
while (size < len(ar)):
size *= 2
tree = [float('inf') for i in range(2 * size)]
for i in range(size, 2 * size):
tree[i] = getvalue(i - size + 1)
for i in range(size // 2, size):
if i * 2 - size + 1 < len(compress) and i * 2 - size + 2 < len(compress):#sparce
for j in range(compress[i * 2 - size + 1], compress[i * 2 - size + 2]):
tree[i] = min(tree[i], getvalue(j))
for i in range(size // 2 - 1, 0, -1):
tree[i] = min(tree[i * 2], tree[i * 2 + 1])
for i in range(m):
if questions[i][0] == 1:
update(d[questions[i][1]], d[questions[i][2]], 1, 1, size, questions[i][-1])
else:
stdout.write(str(get(d[questions[i][1]], d[questions[i][2]], 1, 1, size)) + '\n')
| 32 | 2,823 | 117,350,400 |
230770543
|
import sys
import random
input = sys.stdin.readline
rd = random.randint(10 ** 9, 2 * 10 ** 9)
class SegmentTree():
def __init__(self, init, unitX, f):
self.f = f # (X, X) -> X
self.unitX = unitX
self.f = f
if type(init) == int:
self.n = init
self.n = 1 << (self.n - 1).bit_length()
self.X = [unitX] * (self.n * 2)
else:
self.n = len(init)
self.n = 1 << (self.n - 1).bit_length()
# len(init)が2の累乗ではない時UnitXで埋める
self.X = [unitX] * self.n + init + [unitX] * (self.n - len(init))
# 配列のindex1まで埋める
for i in range(self.n - 1, 0, -1):
self.X[i] = self.f(self.X[i * 2], self.X[i * 2 | 1])
def update(self, i, x):
"""0-indexedのi番目の値をxで置換"""
# 最下段に移動
i += self.n
self.X[i] = x
# 上向に更新
i >>= 1
while i:
self.X[i] = self.f(self.X[i * 2], self.X[i * 2 | 1])
i >>= 1
def getvalue(self, i):
"""元の配列のindexの値を見る"""
return self.X[i + self.n]
def getrange(self, l, r):
"""区間[l, r)でのfを行った値"""
l += self.n
r += self.n
al = self.unitX
ar = self.unitX
while l < r:
# 左端が右子ノードであれば
if l & 1:
al = self.f(al, self.X[l])
l += 1
# 右端が右子ノードであれば
if r & 1:
r -= 1
ar = self.f(self.X[r], ar)
l >>= 1
r >>= 1
return self.f(al, ar)
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
q = int(input())
queries = [list(map(int, input().split())) for _ in range(q)]
min_tree = SegmentTree(a, 10 ** 9, min)
p = []
for qq in queries:
p.append(qq[1] - 1)
p.append(qq[2] - 1)
p.sort()
cur = 0
pre = -1
new = []
d = dict()
for i in range(len(p)):
if p[i] == pre:
continue
if p[i] - pre > 1:
if p[i] - pre > n:
mi = min_tree.getrange(0, n)
elif pre % n < p[i] % n:
mi = min_tree.getrange(pre % n + 1, p[i] % n)
else:
mi = min(min_tree.getrange(pre % n + 1, n), min_tree.getrange(0, p[i] % n))
new.append(mi)
cur += 1
new.append(a[p[i] % n])
d[p[i]] = cur
cur += 1
pre = p[i]
def push_lazy(node, left, right):
if lazy[node] != 0:
tree[node] = lazy[node]
if left != right:
lazy[node * 2] = lazy[node]
lazy[node * 2 + 1] = lazy[node]
lazy[node] = 0
def update(node, left, right, l, r, val):
push_lazy(node, left, right)
if r < left or l > right:
return
if l <= left and r >= right:
lazy[node] = val
push_lazy(node, left, right)
return
mid = (left + right) // 2
update(node * 2, left, mid, l, r, val)
update(node * 2 + 1, mid + 1, right, l, r, val)
tree[node] = min(tree[node * 2], tree[node * 2 + 1])
def query(node, left, right, l, r):
push_lazy(node, left, right)
if r < left or l > right:
return 10**18
if l <= left and r >= right:
return tree[node]
mid = (left + right) // 2
sum_left = query(node * 2, left, mid, l, r)
sum_right = query(node * 2 + 1, mid + 1, right, l, r)
return min(sum_left, sum_right)
tree = [10 ** 18] * (4 * len(new))
lazy = [0] * (4 * len(new))
for i in range(len(new)):
update(1, 0, len(new) - 1, i, i, new[i])
for qq in queries:
if qq[0] == 1:
l, r = qq[1] - 1, qq[2] - 1
update(1, 0, len(new) - 1, d[l], d[r], qq[3])
else:
l, r = qq[1] - 1, qq[2] - 1
print(query(1, 0, len(new) - 1, d[l], d[r]))
|
Educational Codeforces Round 20
|
ICPC
| 2,017 | 4 | 512 |
Periodic RMQ Problem
|
You are given an array a consisting of positive integers and q queries to this array. There are two types of queries:
- 1 l r x — for each index i such that l ≤ i ≤ r set ai = x.
- 2 l r — find the minimum among such ai that l ≤ i ≤ r.
We decided that this problem is too easy. So the array a is given in a compressed form: there is an array b consisting of n elements and a number k in the input, and before all queries a is equal to the concatenation of k arrays b (so the size of a is n·k).
|
The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 104).
The second line contains n integers — elements of the array b (1 ≤ bi ≤ 109).
The third line contains one integer q (1 ≤ q ≤ 105).
Then q lines follow, each representing a query. Each query is given either as 1 l r x — set all elements in the segment from l till r (including borders) to x (1 ≤ l ≤ r ≤ n·k, 1 ≤ x ≤ 109) or as 2 l r — find the minimum among all elements in the segment from l till r (1 ≤ l ≤ r ≤ n·k).
|
For each query of type 2 print the answer to this query — the minimum on the corresponding segment.
| null | null |
[{"input": "3 1\n1 2 3\n3\n2 1 3\n1 1 2 4\n2 1 3", "output": "1\n3"}, {"input": "3 2\n1 2 3\n5\n2 4 4\n1 4 4 5\n2 4 4\n1 1 6 1\n2 6 6", "output": "1\n5\n1"}]
| 2,300 |
["data structures"]
| 32 |
[{"input": "3 1\r\n1 2 3\r\n3\r\n2 1 3\r\n1 1 2 4\r\n2 1 3\r\n", "output": "1\r\n3\r\n"}, {"input": "3 2\r\n1 2 3\r\n5\r\n2 4 4\r\n1 4 4 5\r\n2 4 4\r\n1 1 6 1\r\n2 6 6\r\n", "output": "1\r\n5\r\n1\r\n"}, {"input": "10 10\r\n10 8 10 9 2 2 4 6 10 1\r\n10\r\n1 17 87 5\r\n2 31 94\r\n1 5 56 8\r\n1 56 90 10\r\n1 25 93 6\r\n1 11 32 4\r\n2 20 49\r\n1 46 87 8\r\n2 14 48\r\n2 40 48\r\n", "output": "1\r\n4\r\n4\r\n6\r\n"}, {"input": "10 10\r\n4 2 3 8 1 2 1 7 5 4\r\n10\r\n2 63 87\r\n2 2 48\r\n2 5 62\r\n2 33 85\r\n2 30 100\r\n2 38 94\r\n2 7 81\r\n2 13 16\r\n2 26 36\r\n2 64 96\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n"}, {"input": "10 10\r\n8 7 4 6 8 2 4 9 4 3\r\n10\r\n1 43 67 7\r\n1 13 73 4\r\n1 35 71 6\r\n1 9 18 8\r\n1 3 55 2\r\n1 49 67 10\r\n1 22 49 2\r\n1 66 70 6\r\n1 17 21 10\r\n1 61 77 3\r\n", "output": ""}]
| false |
stdio
| null | true |
617/C
|
617
|
C
|
PyPy 3-64
|
TESTS
| 3 | 46 | 0 |
224668092
|
def app(x,y,a,b,r):
return (abs(x-a)**2 + abs(y-b)**2) <= r
def des(x,y,a,b):
return (abs(x-a)**2 + abs(y-b)**2)
numbers = input().split()
arr = []
for i in range(int(numbers[0])):
x,y = input().split()
arr.append((int(x),int(y)))
con = []
i=1
while i <len(numbers):
con.append((int(numbers[i]),int(numbers[i+1])))
i = i+2
bb=[0]*len(con)
m = 0
for i in arr :
n = 10**9
for index,j in enumerate(con):
if des(i[0],i[1],j[0],j[1]) < n:
n = des(i[0],i[1],j[0],j[1])
k = index
if m < n:
m = n
bb[k] = m
index = 0
elements_to_remove = []
for i in con:
for j in arr:
if (abs(i[0]-j[0])**2 + abs(i[1]-j[1])**2) <= bb[index]:
elements_to_remove.append(j)
index = index + 1
for element in elements_to_remove:
arr.remove(element)
if bb[0] ==0:
for i in arr:
bb[0] = max(bb[0],des(con[0][0],con[0][1],i[0],i[1]))
else:
for i in arr:
bb[1] = max(bb[1],des(con[1][0],con[1][1],i[0],i[1]))
print(sum(bb))
| 31 | 124 | 2,457,600 |
231028527
|
import sys
input=sys.stdin.readline
n,x1,y1,x2,y2=map(int,input().split())
l=[]
for _ in range(n):
x,y=map(int,input().split())
l.append([(x-x1)**2+(y-y1)**2,(x-x2)**2+(y-y2)**2])
l.sort(reverse=True)
s=10**18
q=0
for u,v in l:
s=min(s,u+q)
q=max(q,v)
print(min(s,q))
|
Codeforces Round 340 (Div. 2)
|
CF
| 2,016 | 2 | 256 |
Watering Flowers
|
A flowerbed has many flowers and two fountains.
You can adjust the water pressure and set any values r1(r1 ≥ 0) and r2(r2 ≥ 0), giving the distances at which the water is spread from the first and second fountain respectively. You have to set such r1 and r2 that all the flowers are watered, that is, for each flower, the distance between the flower and the first fountain doesn't exceed r1, or the distance to the second fountain doesn't exceed r2. It's OK if some flowers are watered by both fountains.
You need to decrease the amount of water you need, that is set such r1 and r2 that all the flowers are watered and the r12 + r22 is minimum possible. Find this minimum value.
|
The first line of the input contains integers n, x1, y1, x2, y2 (1 ≤ n ≤ 2000, - 107 ≤ x1, y1, x2, y2 ≤ 107) — the number of flowers, the coordinates of the first and the second fountain.
Next follow n lines. The i-th of these lines contains integers xi and yi ( - 107 ≤ xi, yi ≤ 107) — the coordinates of the i-th flower.
It is guaranteed that all n + 2 points in the input are distinct.
|
Print the minimum possible value r12 + r22. Note, that in this problem optimal answer is always integer.
| null |
The first sample is (r12 = 5, r22 = 1): The second sample is (r12 = 1, r22 = 32):
|
[{"input": "2 -1 0 5 3\n0 2\n5 2", "output": "6"}, {"input": "4 0 0 5 0\n9 4\n8 3\n-1 0\n1 4", "output": "33"}]
| 1,600 |
["implementation"]
| 31 |
[{"input": "2 -1 0 5 3\r\n0 2\r\n5 2\r\n", "output": "6\r\n"}, {"input": "4 0 0 5 0\r\n9 4\r\n8 3\r\n-1 0\r\n1 4\r\n", "output": "33\r\n"}, {"input": "5 -6 -4 0 10\r\n-7 6\r\n-9 7\r\n-5 -1\r\n-2 1\r\n-8 10\r\n", "output": "100\r\n"}, {"input": "10 -68 10 87 22\r\n30 89\r\n82 -97\r\n-52 25\r\n76 -22\r\n-20 95\r\n21 25\r\n2 -3\r\n45 -7\r\n-98 -56\r\n-15 16\r\n", "output": "22034\r\n"}, {"input": "1 -10000000 -10000000 -10000000 -9999999\r\n10000000 10000000\r\n", "output": "799999960000001\r\n"}]
| false |
stdio
| null | true |
504/D
|
504
|
D
|
Python 3
|
TESTS
| 0 | 92 | 204,800 |
97682151
|
m = int(input())
b = []
k = []
for i in range(m):
x = int(input())
c = 0
for i in range(len(b)):
v = b[i]
d = k[i]
if (x ^ v) < x:
x ^= v
c ^= d
if x != 0:
print(0)
c ^= 2 ** i
b.append(x)
k.append(c)
else:
a = []
for j in range(m):
if c & 1 == 1:
a.append(j)
c >>= 1
print(len(a), end='')
for v in a:
print(' ', v, sep='', end='')
print()
| 52 | 1,123 | 13,004,800 |
153501285
|
m = int(input())
basis = []
for i in range(m):
a = int(input())
need=0
for v,bitset in basis:
if a^v<a:
need^=bitset
a^=v
if a:
basis.append((a,need^(1<<i)))
print(0)
else:
res = [d for d in range(i) if 1<<d&need]
print(len(res), *res)
|
Codeforces Round 285 (Div. 1)
|
CF
| 2,015 | 2 | 256 |
Misha and XOR
|
After Misha's birthday he had many large numbers left, scattered across the room. Now it's time to clean up and Misha needs to put them in a basket. He ordered this task to his pet robot that agreed to complete the task at certain conditions. Before the robot puts a number x to the basket, Misha should answer the question: is it possible to choose one or multiple numbers that already are in the basket, such that their XOR sum equals x?
If the answer is positive, you also need to give the indexes of these numbers. If there are multiple options of choosing numbers, you are allowed to choose any correct option. After Misha's answer the robot puts the number to the basket.
Initially the basket is empty. Each integer you put in the basket takes some number. The first integer you put into the basket take number 0, the second integer takes number 1 and so on.
Misha needs to clean up the place as soon as possible but unfortunately, he isn't that good at mathematics. He asks you to help him.
|
The first line contains number m (1 ≤ m ≤ 2000), showing how many numbers are scattered around the room.
The next m lines contain the numbers in the order in which the robot puts them in the basket. Each number is a positive integer strictly less than 10600 that doesn't contain leading zeroes.
|
For each number either print a 0 on the corresponding line, if the number cannot be represented as a XOR sum of numbers that are in the basket, or print integer k showing how many numbers are in the representation and the indexes of these numbers. Separate the numbers by spaces. Each number can occur in the representation at most once.
| null |
The XOR sum of numbers is the result of bitwise sum of numbers modulo 2.
|
[{"input": "7\n7\n6\n5\n4\n3\n2\n1", "output": "0\n0\n0\n3 0 1 2\n2 1 2\n2 0 2\n2 0 1"}, {"input": "2\n5\n5", "output": "0\n1 0"}]
| 2,700 |
["bitmasks"]
| 52 |
[{"input": "7\r\n7\r\n6\r\n5\r\n4\r\n3\r\n2\r\n1\r\n", "output": "0\r\n0\r\n0\r\n3 0 1 2\r\n2 1 2\r\n2 0 2\r\n2 0 1\r\n"}, {"input": "2\r\n5\r\n5\r\n", "output": "0\r\n1 0\r\n"}, {"input": "10\r\n81\r\n97\r\n12\r\n2\r\n16\r\n96\r\n80\r\n99\r\n6\r\n83\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n3 0 1 5\r\n2 1 3\r\n0\r\n2 0 3\r\n"}, {"input": "10\r\n15106\r\n13599\r\n69319\r\n33224\r\n26930\r\n94490\r\n85089\r\n60931\r\n23137\r\n62868\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "10\r\n5059464500\r\n8210395556\r\n3004213265\r\n248593357\r\n5644084048\r\n9359824793\r\n8120649160\r\n4288978422\r\n183848555\r\n8135845959\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "10\r\n4\r\n12\r\n28\r\n29\r\n31\r\n31\r\n31\r\n31\r\n31\r\n31\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n1 4\r\n1 4\r\n1 4\r\n1 4\r\n1 4\r\n"}, {"input": "10\r\n16\r\n24\r\n28\r\n30\r\n31\r\n31\r\n31\r\n31\r\n31\r\n31\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n1 4\r\n1 4\r\n1 4\r\n1 4\r\n1 4\r\n"}, {"input": "10\r\n16\r\n8\r\n4\r\n2\r\n1\r\n31\r\n31\r\n31\r\n31\r\n31\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n"}, {"input": "10\r\n1\r\n2\r\n4\r\n8\r\n16\r\n31\r\n31\r\n31\r\n31\r\n31\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n5 0 1 2 3 4\r\n"}]
| false |
stdio
| null | true |
1006/E
|
1006
|
E
|
Python 3
|
TESTS
| 2 | 717 | 31,948,800 |
151488498
|
import threading
import sys
from sys import stdin
input=stdin.readline
sys.setrecursionlimit(10**7)
from collections import defaultdict
def main():
ans=[]
size=[]
n,q=map(int,input().split())
arr=list(map(int,input().split()))
graph=[[] for _ in range(n)]
for i in range(n-1):
graph[arr[i]-1].append(i+1)
for el in graph:
el.sort()
size=[0]*n
def solve(i):
ans.append(i)
size[i]=1
for nie in graph[i]:
size[i]+=solve(nie)
return size[i]
solve(0)
p={}
for i in range(n):
p[ans[i]]=i
for _ in range(q):
u,k=map(int,map(int,input().split()))
u-=1
if size[u]<k:
print(-1)
else :
print(ans[p[u]+k-1]+1)
threading.stack_size(10 ** 7)
t = threading.Thread(target=main)
t.start()
t.join()
| 31 | 514 | 53,657,600 |
197438558
|
# https://codeforces.com/contest/1006
import sys
from collections import deque
input = lambda: sys.stdin.readline().rstrip() # faster!
def solve_case():
n, q = map(int, input().split())
p = list(map(int, input().split()))
direct_subordinates = [[] for _ in range(n + 1)]
for officer, direct_superior in enumerate(p, 2):
direct_subordinates[direct_superior] += [officer]
order = [0] * (n + 1)
start = [0] * (n + 1)
finish = [0] * (n + 1)
stack = deque([(1, True)])
i = 1
while stack:
officer, go_down = stack.pop()
if not go_down:
finish[officer] = i
continue
order[i] = officer
start[officer] = i
i += 1
stack.append((officer, False))
for direct_subordinate in sorted(direct_subordinates[officer], reverse=True):
stack.append((direct_subordinate, True))
for _ in range(q):
u, k = map(int, input().split())
if start[u] + k - 1 < finish[u]:
print(order[start[u] + k - 1])
else:
print(-1)
solve_case()
|
Codeforces Round 498 (Div. 3)
|
ICPC
| 2,018 | 3 | 256 |
Military Problem
|
In this problem you will have to help Berland army with organizing their command delivery system.
There are $$$n$$$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $$$a$$$ is the direct superior of officer $$$b$$$, then we also can say that officer $$$b$$$ is a direct subordinate of officer $$$a$$$.
Officer $$$x$$$ is considered to be a subordinate (direct or indirect) of officer $$$y$$$ if one of the following conditions holds:
- officer $$$y$$$ is the direct superior of officer $$$x$$$;
- the direct superior of officer $$$x$$$ is a subordinate of officer $$$y$$$.
For example, on the picture below the subordinates of the officer $$$3$$$ are: $$$5, 6, 7, 8, 9$$$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $$$n$$$ vertices, in which vertex $$$u$$$ corresponds to officer $$$u$$$. The parent of vertex $$$u$$$ corresponds to the direct superior of officer $$$u$$$. The root (which has index $$$1$$$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $$$q$$$ queries, the $$$i$$$-th query is given as $$$(u_i, k_i)$$$, where $$$u_i$$$ is some officer, and $$$k_i$$$ is a positive integer.
To process the $$$i$$$-th query imagine how a command from $$$u_i$$$ spreads to the subordinates of $$$u_i$$$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $$$a$$$ and he spreads a command. Officer $$$a$$$ chooses $$$b$$$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $$$a$$$ chooses the one having minimal index. Officer $$$a$$$ gives a command to officer $$$b$$$. Afterwards, $$$b$$$ uses exactly the same algorithm to spread the command to its subtree. After $$$b$$$ finishes spreading the command, officer $$$a$$$ chooses the next direct subordinate again (using the same strategy). When officer $$$a$$$ cannot choose any direct subordinate who still hasn't received this command, officer $$$a$$$ finishes spreading the command.
Let's look at the following example:
If officer $$$1$$$ spreads a command, officers receive it in the following order: $$$[1, 2, 3, 5 ,6, 8, 7, 9, 4]$$$.
If officer $$$3$$$ spreads a command, officers receive it in the following order: $$$[3, 5, 6, 8, 7, 9]$$$.
If officer $$$7$$$ spreads a command, officers receive it in the following order: $$$[7, 9]$$$.
If officer $$$9$$$ spreads a command, officers receive it in the following order: $$$[9]$$$.
To answer the $$$i$$$-th query $$$(u_i, k_i)$$$, construct a sequence which describes the order in which officers will receive the command if the $$$u_i$$$-th officer spreads it. Return the $$$k_i$$$-th element of the constructed list or -1 if there are fewer than $$$k_i$$$ elements in it.
You should process queries independently. A query doesn't affect the following queries.
|
The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$$$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $$$n - 1$$$ integers $$$p_2, p_3, \dots, p_n$$$ ($$$1 \le p_i < i$$$), where $$$p_i$$$ is the index of the direct superior of the officer having the index $$$i$$$. The commander has index $$$1$$$ and doesn't have any superiors.
The next $$$q$$$ lines describe the queries. The $$$i$$$-th query is given as a pair ($$$u_i, k_i$$$) ($$$1 \le u_i, k_i \le n$$$), where $$$u_i$$$ is the index of the officer which starts spreading a command, and $$$k_i$$$ is the index of the required officer in the command spreading sequence.
|
Print $$$q$$$ numbers, where the $$$i$$$-th number is the officer at the position $$$k_i$$$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $$$u_i$$$. Print "-1" if the number of officers which receive the command is less than $$$k_i$$$.
You should process queries independently. They do not affect each other.
| null | null |
[{"input": "9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9", "output": "3\n6\n8\n-1\n9\n4"}]
| 1,600 |
["dfs and similar", "graphs", "trees"]
| 31 |
[{"input": "9 6\r\n1 1 1 3 5 3 5 7\r\n3 1\r\n1 5\r\n3 4\r\n7 3\r\n1 8\r\n1 9\r\n", "output": "3\r\n6\r\n8\r\n-1\r\n9\r\n4\r\n"}, {"input": "2 1\r\n1\r\n1 1\r\n", "output": "1\r\n"}, {"input": "13 12\r\n1 1 1 1 1 1 1 1 1 1 1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n"}]
| false |
stdio
| null | true |
975/E
|
975
|
E
|
Python 3
|
TESTS
| 2 | 77 | 7,065,600 |
37858919
|
from sys import stdin, stdout
from math import atan2, sqrt, sin, cos, pi
def main():
n, q = stdin.readline().split()
n = int(n)
q = int(q)
lines = stdin.readlines()
x = [0] * (n + 1)
y = [0] * (n + 1)
for i in range(n):
x[i], y[i] = lines[i].split()
x[i] = int(x[i])
y[i] = int(y[i])
x[n] = x[0]
y[n] = y[0]
a = 0
cx = 0
cy = 0
for i in range(n):
ii = i + 1
a += x[i] * y[ii] - x[ii] * y[i]
cx += (x[i] + x[ii]) * (x[i] * y[ii] - x[ii] * y[i])
cy += (y[i] + y[ii]) * (x[i] * y[ii] - x[ii] * y[i])
a *= 3
r = [0.0] * n
z = [0.0] * n
for i in range(n):
x[i] = x[i] * a - cx
y[i] = y[i] * a - cy
r[i] = sqrt(x[i] * x[i] + y[i] * y[i]) / a
z[i] = atan2(y[i], x[i])
cx /= a
cy /= a
p = [0, 1]
mt = 0.0
for i in range(q):
tmp = lines[n + i].split()
ta = int(tmp[0])
tb = int(tmp[1])
if ta == 1:
ta = int(ta)
ta -= 1
tb -= 1
if p[0] == tb:
p[0], p[1] = p[1], p[0]
p[1] = ta
tc = z[p[0]] + mt
cx += r[p[0]] * cos(tc)
cy += r[p[0]] * (sin(tc) - 1)
tc = pi / 2 - tc
mt += tc
else:
tb -= 1
tc = z[tb] + mt
stdout.write('%.10lf %.10lf\n' % (cx + r[tb] * cos(tc), cy + r[tb] * sin(tc)))
if __name__ == '__main__':
main()
| 31 | 1,060 | 9,318,400 |
230718925
|
import sys
read=lambda:map(int,sys.stdin.buffer.readline().split())
from math import *
n,q=read()
x,y=[],[]
cx,cy=0,0
deg=0
p1,p2=0,1
def pos(i):
return (cx+x[i]*cos(deg)-y[i]*sin(deg),cy+y[i]*cos(deg)+x[i]*sin(deg))
for i in range(n):
_x,_y=read()
x.append(_x);y.append(_y)
s=0
for i in range(2,n):
_s=(x[i]-x[0])*(y[i-1]-y[0])-(y[i]-y[0])*(x[i-1]-x[0])
cx+=(x[0]+x[i]+x[i-1])/3*_s
cy+=(y[0]+y[i]+y[i-1])/3*_s
s+=_s
cx/=s;cy/=s
for i in range(n): x[i]-=cx;y[i]-=cy
for i in range(q):
v=list(read())
if v[0]==1:
if p1==v[1]-1: p1,p2=p2,None
else: p1,p2=p1,None
cx,cy=pos(p1);cy-=hypot(x[p1],y[p1])
deg=pi/2-atan2(y[p1],x[p1]);p2=v[2]-1
else:
print("%.15f %.15f"%pos(v[1]-1))
|
Codeforces Round 478 (Div. 2)
|
CF
| 2,018 | 3 | 256 |
Hag's Khashba
|
Hag is a very talented person. He has always had an artist inside him but his father forced him to study mechanical engineering.
Yesterday he spent all of his time cutting a giant piece of wood trying to make it look like a goose. Anyway, his dad found out that he was doing arts rather than studying mechanics and other boring subjects. He confronted Hag with the fact that he is a spoiled son that does not care about his future, and if he continues to do arts he will cut his 25 Lira monthly allowance.
Hag is trying to prove to his dad that the wooden piece is a project for mechanics subject. He also told his dad that the wooden piece is a strictly convex polygon with $$$n$$$ vertices.
Hag brought two pins and pinned the polygon with them in the $$$1$$$-st and $$$2$$$-nd vertices to the wall. His dad has $$$q$$$ queries to Hag of two types.
- $$$1$$$ $$$f$$$ $$$t$$$: pull a pin from the vertex $$$f$$$, wait for the wooden polygon to rotate under the gravity force (if it will rotate) and stabilize. And then put the pin in vertex $$$t$$$.
- $$$2$$$ $$$v$$$: answer what are the coordinates of the vertex $$$v$$$.
Please help Hag to answer his father's queries.
You can assume that the wood that forms the polygon has uniform density and the polygon has a positive thickness, same in all points. After every query of the 1-st type Hag's dad tries to move the polygon a bit and watches it stabilize again.
|
The first line contains two integers $$$n$$$ and $$$q$$$ ($$$3\leq n \leq 10\,000$$$, $$$1 \leq q \leq 200000$$$) — the number of vertices in the polygon and the number of queries.
The next $$$n$$$ lines describe the wooden polygon, the $$$i$$$-th line contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$|x_i|, |y_i|\leq 10^8$$$) — the coordinates of the $$$i$$$-th vertex of the polygon. It is guaranteed that polygon is strictly convex and the vertices are given in the counter-clockwise order and all vertices are distinct.
The next $$$q$$$ lines describe the queries, one per line. Each query starts with its type $$$1$$$ or $$$2$$$. Each query of the first type continues with two integers $$$f$$$ and $$$t$$$ ($$$1 \le f, t \le n$$$) — the vertex the pin is taken from, and the vertex the pin is put to and the polygon finishes rotating. It is guaranteed that the vertex $$$f$$$ contains a pin. Each query of the second type continues with a single integer $$$v$$$ ($$$1 \le v \le n$$$) — the vertex the coordinates of which Hag should tell his father.
It is guaranteed that there is at least one query of the second type.
|
The output should contain the answer to each query of second type — two numbers in a separate line. Your answer is considered correct, if its absolute or relative error does not exceed $$$10^{-4}$$$.
Formally, let your answer be $$$a$$$, and the jury's answer be $$$b$$$. Your answer is considered correct if $$$\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-4}$$$
| null |
In the first test note the initial and the final state of the wooden polygon.
Red Triangle is the initial state and the green one is the triangle after rotation around $$$(2,0)$$$.
In the second sample note that the polygon rotates $$$180$$$ degrees counter-clockwise or clockwise direction (it does not matter), because Hag's father makes sure that the polygon is stable and his son does not trick him.
|
[{"input": "3 4\n0 0\n2 0\n2 2\n1 1 2\n2 1\n2 2\n2 3", "output": "3.4142135624 -1.4142135624\n2.0000000000 0.0000000000\n0.5857864376 -1.4142135624"}, {"input": "3 2\n-1 1\n0 0\n1 1\n1 1 2\n2 1", "output": "1.0000000000 -1.0000000000"}]
| 2,600 |
["geometry"]
| 31 |
[{"input": "3 4\r\n0 0\r\n2 0\r\n2 2\r\n1 1 2\r\n2 1\r\n2 2\r\n2 3\r\n", "output": "3.4142135624 -1.4142135624\r\n2.0000000000 0.0000000000\r\n0.5857864376 -1.4142135624\r\n"}, {"input": "3 2\r\n-1 1\r\n0 0\r\n1 1\r\n1 1 2\r\n2 1\r\n", "output": "1.0000000000 -1.0000000000\r\n"}, {"input": "10 10\r\n0 -100000000\r\n1 -100000000\r\n1566 -99999999\r\n2088 -99999997\r\n2610 -99999994\r\n3132 -99999990\r\n3654 -99999985\r\n4176 -99999979\r\n4698 -99999972\r\n5220 -99999964\r\n1 2 5\r\n2 1\r\n1 1 7\r\n2 5\r\n1 5 4\r\n1 4 2\r\n2 8\r\n1 7 9\r\n2 1\r\n1 2 10\r\n", "output": "0.0000000000 -100000000.0000000000\r\n-7.3726558373 -100002609.9964835122\r\n-129.8654413032 -100003125.4302210321\r\n-114.4079442212 -100007299.4544525659\r\n"}, {"input": "4 10\r\n0 0\r\n2 0\r\n2 2\r\n0 2\r\n2 3\r\n2 1\r\n2 1\r\n1 1 1\r\n2 3\r\n1 2 4\r\n1 4 4\r\n2 4\r\n1 1 3\r\n2 3\r\n", "output": "2.0000000000 2.0000000000\r\n0.0000000000 0.0000000000\r\n0.0000000000 0.0000000000\r\n0.5857864376 -1.4142135624\r\n4.8284271247 -2.8284271247\r\n6.2426406871 -4.2426406871\r\n"}, {"input": "3 2\r\n0 0\r\n1 0\r\n1566 1\r\n1 2 1\r\n2 3\r\n", "output": "0.0006381620 -1566.0003192846\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input to count type 2 queries
with open(input_path, 'r') as f:
n, q = map(int, f.readline().split())
for _ in range(n):
f.readline() # skip vertices
type2_count = 0
for _ in range(q):
parts = f.readline().split()
if parts[0] == '2':
type2_count += 1
# Read correct and submission outputs
with open(output_path, 'r') as f:
correct = [line.strip() for line in f]
with open(submission_path, 'r') as f:
submission = [line.strip() for line in f]
# Check line count
if len(correct) != len(submission) or len(correct) != type2_count:
print(0)
return
# Check each line
for c_line, s_line in zip(correct, submission):
try:
c_x, c_y = map(float, c_line.split())
s_x, s_y = map(float, s_line.split())
except:
print(0)
return
# Check x
dx = abs(s_x - c_x)
max_x = max(1.0, abs(c_x))
if dx > 1e-4 * max_x:
print(0)
return
# Check y
dy = abs(s_y - c_y)
max_y = max(1.0, abs(c_y))
if dy > 1e-4 * max_y:
print(0)
return
# All checks passed
print(1)
if __name__ == "__main__":
main()
| true |
803/G
|
803
|
G
|
Python 3
|
TESTS
| 2 | 46 | 204,800 |
26910715
|
from sys import stdin, stdout
n, k = map(int, stdin.readline().split())
values = list(map(int, stdin.readline().split())) * k
m = int(stdin.readline())
n *= k
size = 1
while (size < n):
size *= 2
tree = [float('inf') for i in range(2 * size)]
for i in range(n):
tree[size + i] = values[i]
for i in range(size - 1, 0, -1):
tree[i] = min(tree[i * 2], tree[i * 2 + 1])
def update(l, r, ind, lb, rb, value):
if (l == lb and r == rb):
tree[ind] = value
while ind != 1:
ind //= 2
tree[ind] = min(tree[ind * 2], tree[ind * 2 + 1])
else:
m = (lb + rb) // 2
if (l <= m):
update(l, min(m, r), ind * 2, l, m, value)
if (r > m):
update(max(l, m + 1), r, ind * 2 + 1, m + 1, rb, value)
def get(l, r, ind, lb, rb):
if (l == lb and r == rb):
cnt = tree[ind]
while ind != 1:
ind //= 2
if (tree[ind * 2] != tree[ind] and tree[ind * 2 + 1] != tree[ind]):
cnt = min(tree[ind], cnt)
return cnt
else:
m = (lb + rb) // 2
ans = float('inf')
if (l <= m):
ans = get(l, min(m, r), ind * 2, l, m)
if (r > m):
ans = min(ans, get(max(l, m + 1), r, ind * 2 + 1, m + 1, rb))
return ans
for i in range(m):
questions = tuple(map(int, stdin.readline().split()))
if questions[0] == 1:
update(questions[1], questions[2], 1, 1, size, questions[-1])
else:
stdout.write('\n' + str(get(questions[1], questions[2], 1, 1, size)))
| 32 | 2,823 | 117,350,400 |
230770543
|
import sys
import random
input = sys.stdin.readline
rd = random.randint(10 ** 9, 2 * 10 ** 9)
class SegmentTree():
def __init__(self, init, unitX, f):
self.f = f # (X, X) -> X
self.unitX = unitX
self.f = f
if type(init) == int:
self.n = init
self.n = 1 << (self.n - 1).bit_length()
self.X = [unitX] * (self.n * 2)
else:
self.n = len(init)
self.n = 1 << (self.n - 1).bit_length()
# len(init)が2の累乗ではない時UnitXで埋める
self.X = [unitX] * self.n + init + [unitX] * (self.n - len(init))
# 配列のindex1まで埋める
for i in range(self.n - 1, 0, -1):
self.X[i] = self.f(self.X[i * 2], self.X[i * 2 | 1])
def update(self, i, x):
"""0-indexedのi番目の値をxで置換"""
# 最下段に移動
i += self.n
self.X[i] = x
# 上向に更新
i >>= 1
while i:
self.X[i] = self.f(self.X[i * 2], self.X[i * 2 | 1])
i >>= 1
def getvalue(self, i):
"""元の配列のindexの値を見る"""
return self.X[i + self.n]
def getrange(self, l, r):
"""区間[l, r)でのfを行った値"""
l += self.n
r += self.n
al = self.unitX
ar = self.unitX
while l < r:
# 左端が右子ノードであれば
if l & 1:
al = self.f(al, self.X[l])
l += 1
# 右端が右子ノードであれば
if r & 1:
r -= 1
ar = self.f(self.X[r], ar)
l >>= 1
r >>= 1
return self.f(al, ar)
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
q = int(input())
queries = [list(map(int, input().split())) for _ in range(q)]
min_tree = SegmentTree(a, 10 ** 9, min)
p = []
for qq in queries:
p.append(qq[1] - 1)
p.append(qq[2] - 1)
p.sort()
cur = 0
pre = -1
new = []
d = dict()
for i in range(len(p)):
if p[i] == pre:
continue
if p[i] - pre > 1:
if p[i] - pre > n:
mi = min_tree.getrange(0, n)
elif pre % n < p[i] % n:
mi = min_tree.getrange(pre % n + 1, p[i] % n)
else:
mi = min(min_tree.getrange(pre % n + 1, n), min_tree.getrange(0, p[i] % n))
new.append(mi)
cur += 1
new.append(a[p[i] % n])
d[p[i]] = cur
cur += 1
pre = p[i]
def push_lazy(node, left, right):
if lazy[node] != 0:
tree[node] = lazy[node]
if left != right:
lazy[node * 2] = lazy[node]
lazy[node * 2 + 1] = lazy[node]
lazy[node] = 0
def update(node, left, right, l, r, val):
push_lazy(node, left, right)
if r < left or l > right:
return
if l <= left and r >= right:
lazy[node] = val
push_lazy(node, left, right)
return
mid = (left + right) // 2
update(node * 2, left, mid, l, r, val)
update(node * 2 + 1, mid + 1, right, l, r, val)
tree[node] = min(tree[node * 2], tree[node * 2 + 1])
def query(node, left, right, l, r):
push_lazy(node, left, right)
if r < left or l > right:
return 10**18
if l <= left and r >= right:
return tree[node]
mid = (left + right) // 2
sum_left = query(node * 2, left, mid, l, r)
sum_right = query(node * 2 + 1, mid + 1, right, l, r)
return min(sum_left, sum_right)
tree = [10 ** 18] * (4 * len(new))
lazy = [0] * (4 * len(new))
for i in range(len(new)):
update(1, 0, len(new) - 1, i, i, new[i])
for qq in queries:
if qq[0] == 1:
l, r = qq[1] - 1, qq[2] - 1
update(1, 0, len(new) - 1, d[l], d[r], qq[3])
else:
l, r = qq[1] - 1, qq[2] - 1
print(query(1, 0, len(new) - 1, d[l], d[r]))
|
Educational Codeforces Round 20
|
ICPC
| 2,017 | 4 | 512 |
Periodic RMQ Problem
|
You are given an array a consisting of positive integers and q queries to this array. There are two types of queries:
- 1 l r x — for each index i such that l ≤ i ≤ r set ai = x.
- 2 l r — find the minimum among such ai that l ≤ i ≤ r.
We decided that this problem is too easy. So the array a is given in a compressed form: there is an array b consisting of n elements and a number k in the input, and before all queries a is equal to the concatenation of k arrays b (so the size of a is n·k).
|
The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 104).
The second line contains n integers — elements of the array b (1 ≤ bi ≤ 109).
The third line contains one integer q (1 ≤ q ≤ 105).
Then q lines follow, each representing a query. Each query is given either as 1 l r x — set all elements in the segment from l till r (including borders) to x (1 ≤ l ≤ r ≤ n·k, 1 ≤ x ≤ 109) or as 2 l r — find the minimum among all elements in the segment from l till r (1 ≤ l ≤ r ≤ n·k).
|
For each query of type 2 print the answer to this query — the minimum on the corresponding segment.
| null | null |
[{"input": "3 1\n1 2 3\n3\n2 1 3\n1 1 2 4\n2 1 3", "output": "1\n3"}, {"input": "3 2\n1 2 3\n5\n2 4 4\n1 4 4 5\n2 4 4\n1 1 6 1\n2 6 6", "output": "1\n5\n1"}]
| 2,300 |
["data structures"]
| 32 |
[{"input": "3 1\r\n1 2 3\r\n3\r\n2 1 3\r\n1 1 2 4\r\n2 1 3\r\n", "output": "1\r\n3\r\n"}, {"input": "3 2\r\n1 2 3\r\n5\r\n2 4 4\r\n1 4 4 5\r\n2 4 4\r\n1 1 6 1\r\n2 6 6\r\n", "output": "1\r\n5\r\n1\r\n"}, {"input": "10 10\r\n10 8 10 9 2 2 4 6 10 1\r\n10\r\n1 17 87 5\r\n2 31 94\r\n1 5 56 8\r\n1 56 90 10\r\n1 25 93 6\r\n1 11 32 4\r\n2 20 49\r\n1 46 87 8\r\n2 14 48\r\n2 40 48\r\n", "output": "1\r\n4\r\n4\r\n6\r\n"}, {"input": "10 10\r\n4 2 3 8 1 2 1 7 5 4\r\n10\r\n2 63 87\r\n2 2 48\r\n2 5 62\r\n2 33 85\r\n2 30 100\r\n2 38 94\r\n2 7 81\r\n2 13 16\r\n2 26 36\r\n2 64 96\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n"}, {"input": "10 10\r\n8 7 4 6 8 2 4 9 4 3\r\n10\r\n1 43 67 7\r\n1 13 73 4\r\n1 35 71 6\r\n1 9 18 8\r\n1 3 55 2\r\n1 49 67 10\r\n1 22 49 2\r\n1 66 70 6\r\n1 17 21 10\r\n1 61 77 3\r\n", "output": ""}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 124 | 6,963,200 |
127025057
|
n=int(input())
a=[]
ans=0
for i in range(n):
t=int(input())
a.append(t)
ans+=a[0]+1
for i in range(1,n):
if(a[i]<a[i-1]):
ans+=a[i]+1
else:
ans+=(a[i]-a[i-1]+1)
ans+=1
print(ans)
| 15 | 248 | 3,072,000 |
106210111
|
from sys import stdin
t = int(stdin.readline())
ans = 0
a = 1
for k in range(t):
b = int(stdin.readline())
ans += abs(a-b) + 2
a = b
print(ans)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
PyPy 3
|
TESTS
| 2 | 124 | 1,433,600 |
129913770
|
n=int(input())
l=[]
for i in range(n):
l.append(i)
t=l[0]+2
for i in range(len(l)-1):
if l[i+1]>=l[i]:
t+=l[i+1]-l[i]+2
else:
t+=l[i+1]+2
print(t)
| 15 | 248 | 7,065,600 |
137094837
|
#!/usr/bin/env pypy3
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def read_int_list(): return list(map(int, input().split()))
def read_int_tuple(): return tuple(map(int, input().split()))
def read_int(): return int(input())
### CODE HERE
arr = [read_int() for _ in range(read_int())]
y = 0
ret = 0
for i,a in enumerate(arr):
if y > a:
# print(f"move to jump height in {a-y} moves")
ret += y - a
y = a
else:
pass
# print("move to jump height in 0 moves")
if i > 0:
# print("jump")
ret += 1
# print(f"move to top in {a-y} moves")
ret += a-y
y = a
# print(f"eat in 1 moves")
ret += 1
print(ret)
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
265/B
|
265
|
B
|
Python 3
|
TESTS
| 2 | 124 | 409,600 |
107040910
|
n=int(input())
arr=[]
for i in range(n):
arr.append(int(input()))
t1=arr[0]+1
i=0
while(i<n-1):
if arr[i]>arr[i+1]:
t1=t1+2
i=i+1
else:
t1=t1+arr[i]+arr[i+1]+1
i=i+1
arr=arr[::-1]
t2=arr[0]+1
i=0
while(i<n-1):
if arr[i]>arr[i+1]:
t2=t2+2
i=i+1
else:
t2=t2+arr[i]+arr[i+1]+1
i=i+1
print(min([t1,t2]))
| 15 | 248 | 9,113,600 |
146829397
|
def solve(n,arr):
t=0
pos=0
for i in range(n-1) :
t+= arr[i]-pos
t+=1
if arr[i]<=arr[i+1] :
t+=1
pos = arr[i]
continue
else:
t+= arr[i]-arr[i+1]
pos = arr[i+1]
t+=1
return t+arr[-1]-pos+1
from sys import stdin
input = stdin.readline
l=[]
n=int(input())
for i in range(n):
l.append(int(input()))
print(solve(n,l))
|
Codeforces Round 162 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Roadside Trees (Simplified Edition)
|
Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree.
- Eat a nut on the top of the current tree.
- Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
| null | null |
[{"input": "2\n1\n2", "output": "5"}, {"input": "5\n2\n1\n2\n1\n1", "output": "14"}]
| 1,000 |
["greedy", "implementation"]
| 15 |
[{"input": "2\r\n1\r\n2\r\n", "output": "5\r\n"}, {"input": "5\r\n2\r\n1\r\n2\r\n1\r\n1\r\n", "output": "14\r\n"}, {"input": "1\r\n1\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
1006/E
|
1006
|
E
|
Python 3
|
TESTS
| 2 | 702 | 154,419,200 |
140337141
|
from collections import defaultdict
import threading
from sys import stdin,setrecursionlimit
setrecursionlimit(10**5)
input=stdin.readline
path=[]
g=defaultdict(list)
def dfs(node,g,dp):
global path
path.append(node)
for i in g[node]:
dp[node]+=dfs(i,g,dp)
return dp[node]+1
def main():
global path
n,m=map(int,input().strip().split())
par=[*map(lambda s :int(s)-1,input().strip().split())]
par=[-1]+par
path=[]
for id,i in enumerate(par):
if id>=1:
g[i].append(id)
dp=[0]*n
path=[]
dfs(0,g,dp)
dic=defaultdict(int)
for id,i in enumerate(path):
dic[i]=id
for _ in range(m):
node,size=map(int,input().strip().split())
node-=1
size-=1
if dp[node]<size:
print(-1)
continue
print(path[dic[node]+size]+1)
# print(dic[node],dic[node]+size,dp[node])
# print(dp,path)
threading.stack_size(10 ** 8)
t = threading.Thread(target=main)
t.start()
t.join()
| 31 | 529 | 70,656,000 |
206735860
|
from sys import stdin
input=lambda :stdin.readline()[:-1]
n,q=map(int,input().split())
p=[-1]+list(map(lambda x:int(x)-1,input().split()))
edge=[[] for i in range(n)]
for i in range(1,n):
edge[i].append(p[i])
edge[p[i]].append(i)
res=[]
size=[1]*n
todo=[(0,-1)]
while todo:
v,p=todo.pop()
if v>=0:
res.append(v)
edge[v].sort(reverse=True)
for u in edge[v]:
if u!=p:
todo.append((~u,v))
todo.append((u,v))
else:
v=~v
size[p]+=size[v]
invres=[0]*n
for i in range(n):
invres[res[i]]=i
for _ in range(q):
u,k=map(int,input().split())
u-=1
if size[u]<k:
print(-1)
else:
print(res[invres[u]+k-1]+1)
|
Codeforces Round 498 (Div. 3)
|
ICPC
| 2,018 | 3 | 256 |
Military Problem
|
In this problem you will have to help Berland army with organizing their command delivery system.
There are $$$n$$$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $$$a$$$ is the direct superior of officer $$$b$$$, then we also can say that officer $$$b$$$ is a direct subordinate of officer $$$a$$$.
Officer $$$x$$$ is considered to be a subordinate (direct or indirect) of officer $$$y$$$ if one of the following conditions holds:
- officer $$$y$$$ is the direct superior of officer $$$x$$$;
- the direct superior of officer $$$x$$$ is a subordinate of officer $$$y$$$.
For example, on the picture below the subordinates of the officer $$$3$$$ are: $$$5, 6, 7, 8, 9$$$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $$$n$$$ vertices, in which vertex $$$u$$$ corresponds to officer $$$u$$$. The parent of vertex $$$u$$$ corresponds to the direct superior of officer $$$u$$$. The root (which has index $$$1$$$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $$$q$$$ queries, the $$$i$$$-th query is given as $$$(u_i, k_i)$$$, where $$$u_i$$$ is some officer, and $$$k_i$$$ is a positive integer.
To process the $$$i$$$-th query imagine how a command from $$$u_i$$$ spreads to the subordinates of $$$u_i$$$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $$$a$$$ and he spreads a command. Officer $$$a$$$ chooses $$$b$$$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $$$a$$$ chooses the one having minimal index. Officer $$$a$$$ gives a command to officer $$$b$$$. Afterwards, $$$b$$$ uses exactly the same algorithm to spread the command to its subtree. After $$$b$$$ finishes spreading the command, officer $$$a$$$ chooses the next direct subordinate again (using the same strategy). When officer $$$a$$$ cannot choose any direct subordinate who still hasn't received this command, officer $$$a$$$ finishes spreading the command.
Let's look at the following example:
If officer $$$1$$$ spreads a command, officers receive it in the following order: $$$[1, 2, 3, 5 ,6, 8, 7, 9, 4]$$$.
If officer $$$3$$$ spreads a command, officers receive it in the following order: $$$[3, 5, 6, 8, 7, 9]$$$.
If officer $$$7$$$ spreads a command, officers receive it in the following order: $$$[7, 9]$$$.
If officer $$$9$$$ spreads a command, officers receive it in the following order: $$$[9]$$$.
To answer the $$$i$$$-th query $$$(u_i, k_i)$$$, construct a sequence which describes the order in which officers will receive the command if the $$$u_i$$$-th officer spreads it. Return the $$$k_i$$$-th element of the constructed list or -1 if there are fewer than $$$k_i$$$ elements in it.
You should process queries independently. A query doesn't affect the following queries.
|
The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$$$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $$$n - 1$$$ integers $$$p_2, p_3, \dots, p_n$$$ ($$$1 \le p_i < i$$$), where $$$p_i$$$ is the index of the direct superior of the officer having the index $$$i$$$. The commander has index $$$1$$$ and doesn't have any superiors.
The next $$$q$$$ lines describe the queries. The $$$i$$$-th query is given as a pair ($$$u_i, k_i$$$) ($$$1 \le u_i, k_i \le n$$$), where $$$u_i$$$ is the index of the officer which starts spreading a command, and $$$k_i$$$ is the index of the required officer in the command spreading sequence.
|
Print $$$q$$$ numbers, where the $$$i$$$-th number is the officer at the position $$$k_i$$$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $$$u_i$$$. Print "-1" if the number of officers which receive the command is less than $$$k_i$$$.
You should process queries independently. They do not affect each other.
| null | null |
[{"input": "9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9", "output": "3\n6\n8\n-1\n9\n4"}]
| 1,600 |
["dfs and similar", "graphs", "trees"]
| 31 |
[{"input": "9 6\r\n1 1 1 3 5 3 5 7\r\n3 1\r\n1 5\r\n3 4\r\n7 3\r\n1 8\r\n1 9\r\n", "output": "3\r\n6\r\n8\r\n-1\r\n9\r\n4\r\n"}, {"input": "2 1\r\n1\r\n1 1\r\n", "output": "1\r\n"}, {"input": "13 12\r\n1 1 1 1 1 1 1 1 1 1 1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n"}]
| false |
stdio
| null | true |
975/E
|
975
|
E
|
PyPy 3
|
TESTS
| 2 | 140 | 307,200 |
66399535
|
# -*- coding: utf-8 -*-
"""
Created on Tue Dec 3 11:47:37 2019
@author: CronuS
"""
def mult (m1, m2):
return [[sum([m1[i][k] * m2[k][l] for k in range(3)]) for l in range(3)] for i in range(len(m1))]
def getmatr (xc, yc, xi, yi, matr):
xc_ = matr[0][0] * xc + matr[1][0] * yc + matr[2][0]
yc_ = matr[0][1] * xc + matr[1][1] * yc + matr[2][1]
xi_ = matr[0][0] * xi + matr[1][0] * yi + matr[2][0]
yi_ = matr[0][1] * xi + matr[1][1] * yi + matr[2][1]
s_ = ((xc_ - xi_) ** 2 + (yc_ - yi_) ** 2) ** 0.5
cosa_ = (yi_ - yc_) / s_
sina_ = -(xi_ - xc_) / s_
matrnew = [[cosa_, -sina_, 0], [sina_, cosa_, 0], [-xi_ * cosa_ - yi_ * sina_ + xi_, xi_ * sina_ + yi_ * cosa_ + yi_, 1]]
matrnew = mult(matr, matrnew)
return matrnew
#print(mult(matr0, matr0))
s = list(map(int, input().split()))
n = s[0]; q = s[1]
p = []
mx = 0; my = 0; m = 0
k1 = 1; k2 = 2
for i in range(n):
p = p + [list(map(int, input().split()))]
if (i > 2):
mxi = (p[0][0] + p[i - 1][0] + p[i][0]) / 3
myi = (p[0][1] + p[i - 1][1] + p[i][1]) / 3
mi = abs((p[i - 1][0] - p[0][0]) * (p[i][1] - p[0][1]) - (p[i][0] - p[0][0]) * (p[i - 1][1] - p[0][1]))
mx = (mx * m + mxi * mi) / (m + mi)
my = (my * m + myi * mi) / (m + mi)
m = m + mi
elif (i == 2):
mx = (p[0][0] + p[1][0] + p[2][0]) / 3
my = (p[0][1] + p[1][1] + p[2][1]) / 3
m = abs((p[1][0] - p[0][0]) * (p[2][1] - p[0][1]) - (p[2][0] - p[0][0]) * (p[1][1] - p[0][1]))
p1, p2 = 1, 2
matr = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]
ii = 0
for i in range(q):
z = list(map(int, input().split()))
if (z[0] == 1):
f = z[1]; t = z[2]
if (f == p1):
p1 = t
ii = p2 - 1
pass
else:
p2 = t
ii = p1 - 1
pass
matr = getmatr(mx, my, p[ii][0], p[ii][1], matr)
else:
ii = z[1] - 1
m = mult([[p[ii][0], p[ii][1], 1]], matr)
print(m[0][0], m[0][1])
#print(matr)
| 31 | 2,074 | 8,089,600 |
37912848
|
from math import hypot, atan2, cos, sin, sqrt, pi
import sys
mapInt = lambda:map(int,sys.stdin.buffer.readline().split())
N, Q = mapInt()
# Get points from input
points = {}
for n in range(N) :
x, y = mapInt()
points[n+1] = [x,y]
# Calculate COM (centroid)
centroid = [0,0]
A = 0
for n in range(1,N+1) :
w = points[n][0]*points[n%N+1][1] - points[n%N+1][0]*points[n][1]
centroid[0] += (points[n][0] + points[n%N+1][0])*w
centroid[1] += (points[n][1] + points[n%N+1][1])*w
A += w
centroid[0] /= 3*A
centroid[1] /= 3*A
# Move all points such that COM is at origin
for i in points.keys() :
points[i][0] -= centroid[0]
points[i][1] -= centroid[1]
# atan2(x, -y) to use negative y as reference
totAng = 0
pins = [1,2]
for q in range(Q) :
inp = list(mapInt())
if inp[0] == 1 :
f, t = inp[1],inp[2]
if pins[0] == pins[1] == f :
pins[0] = t
continue
elif pins[0] == f :
pivot = pins[1]
pins[0] = t
else :
pivot = pins[0]
pins[1] = t
x, y = points[pivot]
centroid[0] += sin(totAng + atan2(x,-y))*hypot(x,y)
centroid[1] += (-cos(totAng + atan2(x,-y)) - 1)*hypot(x,y)
totAng = pi-atan2(x,-y)
elif inp[0] == 2 :
x, y = points[inp[1]]
pointX = sin(totAng + atan2(x,-y))*hypot(x,y) + centroid[0]
pointY = -cos(totAng + atan2(x,-y))*hypot(x,y) + centroid[1]
print ("%0.20f %0.20f" % (pointX, pointY))
|
Codeforces Round 478 (Div. 2)
|
CF
| 2,018 | 3 | 256 |
Hag's Khashba
|
Hag is a very talented person. He has always had an artist inside him but his father forced him to study mechanical engineering.
Yesterday he spent all of his time cutting a giant piece of wood trying to make it look like a goose. Anyway, his dad found out that he was doing arts rather than studying mechanics and other boring subjects. He confronted Hag with the fact that he is a spoiled son that does not care about his future, and if he continues to do arts he will cut his 25 Lira monthly allowance.
Hag is trying to prove to his dad that the wooden piece is a project for mechanics subject. He also told his dad that the wooden piece is a strictly convex polygon with $$$n$$$ vertices.
Hag brought two pins and pinned the polygon with them in the $$$1$$$-st and $$$2$$$-nd vertices to the wall. His dad has $$$q$$$ queries to Hag of two types.
- $$$1$$$ $$$f$$$ $$$t$$$: pull a pin from the vertex $$$f$$$, wait for the wooden polygon to rotate under the gravity force (if it will rotate) and stabilize. And then put the pin in vertex $$$t$$$.
- $$$2$$$ $$$v$$$: answer what are the coordinates of the vertex $$$v$$$.
Please help Hag to answer his father's queries.
You can assume that the wood that forms the polygon has uniform density and the polygon has a positive thickness, same in all points. After every query of the 1-st type Hag's dad tries to move the polygon a bit and watches it stabilize again.
|
The first line contains two integers $$$n$$$ and $$$q$$$ ($$$3\leq n \leq 10\,000$$$, $$$1 \leq q \leq 200000$$$) — the number of vertices in the polygon and the number of queries.
The next $$$n$$$ lines describe the wooden polygon, the $$$i$$$-th line contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$|x_i|, |y_i|\leq 10^8$$$) — the coordinates of the $$$i$$$-th vertex of the polygon. It is guaranteed that polygon is strictly convex and the vertices are given in the counter-clockwise order and all vertices are distinct.
The next $$$q$$$ lines describe the queries, one per line. Each query starts with its type $$$1$$$ or $$$2$$$. Each query of the first type continues with two integers $$$f$$$ and $$$t$$$ ($$$1 \le f, t \le n$$$) — the vertex the pin is taken from, and the vertex the pin is put to and the polygon finishes rotating. It is guaranteed that the vertex $$$f$$$ contains a pin. Each query of the second type continues with a single integer $$$v$$$ ($$$1 \le v \le n$$$) — the vertex the coordinates of which Hag should tell his father.
It is guaranteed that there is at least one query of the second type.
|
The output should contain the answer to each query of second type — two numbers in a separate line. Your answer is considered correct, if its absolute or relative error does not exceed $$$10^{-4}$$$.
Formally, let your answer be $$$a$$$, and the jury's answer be $$$b$$$. Your answer is considered correct if $$$\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-4}$$$
| null |
In the first test note the initial and the final state of the wooden polygon.
Red Triangle is the initial state and the green one is the triangle after rotation around $$$(2,0)$$$.
In the second sample note that the polygon rotates $$$180$$$ degrees counter-clockwise or clockwise direction (it does not matter), because Hag's father makes sure that the polygon is stable and his son does not trick him.
|
[{"input": "3 4\n0 0\n2 0\n2 2\n1 1 2\n2 1\n2 2\n2 3", "output": "3.4142135624 -1.4142135624\n2.0000000000 0.0000000000\n0.5857864376 -1.4142135624"}, {"input": "3 2\n-1 1\n0 0\n1 1\n1 1 2\n2 1", "output": "1.0000000000 -1.0000000000"}]
| 2,600 |
["geometry"]
| 31 |
[{"input": "3 4\r\n0 0\r\n2 0\r\n2 2\r\n1 1 2\r\n2 1\r\n2 2\r\n2 3\r\n", "output": "3.4142135624 -1.4142135624\r\n2.0000000000 0.0000000000\r\n0.5857864376 -1.4142135624\r\n"}, {"input": "3 2\r\n-1 1\r\n0 0\r\n1 1\r\n1 1 2\r\n2 1\r\n", "output": "1.0000000000 -1.0000000000\r\n"}, {"input": "10 10\r\n0 -100000000\r\n1 -100000000\r\n1566 -99999999\r\n2088 -99999997\r\n2610 -99999994\r\n3132 -99999990\r\n3654 -99999985\r\n4176 -99999979\r\n4698 -99999972\r\n5220 -99999964\r\n1 2 5\r\n2 1\r\n1 1 7\r\n2 5\r\n1 5 4\r\n1 4 2\r\n2 8\r\n1 7 9\r\n2 1\r\n1 2 10\r\n", "output": "0.0000000000 -100000000.0000000000\r\n-7.3726558373 -100002609.9964835122\r\n-129.8654413032 -100003125.4302210321\r\n-114.4079442212 -100007299.4544525659\r\n"}, {"input": "4 10\r\n0 0\r\n2 0\r\n2 2\r\n0 2\r\n2 3\r\n2 1\r\n2 1\r\n1 1 1\r\n2 3\r\n1 2 4\r\n1 4 4\r\n2 4\r\n1 1 3\r\n2 3\r\n", "output": "2.0000000000 2.0000000000\r\n0.0000000000 0.0000000000\r\n0.0000000000 0.0000000000\r\n0.5857864376 -1.4142135624\r\n4.8284271247 -2.8284271247\r\n6.2426406871 -4.2426406871\r\n"}, {"input": "3 2\r\n0 0\r\n1 0\r\n1566 1\r\n1 2 1\r\n2 3\r\n", "output": "0.0006381620 -1566.0003192846\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input to count type 2 queries
with open(input_path, 'r') as f:
n, q = map(int, f.readline().split())
for _ in range(n):
f.readline() # skip vertices
type2_count = 0
for _ in range(q):
parts = f.readline().split()
if parts[0] == '2':
type2_count += 1
# Read correct and submission outputs
with open(output_path, 'r') as f:
correct = [line.strip() for line in f]
with open(submission_path, 'r') as f:
submission = [line.strip() for line in f]
# Check line count
if len(correct) != len(submission) or len(correct) != type2_count:
print(0)
return
# Check each line
for c_line, s_line in zip(correct, submission):
try:
c_x, c_y = map(float, c_line.split())
s_x, s_y = map(float, s_line.split())
except:
print(0)
return
# Check x
dx = abs(s_x - c_x)
max_x = max(1.0, abs(c_x))
if dx > 1e-4 * max_x:
print(0)
return
# Check y
dy = abs(s_y - c_y)
max_y = max(1.0, abs(c_y))
if dy > 1e-4 * max_y:
print(0)
return
# All checks passed
print(1)
if __name__ == "__main__":
main()
| true |
852/E
|
852
|
E
|
Python 3
|
TESTS
| 3 | 358 | 512,000 |
66270400
|
n = int(input())
degree = n*[0]
for i in range(0, n-1):
u, v = input().split()
u = int(u)
v = int(v)
u -= 1
v -= 1
degree[u] += 1
degree[v] += 1
numL = 0
for i in range(0, n):
if degree[i] == 1:
numL += 1
ans = (n-numL)*2**(n-numL) + numL*2**(n+1-numL)
print(ans)
| 12 | 93 | 3,686,400 |
214753050
|
import sys
input = sys.stdin.readline
n = int(input())
mod = 10 ** 9 + 7
degs = [0] * n
for _ in range(n-1):
u, v = map(int, input().split())
u -= 1
v -= 1
degs[u] += 1
degs[v] += 1
leaves_count = degs.count(1)
print((n + leaves_count) * pow(2, n - leaves_count, mod) % mod)
|
Bubble Cup X - Finals [Online Mirror]
|
ICPC
| 2,017 | 1 | 256 |
Casinos and travel
|
John has just bought a new car and is planning a journey around the country. Country has N cities, some of which are connected by bidirectional roads. There are N - 1 roads and every city is reachable from any other city. Cities are labeled from 1 to N.
John first has to select from which city he will start his journey. After that, he spends one day in a city and then travels to a randomly choosen city which is directly connected to his current one and which he has not yet visited. He does this until he can't continue obeying these rules.
To select the starting city, he calls his friend Jack for advice. Jack is also starting a big casino business and wants to open casinos in some of the cities (max 1 per city, maybe nowhere). Jack knows John well and he knows that if he visits a city with a casino, he will gamble exactly once before continuing his journey.
He also knows that if John enters a casino in a good mood, he will leave it in a bad mood and vice versa. Since he is John's friend, he wants him to be in a good mood at the moment when he finishes his journey. John is in a good mood before starting the journey.
In how many ways can Jack select a starting city for John and cities where he will build casinos such that no matter how John travels, he will be in a good mood at the end? Print answer modulo 109 + 7.
|
In the first line, a positive integer N (1 ≤ N ≤ 100000), the number of cities.
In the next N - 1 lines, two numbers a, b (1 ≤ a, b ≤ N) separated by a single space meaning that cities a and b are connected by a bidirectional road.
|
Output one number, the answer to the problem modulo 109 + 7.
| null |
Example 1: If Jack selects city 1 as John's starting city, he can either build 0 casinos, so John will be happy all the time, or build a casino in both cities, so John would visit a casino in city 1, become unhappy, then go to city 2, visit a casino there and become happy and his journey ends there because he can't go back to city 1. If Jack selects city 2 for start, everything is symmetrical, so the answer is 4.
Example 2: If Jack tells John to start from city 1, he can either build casinos in 0 or 2 cities (total 4 possibilities). If he tells him to start from city 2, then John's journey will either contain cities 2 and 1 or 2 and 3. Therefore, Jack will either have to build no casinos, or build them in all three cities. With other options, he risks John ending his journey unhappy. Starting from 3 is symmetric to starting from 1, so in total we have 4 + 2 + 4 = 10 options.
|
[{"input": "2\n1 2", "output": "4"}, {"input": "3\n1 2\n2 3", "output": "10"}]
| 2,100 |
["dp"]
| 12 |
[{"input": "2\r\n1 2\r\n", "output": "4\r\n"}, {"input": "3\r\n1 2\r\n2 3\r\n", "output": "10\r\n"}, {"input": "4\r\n1 2\r\n2 3\r\n3 4\r\n", "output": "24\r\n"}]
| false |
stdio
| null | true |
898/C
|
898
|
C
|
Python 3
|
TESTS
| 2 | 31 | 0 |
158634578
|
d = {}
for s in [*open(0)][1:]:
s = s.split()
d.setdefault(s[0],[])
for i in s[2:]:d[s[0]].append(i)
for i in d:
x = d[i]
if len(x)>1:
for j in range (len(x)):
for k in range (len(x)):
if x[j] == x[k]:continue
a,b = x[j],x[k]
a,b = str(a),str(b)
if len(a)<len(b):
if b[-len(a):] == a and len(a)!=1:x[j] = -1
elif len(b) < len(a):
if a[-len(b):] == b and len(b)!=1:x[k] = -1
x = set(x)
x.discard(-1)
d[i] = x
print(len(d))
for i in d:print(f'{i} {len(d[i])} {" ".join(d[i])}')
| 59 | 62 | 5,529,600 |
33298133
|
phones = dict()
n = int(input())
for i in range(n):
name, num, *numbers = input().split()
for number in numbers:
try:
for el in phones[name]:
if el.endswith(number):
break
if number.endswith(el):
phones[name].remove(el)
phones[name].add(number)
break
else:
phones[name].add(number)
except KeyError:
phones[name] = {number}
print(len(phones))
print('\n'.join([' '.join([name, ' '.join([str(len(numbers)), ' '.join(numbers)])]) for name, numbers in phones.items()]))
|
Codeforces Round 451 (Div. 2)
|
CF
| 2,017 | 2 | 256 |
Phone Numbers
|
Vasya has several phone books, in which he recorded the telephone numbers of his friends. Each of his friends can have one or several phone numbers.
Vasya decided to organize information about the phone numbers of friends. You will be given n strings — all entries from Vasya's phone books. Each entry starts with a friend's name. Then follows the number of phone numbers in the current entry, and then the phone numbers themselves. It is possible that several identical phones are recorded in the same record.
Vasya also believes that if the phone number a is a suffix of the phone number b (that is, the number b ends up with a), and both numbers are written by Vasya as the phone numbers of the same person, then a is recorded without the city code and it should not be taken into account.
The task is to print organized information about the phone numbers of Vasya's friends. It is possible that two different people have the same number. If one person has two numbers x and y, and x is a suffix of y (that is, y ends in x), then you shouldn't print number x. If the number of a friend in the Vasya's phone books is recorded several times in the same format, it is necessary to take it into account exactly once.
Read the examples to understand statement and format of the output better.
|
First line contains the integer n (1 ≤ n ≤ 20) — number of entries in Vasya's phone books.
The following n lines are followed by descriptions of the records in the format described in statement. Names of Vasya's friends are non-empty strings whose length does not exceed 10. They consists only of lowercase English letters. Number of phone numbers in one entry is not less than 1 is not more than 10. The telephone numbers consist of digits only. If you represent a phone number as a string, then its length will be in range from 1 to 10. Phone numbers can contain leading zeros.
|
Print out the ordered information about the phone numbers of Vasya's friends. First output m — number of friends that are found in Vasya's phone books.
The following m lines must contain entries in the following format "name number_of_phone_numbers phone_numbers". Phone numbers should be separated by a space. Each record must contain all the phone numbers of current friend.
Entries can be displayed in arbitrary order, phone numbers for one record can also be printed in arbitrary order.
| null | null |
[{"input": "2\nivan 1 00123\nmasha 1 00123", "output": "2\nmasha 1 00123\nivan 1 00123"}, {"input": "3\nkarl 2 612 12\npetr 1 12\nkatya 1 612", "output": "3\nkatya 1 612\npetr 1 12\nkarl 1 612"}, {"input": "4\nivan 3 123 123 456\nivan 2 456 456\nivan 8 789 3 23 6 56 9 89 2\ndasha 2 23 789", "output": "2\ndasha 2 23 789\nivan 4 789 123 2 456"}]
| 1,400 |
["implementation", "strings"]
| 59 |
[{"input": "2\r\nivan 1 00123\r\nmasha 1 00123\r\n", "output": "2\r\nmasha 1 00123 \r\nivan 1 00123 \r\n"}, {"input": "3\r\nkarl 2 612 12\r\npetr 1 12\r\nkatya 1 612\r\n", "output": "3\r\nkatya 1 612 \r\npetr 1 12 \r\nkarl 1 612 \r\n"}, {"input": "4\r\nivan 3 123 123 456\r\nivan 2 456 456\r\nivan 8 789 3 23 6 56 9 89 2\r\ndasha 2 23 789\r\n", "output": "2\r\ndasha 2 789 23 \r\nivan 4 789 123 456 2 \r\n"}, {"input": "20\r\nnxj 6 7 6 6 7 7 7\r\nnxj 10 8 5 1 7 6 1 0 7 0 6\r\nnxj 2 6 5\r\nnxj 10 6 7 6 6 5 8 3 6 6 8\r\nnxj 10 6 1 7 6 7 1 8 7 8 6\r\nnxj 10 8 5 8 6 5 6 1 9 6 3\r\nnxj 10 8 1 6 4 8 0 4 6 0 1\r\nnxj 9 2 6 6 8 1 1 3 6 6\r\nnxj 10 8 9 0 9 1 3 2 3 2 3\r\nnxj 6 6 7 0 8 1 2\r\nnxj 7 7 7 8 1 3 6 9\r\nnxj 10 2 7 0 1 5 1 9 1 2 6\r\nnxj 6 9 6 9 6 3 7\r\nnxj 9 0 1 7 8 2 6 6 5 6\r\nnxj 4 0 2 3 7\r\nnxj 10 0 4 0 6 1 1 8 8 4 7\r\nnxj 8 4 6 2 6 6 1 2 7\r\nnxj 10 5 3 4 2 1 0 7 0 7 6\r\nnxj 10 9 6 0 6 1 6 2 1 9 6\r\nnxj 4 2 9 0 1\r\n", "output": "1\r\nnxj 10 0 3 2 1 4 7 8 5 9 6 \r\n"}, {"input": "20\r\nl 6 02 02 2 02 02 2\r\nl 8 8 8 8 2 62 13 31 3\r\ne 9 0 91 0 0 60 91 60 2 44\r\ne 9 69 2 1 44 2 91 66 1 70\r\nl 9 7 27 27 3 1 3 7 80 81\r\nl 9 2 1 13 7 2 10 02 3 92\r\ne 9 0 15 3 5 5 15 91 09 44\r\nl 7 2 50 4 5 98 31 98\r\nl 3 26 7 3\r\ne 6 7 5 0 62 65 91\r\nl 8 80 0 4 0 2 2 0 13\r\nl 9 19 13 02 2 1 4 19 26 02\r\nl 10 7 39 7 9 22 22 26 2 90 4\r\ne 7 65 2 36 0 34 57 9\r\ne 8 13 02 09 91 73 5 36 62\r\nl 9 75 0 10 8 76 7 82 8 34\r\nl 7 34 0 19 80 6 4 7\r\ne 5 4 2 5 7 2\r\ne 7 4 02 69 7 07 20 2\r\nl 4 8 2 1 63\r\n", "output": "2\r\ne 18 15 62 07 70 91 57 02 66 65 69 09 13 20 44 73 34 60 36 \r\nl 21 27 50 22 63 75 19 26 90 02 92 62 31 10 76 82 80 98 81 39 34 13 \r\n"}, {"input": "20\r\no 10 6 6 97 45 6 6 6 6 5 6\r\nl 8 5 5 5 19 59 5 8 5\r\nj 9 2 30 58 2 2 1 0 30 4\r\nc 10 1 1 7 51 7 7 51 1 1 1\r\no 9 7 97 87 70 2 19 2 14 6\r\ne 6 26 6 6 6 26 5\r\ng 9 3 3 3 3 3 78 69 8 9\r\nl 8 8 01 1 5 8 41 72 3\r\nz 10 1 2 2 2 9 1 9 1 6 7\r\ng 8 7 78 05 36 7 3 67 9\r\no 5 6 9 9 7 7\r\ne 10 30 2 1 1 2 5 04 0 6 6\r\ne 9 30 30 2 2 0 26 30 79 8\r\nt 10 2 2 9 29 7 7 7 9 2 9\r\nc 7 7 51 1 31 2 7 4\r\nc 9 83 1 6 78 94 74 54 8 32\r\ng 8 4 1 01 9 39 28 6 6\r\nt 7 9 2 01 4 4 9 58\r\nj 5 0 1 58 02 4\r\nw 10 80 0 91 91 06 91 9 9 27 7\r\n", "output": "9\r\nw 5 91 27 06 9 80 \r\nt 6 7 58 4 29 2 01 \r\ne 8 8 79 5 30 2 26 04 1 \r\nl 8 72 3 19 59 41 01 5 8 \r\nj 5 4 30 58 1 02 \r\nz 5 6 1 9 2 7 \r\ng 10 05 39 4 3 36 01 67 69 28 78 \r\no 8 14 87 97 6 19 70 45 2 \r\nc 10 83 51 31 54 74 32 7 94 78 6 \r\n"}, {"input": "1\r\negew 5 3 123 23 1234 134\r\n", "output": "1\r\negew 3 123 134 1234 \r\n"}]
| false |
stdio
|
import sys
from collections import defaultdict
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input and build friends data
friends = defaultdict(set)
with open(input_path, 'r') as f:
n = int(f.readline())
for _ in range(n):
parts = f.readline().strip().split()
name = parts[0]
numbers = parts[2:]
friends[name].update(numbers) # Add all numbers, duplicates in same line are handled via set
# Read submission output
submission_friends = set()
try:
with open(submission_path, 'r') as f:
lines = f.read().splitlines()
except:
print(0)
return
if not lines:
print(0)
return
# Check m line
try:
m_line = lines[0]
m = int(m_line)
except:
print(0)
return
if m != len(friends):
print(0)
return
submission_lines = lines[1:]
if len(submission_lines) != m:
print(0)
return
for line in submission_lines:
parts = line.strip().split()
if len(parts) < 2:
print(0)
return
name = parts[0]
if name not in friends:
print(0)
return
submission_friends.add(name)
try:
k = int(parts[1])
except:
print(0)
return
numbers = parts[2:]
if len(numbers) != k:
print(0)
return
# Check for duplicates in submission's numbers
if len(set(numbers)) != k:
print(0)
return
# Check all numbers are present in friends' merged set
merged_numbers = friends[name]
for num in numbers:
if num not in merged_numbers:
print(0)
return
# Check no two numbers in submission are suffix of each other
for i in range(len(numbers)):
for j in range(i+1, len(numbers)):
a = numbers[i]
b = numbers[j]
if (len(a) >= len(b) and a.endswith(b)) or (len(b) >= len(a) and b.endswith(a)):
print(0)
return
# Check every merged number is either in submission or is a suffix of some submission number
for merged_num in merged_numbers:
if merged_num in numbers:
continue
found = False
for num in numbers:
if len(merged_num) <= len(num) and num.endswith(merged_num):
found = True
break
if not found:
print(0)
return
# Check all friends are present in submission
if submission_friends != set(friends.keys()):
print(0)
return
# All checks passed
print(100)
if __name__ == "__main__":
main()
| true |
840/B
|
840
|
B
|
Python 3
|
TESTS
| 2 | 46 | 0 |
29675269
|
n, m = map(int, input().split())
d = [-123] + list(map(int, input().split()))
edges = []
for i in range(m):
u, v = map(int, input().split())
edges.append((u, v))
def gao():
if 0 in d or -1 in d:
return 0
for i, edge in enumerate(edges):
u, v = edge
if d[u] == d[v]:
return i
return -1
print(gao())
| 73 | 2,979 | 87,756,800 |
167362143
|
import bisect
import copy
import decimal
import fractions
import heapq
import itertools
import math
import random
import sys
import time
from collections import Counter,deque,defaultdict
from functools import lru_cache,reduce
from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
def _heappush_max(heap,item):
heap.append(item)
heapq._siftdown_max(heap, 0, len(heap)-1)
def _heappushpop_max(heap, item):
if heap and item < heap[0]:
item, heap[0] = heap[0], item
heapq._siftup_max(heap, 0)
return item
from math import gcd as GCD
read=sys.stdin.read
readline=sys.stdin.readline
readlines=sys.stdin.readlines
class Graph:
def __init__(self,V,edges=False,graph=False,directed=False,weighted=False,inf=float("inf")):
self.V=V
self.directed=directed
self.weighted=weighted
self.inf=inf
if graph:
self.graph=graph
self.edges=[]
for i in range(self.V):
if self.weighted:
for j,d in self.graph[i]:
if self.directed or not self.directed and i<=j:
self.edges.append((i,j,d))
else:
for j in self.graph[i]:
if self.directed or not self.directed and i<=j:
self.edges.append((i,j))
else:
self.edges=edges
self.graph=[[] for i in range(self.V)]
if weighted:
for i,j,d in self.edges:
self.graph[i].append((j,d))
if not self.directed:
self.graph[j].append((i,d))
else:
for i,j in self.edges:
self.graph[i].append(j)
if not self.directed:
self.graph[j].append(i)
def SIV_DFS(self,s,bipartite_graph=False,cycle_detection=False,directed_acyclic=False,euler_tour=False,linked_components=False,lowlink=False,parents=False,postorder=False,preorder=False,subtree_size=False,topological_sort=False,unweighted_dist=False,weighted_dist=False):
seen=[False]*self.V
finished=[False]*self.V
if directed_acyclic or cycle_detection or topological_sort:
dag=True
if euler_tour:
et=[]
if linked_components:
lc=[]
if lowlink:
order=[None]*self.V
ll=[None]*self.V
idx=0
if parents or cycle_detection or lowlink or subtree_size:
ps=[None]*self.V
if postorder or topological_sort:
post=[]
if preorder:
pre=[]
if subtree_size:
ss=[1]*self.V
if unweighted_dist or bipartite_graph:
uwd=[self.inf]*self.V
uwd[s]=0
if weighted_dist:
wd=[self.inf]*self.V
wd[s]=0
stack=[(s,0)] if self.weighted else [s]
while stack:
if self.weighted:
x,d=stack.pop()
else:
x=stack.pop()
if not seen[x]:
seen[x]=True
stack.append((x,d) if self.weighted else x)
if euler_tour:
et.append(x)
if linked_components:
lc.append(x)
if lowlink:
order[x]=idx
ll[x]=idx
idx+=1
if preorder:
pre.append(x)
for y in self.graph[x]:
if self.weighted:
y,d=y
if not seen[y]:
stack.append((y,d) if self.weighted else y)
if parents or cycle_detection or lowlink or subtree_size:
ps[y]=x
if unweighted_dist or bipartite_graph:
uwd[y]=uwd[x]+1
if weighted_dist:
wd[y]=wd[x]+d
elif not finished[y]:
if (directed_acyclic or cycle_detection or topological_sort) and dag:
dag=False
if cycle_detection:
cd=(y,x)
elif not finished[x]:
finished[x]=True
if euler_tour:
et.append(~x)
if lowlink:
bl=True
for y in self.graph[x]:
if self.weighted:
y,d=y
if ps[x]==y and bl:
bl=False
continue
ll[x]=min(ll[x],order[y])
if x!=s:
ll[ps[x]]=min(ll[ps[x]],ll[x])
if postorder or topological_sort:
post.append(x)
if subtree_size:
for y in self.graph[x]:
if self.weighted:
y,d=y
if y==ps[x]:
continue
ss[x]+=ss[y]
if bipartite_graph:
bg=[[],[]]
for tpl in self.edges:
x,y=tpl[:2] if self.weighted else tpl
if uwd[x]==self.inf or uwd[y]==self.inf:
continue
if not uwd[x]%2^uwd[y]%2:
bg=False
break
else:
for x in range(self.V):
if uwd[x]==self.inf:
continue
bg[uwd[x]%2].append(x)
retu=()
if bipartite_graph:
retu+=(bg,)
if cycle_detection:
if dag:
cd=[]
else:
y,x=cd
cd=self.Route_Restoration(y,x,ps)
retu+=(cd,)
if directed_acyclic:
retu+=(dag,)
if euler_tour:
retu+=(et,)
if linked_components:
retu+=(lc,)
if lowlink:
retu=(ll,)
if parents:
retu+=(ps,)
if postorder:
retu+=(post,)
if preorder:
retu+=(pre,)
if subtree_size:
retu+=(ss,)
if topological_sort:
if dag:
tp_sort=post[::-1]
else:
tp_sort=[]
retu+=(tp_sort,)
if unweighted_dist:
retu+=(uwd,)
if weighted_dist:
retu+=(wd,)
if len(retu)==1:
retu=retu[0]
return retu
N,M=map(int,readline().split())
D=list(map(int,readline().split()))
edges=[]
idx={}
for i in range(M):
u,v=map(int,readline().split())
u-=1;v-=1
edges.append((u,v))
idx[(u,v)]=i
idx[(v,u)]=i
G=Graph(N,edges=edges)
parents,tour=G.SIV_DFS(0,parents=True,postorder=True)
if D.count(1)%2:
for i in range(N):
if D[i]==-1:
D[i]=1
break
for i in range(N):
if D[i]==-1:
D[i]=0
if D.count(1)%2:
print(-1)
exit()
ans_lst=[]
for x in tour[:-1]:
if D[x]:
ans_lst.append(idx[(x,parents[x])]+1)
D[x]^=1
D[parents[x]]^=1
print(len(ans_lst))
if ans_lst:
print(*ans_lst,sep="\n")
|
Codeforces Round 429 (Div. 1)
|
CF
| 2,017 | 3 | 256 |
Leha and another game about graph
|
Leha plays a computer game, where is on each level is given a connected graph with n vertices and m edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer di, which can be equal to 0, 1 or - 1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, di = - 1 or it's degree modulo 2 is equal to di. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them.
|
The first line contains two integers n, m (1 ≤ n ≤ 3·105, n - 1 ≤ m ≤ 3·105) — number of vertices and edges.
The second line contains n integers d1, d2, ..., dn ( - 1 ≤ di ≤ 1) — numbers on the vertices.
Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n) — edges. It's guaranteed, that graph in the input is connected.
|
Print - 1 in a single line, if solution doesn't exist. Otherwise in the first line k — number of edges in a subset. In the next k lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1.
| null |
In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1.
|
[{"input": "1 0\n1", "output": "-1"}, {"input": "4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4", "output": "0"}, {"input": "2 1\n1 1\n1 2", "output": "1\n1"}, {"input": "3 3\n0 -1 1\n1 2\n2 3\n1 3", "output": "1\n2"}]
| 2,100 |
["constructive algorithms", "data structures", "dfs and similar", "dp", "graphs"]
| 73 |
[{"input": "1 0\r\n1\r\n", "output": "-1\r\n"}, {"input": "4 5\r\n0 0 0 -1\r\n1 2\r\n2 3\r\n3 4\r\n1 4\r\n2 4\r\n", "output": "0\r\n"}, {"input": "2 1\r\n1 1\r\n1 2\r\n", "output": "1\r\n1\r\n"}, {"input": "3 3\r\n0 -1 1\r\n1 2\r\n2 3\r\n1 3\r\n", "output": "1\r\n2\r\n"}, {"input": "10 10\r\n-1 -1 -1 -1 -1 -1 -1 -1 -1 -1\r\n6 7\r\n8 3\r\n6 4\r\n4 2\r\n9 2\r\n5 10\r\n9 8\r\n10 7\r\n5 1\r\n6 2\r\n", "output": "0\r\n"}, {"input": "3 2\r\n1 0 1\r\n1 2\r\n2 3\r\n", "output": "2\r\n2\r\n1\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_output_path):
# Read input
with open(input_path, 'r') as f:
n, m = map(int, f.readline().split())
d = list(map(int, f.readline().split()))
edges = [tuple(map(int, f.readline().split())) for _ in range(m)]
# Read submission output
with open(submission_output_path, 'r') as f:
lines = f.read().strip().split('\n')
if not lines:
print(0)
return
first_line = lines[0].strip()
if first_line == '-1':
sum_d = sum(x for x in d if x != -1)
has_neg1 = any(x == -1 for x in d)
exists = (sum_d % 2 == 0) or (has_neg1 and (sum_d % 2 == 1))
print(0 if exists else 1)
return
else:
try:
k = int(first_line)
if len(lines) < k + 1:
print(0)
return
selected = list(map(int, lines[1:1 + k]))
except:
print(0)
return
# Check edge indices
if any(e < 1 or e > m for e in selected):
print(0)
return
if len(set(selected)) != len(selected):
print(0)
return
# Compute degrees
degree = [0] * (n + 1) # 1-based
for e in selected:
u, v = edges[e - 1]
degree[u] += 1
degree[v] += 1
# Check each vertex
valid = True
for i in range(n):
di = d[i]
if di == -1:
continue
if degree[i + 1] % 2 != di:
valid = False
break
print(1 if valid else 0)
if __name__ == '__main__':
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_output_path = sys.argv[3]
main(input_path, output_path, submission_output_path)
| true |
925/B
|
925
|
B
|
Python 3
|
TESTS
| 0 | 109 | 102,400 |
47186519
|
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 12/18/18
"""
import collections
import time
import os
import sys
import bisect
import heapq
n, x1, x2 = map(int, input().split())
C = [int(x) for x in input().split()]
C = [(v, i+1) for i, v in enumerate(C)]
C.sort()
def check(x1, x2, rev):
r = n
p1, p2 = [], []
for i in range(n-1, -1, -1):
if C[i][0] * (n-i) >= x1:
r = i
p1 = [v[1] for v in C[i:]]
break
if p1:
for i in range(r, -1, -1):
if C[i][0] * (r-i) >= x2:
p2 = [v[1] for v in C[i: r]]
if rev:
p1, p2 = p2, p1
print('YES')
print('{} {}'.format(len(p1), len(p2)))
print(' '.join(map(str, p1)))
print(' '.join(map(str, p2)))
exit(0)
check(x1, x2, False)
check(x2, x1, True)
print('No')
| 40 | 1,341 | 31,846,400 |
197614973
|
def fin(c, x):
return (x + c - 1) // c
def ck(x, b):
r = (n, n)
for i in range(b, n):
r = min(r, (i + fin(c[i][0], x), i))
return r
def sol(r, l):
if r[0] <= n and l[0] <= n and r[1] < n and l[1] < n :
print("Yes")
print(r[0] - r[1], l[0]- l[1])
print(' '.join([str(x[1]) for x in c[r[1]:r[0]]]))
print(' '.join([str(x[1]) for x in c[l[1]:l[0]]]))
return True
else:
return False
n, x1, x2 = [int(x) for x in input().split()]
c = sorted([(int(x), i + 1) for i, x in enumerate(input().split())])
r1 = ck(x1, 0)
l1 = ck(x2, r1[0])
r2 = ck(x2, 0)
l2 = ck(x1, r2[0])
if not sol(r1, l1) and not sol(l2, r2):
print("No")
# 6 8 16
# 3 5 2 9 8 7
|
VK Cup 2018 - Round 3
|
CF
| 2,018 | 2 | 256 |
Resource Distribution
|
One department of some software company has $$$n$$$ servers of different specifications. Servers are indexed with consecutive integers from $$$1$$$ to $$$n$$$. Suppose that the specifications of the $$$j$$$-th server may be expressed with a single integer number $$$c_j$$$ of artificial resource units.
In order for production to work, it is needed to deploy two services $$$S_1$$$ and $$$S_2$$$ to process incoming requests using the servers of the department. Processing of incoming requests of service $$$S_i$$$ takes $$$x_i$$$ resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service $$$S_i$$$ is deployed using $$$k_i$$$ servers, then the load is divided equally between these servers and each server requires only $$$x_i / k_i$$$ (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
|
The first line contains three integers $$$n$$$, $$$x_1$$$, $$$x_2$$$ ($$$2 \leq n \leq 300\,000$$$, $$$1 \leq x_1, x_2 \leq 10^9$$$) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains $$$n$$$ space-separated integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \leq c_i \leq 10^9$$$) — the number of resource units provided by each of the servers.
|
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers $$$k_1$$$ and $$$k_2$$$ ($$$1 \leq k_1, k_2 \leq n$$$) — the number of servers used for each of the services.
In the third line print $$$k_1$$$ integers, the indices of the servers that will be used for the first service.
In the fourth line print $$$k_2$$$ integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
| null |
In the first sample test each of the servers 1, 2 and 6 will will provide $$$8 / 3 = 2.(6)$$$ resource units and each of the servers 5, 4 will provide $$$16 / 2 = 8$$$ resource units.
In the second sample test the first server will provide $$$20$$$ resource units and each of the remaining servers will provide $$$32 / 3 = 10.(6)$$$ resource units.
|
[{"input": "6 8 16\n3 5 2 9 8 7", "output": "Yes\n3 2\n1 2 6\n5 4"}, {"input": "4 20 32\n21 11 11 12", "output": "Yes\n1 3\n1\n2 3 4"}, {"input": "4 11 32\n5 5 16 16", "output": "No"}, {"input": "5 12 20\n7 8 4 11 9", "output": "No"}]
| 1,700 |
["binary search", "implementation", "sortings"]
| 40 |
[{"input": "6 8 16\r\n3 5 2 9 8 7\r\n", "output": "Yes\r\n4 2\r\n3 1 2 6\r\n5 4\r\n"}, {"input": "4 20 32\r\n21 11 11 12\r\n", "output": "Yes\r\n1 3\r\n1\r\n2 3 4\r\n"}, {"input": "4 11 32\r\n5 5 16 16\r\n", "output": "No\r\n"}, {"input": "5 12 20\r\n7 8 4 11 9\r\n", "output": "No\r\n"}, {"input": "2 1 1\r\n1 1\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 1\r\n1 1000000\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 1\r\n1000000000 1000000000\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 2\r\n1 1\r\n", "output": "No\r\n"}, {"input": "15 250 200\r\n71 2 77 69 100 53 54 40 73 32 82 58 24 82 41\r\n", "output": "Yes\r\n11 3\r\n13 10 8 15 6 7 12 4 1 9 3\r\n11 14 5\r\n"}, {"input": "4 12 11\r\n4 4 6 11\r\n", "output": "Yes\r\n3 1\r\n1 2 3\r\n4\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, x1, x2 = map(int, f.readline().split())
c = list(map(int, f.readline().split()))
with open(output_path) as f:
ref_lines = [line.strip() for line in f]
with open(submission_path) as f:
sub_lines = [line.strip() for line in f]
non_empty_ref = [line for line in ref_lines if line]
non_empty_sub = [line for line in sub_lines if line]
if not non_empty_ref:
print(0)
return
ref_first = non_empty_ref[0]
if ref_first == 'No':
if len(non_empty_sub) == 1 and non_empty_sub[0] == 'No':
print(1)
else:
print(0)
return
else:
if len(non_empty_sub) != 4 or non_empty_sub[0] != 'Yes':
print(0)
return
try:
k1, k2 = map(int, non_empty_sub[1].split())
except:
print(0)
return
if k1 <= 0 or k2 <= 0 or k1 + k2 > n:
print(0)
return
try:
s1 = list(map(int, non_empty_sub[2].split()))
s2 = list(map(int, non_empty_sub[3].split()))
except:
print(0)
return
if len(s1) != k1 or len(s2) != k2:
print(0)
return
all_indices = set()
for s in s1:
if not (1 <= s <= n) or s in all_indices:
print(0)
return
all_indices.add(s)
for s in s2:
if not (1 <= s <= n) or s in all_indices:
print(0)
return
all_indices.add(s)
for s in s1:
if c[s-1] * k1 < x1:
print(0)
return
for s in s2:
if c[s-1] * k2 < x2:
print(0)
return
print(1)
return
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
723/D
|
723
|
D
|
Python 3
|
TESTS
| 2 | 31 | 0 |
195520964
|
dr = [-1, 0, 0, 1]
dc = [0, -1, 1, 0]
def isValid(r, c):
return r in range(n) and c in range(m)
def DFS(sr, sc):
is_sea = False
count = 1
stack = [(sr, sc)]
visited[sr][sc] = True
while stack:
r, c = stack.pop()
for i in range(4):
nr = r + dr[i]
nc = c + dc[i]
if isValid(nr, nc) and not visited[nr][nc] and terrain[nr][nc] == '.':
if nr == 0 or nc == 0 or nr == n - 1 or nc == m - 1:
is_sea = True
else:
count += 1
visited[nr][nc] = True
stack.append((nr, nc))
if not is_sea:
size.append((count, sr, sc))
def fix(sr, sc):
stack = [(sr, sc)]
terrain[sr][sc] = '*'
while stack:
r, c = stack.pop()
for i in range(4):
nr = r + dr[i]
nc = c + dc[i]
if isValid(nr, nc) and terrain[nr][nc] == '.':
terrain[nr][nc] = '*'
stack.append((nr, nc))
n, m, k = map(int, input().split())
size = []
visited = [[False] * m for i in range(n)]
terrain = []
for i in range(n):
terrain.append(list(input()))
for i in range(n):
for j in range(m):
if terrain[i][j] == '.' and not visited[i][j]:
DFS(i, j)
size.sort()
ans = 0
for i in range(len(size) - k):
ans += size[i][0]
for i in range(len(size) - k):
fix(size[i][1], size[i][2])
print(ans)
for row in terrain:
print(''.join(row))
| 26 | 77 | 1,433,600 |
21180350
|
import sys
sys.setrecursionlimit(5000)
n, m, k = [int(x) for x in input().split()]
land = []
cc = 0
visit = [[0 for _ in range(m)] for _ in range(n)]
for i in range(n):
land.append(list(input()))
edge = False
def DFS(i, j, c):
global land, n, m, k, cc, visit, edge
if i <= 0 or j <= 0 or i >= n-1 or j >= m-1:
edge = True
return 0
if visit[i][j] != 0: return 0
if land[i][j] == '*': return 0
visit[i][j] = 1
cc += 1
if land[i + 1][j] == '.': DFS(i + 1, j, c + 1)
if land[i - 1][j] == '.': DFS(i - 1, j, c + 1)
if land[i][j + 1] == '.': DFS(i, j + 1, c + 1)
if land[i][j - 1] == '.': DFS(i, j - 1, c + 1)
def DFS2(i, j):
global land, n, m, k, cc
if i < 0 or j < 0 or i >= n or j >= m: return 0
if land[i][j] != '.': return 0
land[i][j] = '*'
DFS2(i+1,j)
DFS2(i - 1, j)
DFS2(i, j+1)
DFS2(i, j-1)
return 0
ccc = []
pos = []
for i in range(n):
for j in range(m):
cc = 0
edge = False
if visit[i][j] == 0:
DFS(i,j,0)
if edge is True: continue
if cc > 0:
ccc.append([cc, i, j])
ccc.sort()
red =len(ccc) - k
if red == 0:
print(0)
for i in range(n):
print(''.join(land[i]))
exit()
red_list = ccc[0:red]
p = [x[0] for x in red_list]
print(sum(p))
for i in range(len(red_list)):
DFS2(red_list[i][1], red_list[i][2])
for i in range(n):
print(''.join(land[i]))
|
Codeforces Round 375 (Div. 2)
|
CF
| 2,016 | 2 | 256 |
Lakes in Berland
|
The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cell is either land or water. The map is surrounded by the ocean.
Lakes are the maximal regions of water cells, connected by sides, which are not connected with the ocean. Formally, lake is a set of water cells, such that it's possible to get from any cell of the set to any other without leaving the set and moving only to cells adjacent by the side, none of them is located on the border of the rectangle, and it's impossible to add one more water cell to the set such that it will be connected with any other cell.
You task is to fill up with the earth the minimum number of water cells so that there will be exactly k lakes in Berland. Note that the initial number of lakes on the map is not less than k.
|
The first line of the input contains three integers n, m and k (1 ≤ n, m ≤ 50, 0 ≤ k ≤ 50) — the sizes of the map and the number of lakes which should be left on the map.
The next n lines contain m characters each — the description of the map. Each of the characters is either '.' (it means that the corresponding cell is water) or '*' (it means that the corresponding cell is land).
It is guaranteed that the map contain at least k lakes.
|
In the first line print the minimum number of cells which should be transformed from water to land.
In the next n lines print m symbols — the map after the changes. The format must strictly follow the format of the map in the input data (there is no need to print the size of the map). If there are several answers, print any of them.
It is guaranteed that the answer exists on the given data.
| null |
In the first example there are only two lakes — the first consists of the cells (2, 2) and (2, 3), the second consists of the cell (4, 3). It is profitable to cover the second lake because it is smaller. Pay attention that the area of water in the lower left corner is not a lake because this area share a border with the ocean.
|
[{"input": "5 4 1\n****\n*..*\n****\n**.*\n..**", "output": "1\n****\n*..*\n****\n****\n..**"}, {"input": "3 3 0\n***\n*.*\n***", "output": "1\n***\n***\n***"}]
| 1,600 |
["dfs and similar", "dsu", "graphs", "greedy", "implementation"]
| 26 |
[{"input": "5 4 1\r\n****\r\n*..*\r\n****\r\n**.*\r\n..**\r\n", "output": "1\r\n****\r\n*..*\r\n****\r\n****\r\n..**\r\n"}, {"input": "3 3 0\r\n***\r\n*.*\r\n***\r\n", "output": "1\r\n***\r\n***\r\n***\r\n"}, {"input": "3 5 1\r\n.**.*\r\n*.*.*\r\n***..\r\n", "output": "0\r\n.**.*\r\n*.*.*\r\n***..\r\n"}, {"input": "3 5 0\r\n.**.*\r\n*.*.*\r\n***..\r\n", "output": "1\r\n.**.*\r\n***.*\r\n***..\r\n"}, {"input": "3 50 7\r\n***.********.*********************.**********.****\r\n*...**..*.**.*.*.*.*.*.*.*..*.*.*.*.*.*.*.*.*.*..*\r\n****************.*.********.**********************\r\n", "output": "8\r\n***.********.*********************.**********.****\r\n*...**..****.***.*.*******..*******.*.*.*.*.*.*..*\r\n****************.*.********.**********************\r\n"}, {"input": "50 3 4\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n*.*\r\n***\r\n.**\r\n***\r\n..*\r\n***\r\n***\r\n*.*\r\n***\r\n*.*\r\n***\r\n***\r\n*.*\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n***\r\n*.*\r\n***\r\n***\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n***\r\n***\r\n", "output": "8\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n.**\r\n***\r\n..*\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n***\r\n*.*\r\n*.*\r\n*.*\r\n***\r\n***\r\n***\r\n***\r\n"}, {"input": "1 1 0\r\n.\r\n", "output": "0\r\n.\r\n"}, {"input": "1 1 0\r\n*\r\n", "output": "0\r\n*\r\n"}]
| false |
stdio
|
import sys
from sys import argv
def main():
input_path = argv[1]
output_path = argv[2]
submission_path = argv[3]
with open(input_path, 'r') as f:
input_lines = f.read().splitlines()
n, m, k = map(int, input_lines[0].split())
original_map = input_lines[1:n+1]
with open(output_path, 'r') as f:
ref_lines = f.read().splitlines()
if not ref_lines:
print(0)
return
ref_min = int(ref_lines[0])
with open(submission_path, 'r') as f:
sub_lines = f.read().splitlines()
if not sub_lines:
print(0)
return
# Check first line
try:
sub_min = int(sub_lines[0])
except:
print(0)
return
if sub_min != ref_min:
print(0)
return
# Check map lines
if len(sub_lines) -1 != n:
print(0)
return
sub_map = sub_lines[1:n+1]
for line in sub_map:
if len(line) != m:
print(0)
return
# Check original land preserved and count transformed
transformed = 0
for i in range(n):
orig_line = original_map[i]
sub_line = sub_map[i]
for j in range(m):
oc = orig_line[j]
sc = sub_line[j]
if oc == '*' and sc != '*':
print(0)
return
if oc == '.' and sc == '*':
transformed += 1
if transformed != sub_min:
print(0)
return
# Check number of lakes
visited = [[False]*m for _ in range(n)]
lakes = 0
dirs = [(-1,0), (1,0), (0,-1), (0,1)]
def is_border(x, y):
return x == 0 or x == n-1 or y ==0 or y == m-1
for i in range(n):
for j in range(m):
if sub_map[i][j] == '.' and not visited[i][j]:
stack = [(i,j)]
visited[i][j] = True
has_border = False
for x, y in stack:
if is_border(x, y):
has_border = True
for dx, dy in dirs:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < m and sub_map[nx][ny] == '.' and not visited[nx][ny]:
visited[nx][ny] = True
stack.append((nx, ny))
if not has_border:
lakes += 1
if lakes != k:
print(0)
return
print(1)
if __name__ == '__main__':
main()
| true |
474/D
|
474
|
D
|
Python 3
|
TESTS
| 2 | 124 | 3,276,800 |
190986436
|
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log,comb,perm,factorial
mod = int(1e9 + 7)
inf = int(1e20)
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
t,k=li()
quer=[]
maxq=0
for _ in range(t):
a,b=li()
if b>maxq:maxq=b
quer.append([a,b])
pre=[1]+[0]*maxq
for i in range(1,maxq+1):
pre[i]=(pre[i-1]+pre[i-k])%mod if i>=k else pre[i-1]
pre[0]=0
for i in range(1,maxq+1):
pre[i]=(pre[i]+pre[i-1])%mod
for a,b in quer:
print(pre[b]-pre[a-1])
| 43 | 171 | 10,240,000 |
198619907
|
import sys
input = lambda: sys.stdin.readline().rstrip()
MOD = 1000000007
t , k = map( int , input( ).split( ) )
f = [0] * 100005
f[0] = 1
for i in range( 1 , 100005 ) :
f[i] = f[i-1]
if i >= k : f[i] += f[i-k]
f[i] %= MOD
f[0] = 0
for i in range( 1 , 100001 ) :
f[i] += f[i-1]
f[i] %= MOD
for _ in range( t ) :
l , r = map( int , input( ).split( ) )
print( ( f[r] - f[l-1] + MOD ) % MOD )
|
Codeforces Round 271 (Div. 2)
|
CF
| 2,014 | 1.5 | 256 |
Flowers
|
We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.
But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.
Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (109 + 7).
|
Input contains several test cases.
The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.
The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.
|
Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).
| null |
- For K = 2 and length 1 Marmot can eat (R).
- For K = 2 and length 2 Marmot can eat (RR) and (WW).
- For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).
- For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).
|
[{"input": "3 2\n1 3\n2 3\n4 4", "output": "6\n5\n5"}]
| 1,700 |
["dp"]
| 43 |
[{"input": "3 2\r\n1 3\r\n2 3\r\n4 4\r\n", "output": "6\r\n5\r\n5\r\n"}, {"input": "1 1\r\n1 3\r\n", "output": "14\r\n"}, {"input": "1 2\r\n64329 79425\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
474/D
|
474
|
D
|
Python 3
|
TESTS
| 2 | 109 | 137,318,400 |
187213387
|
from collections import defaultdict, Counter, deque
from math import inf
from functools import lru_cache
from heapq import heappop, heappush
import sys
#input=sys.stdin.readline
Mod = 10**9 + 7
def solution():
t, k = map(int, input().split())
N = 10**5+1
dp = [0]*N
for i in range(1, N):
dp[i] += dp[i-1] + 1
if i - k >= 0:
dp[i] += dp[i-k] + 1
dp[i] %= Mod
for _ in range(t):
a,b = map(int, input().split())
print(dp[b] - dp[a - 1])
def main():
t = 1
#t = int(input())
for _ in range(t):
solution()
import sys
import threading
sys.setrecursionlimit(10**6)
threading.stack_size(1 << 27)
thread = threading.Thread(target=main)
thread.start(); thread.join()
#main()
| 43 | 186 | 10,752,000 |
216377371
|
import sys
input = sys.stdin.readline
M = 10**9+7
t, k = map(int, input().split())
d = [1]*k
for i in range(10**5):
d.append((d[-1]+d[-k])%M)
q = [0]
for i in range(1, 10**5+1):
q.append((q[-1]+d[i])%M)
for i in range(t):
a, b = map(int, input().split())
print((q[b]-q[a-1])%M)
|
Codeforces Round 271 (Div. 2)
|
CF
| 2,014 | 1.5 | 256 |
Flowers
|
We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.
But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.
Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (109 + 7).
|
Input contains several test cases.
The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.
The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.
|
Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).
| null |
- For K = 2 and length 1 Marmot can eat (R).
- For K = 2 and length 2 Marmot can eat (RR) and (WW).
- For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).
- For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).
|
[{"input": "3 2\n1 3\n2 3\n4 4", "output": "6\n5\n5"}]
| 1,700 |
["dp"]
| 43 |
[{"input": "3 2\r\n1 3\r\n2 3\r\n4 4\r\n", "output": "6\r\n5\r\n5\r\n"}, {"input": "1 1\r\n1 3\r\n", "output": "14\r\n"}, {"input": "1 2\r\n64329 79425\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
925/B
|
925
|
B
|
Python 3
|
TESTS
| 0 | 61 | 7,065,600 |
38069595
|
from math import ceil
def merge(arr, l, m, r):
n1 = m - l + 1
n2 = r- m
# create temp arrays
L = [0] * (n1)
R = [0] * (n2)
# Copy data to temp arrays L[] and R[]
for i in range(0 , n1):
L[i] = arr[l + i]
for j in range(0 , n2):
R[j] = arr[m + 1 + j]
# Merge the temp arrays back into arr[l..r]
i = 0 # Initial index of first subarray
j = 0 # Initial index of second subarray
k = l # Initial index of merged subarray
while i < n1 and j < n2 :
if L[i][0] <= R[j][0]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Copy the remaining elements of L[], if there
# are any
while i < n1:
arr[k] = L[i]
i += 1
k += 1
# Copy the remaining elements of R[], if there
# are any
while j < n2:
arr[k] = R[j]
j += 1
k += 1
# l is for left index and r is right index of the
# sub-array of arr to be sorted
def mergeSort(arr,l,r):
if l < r:
# Same as (l+r)/2, but avoids overflow for
# large l and h
m = (l+(r))//2
# Sort first and second halves
mergeSort(arr, l, m)
mergeSort(arr, m+1, r)
merge(arr, l, m, r)
n,x1,x2=[int(x) for x in input().split()]
a=[[int(x),i] for i,x in enumerate(input().split())]
mergeSort(a,0,n-1)
x1,x2=min(x1,x2),max(x1,x2)
c=0
for i in range(n):
if c==2:
break
elif ceil(x1/a[i][0])<n-i:
c=1
j=i+ceil(x1/a[i][0])
ans1=[a[x][1] for x in range(i,j)]
for k in range(j,n):
if c==2:
break
elif k+ceil(x2/a[k][0])<=n:
c=2
ans2=[a[x][1] for x in range(k,k+ceil(x2/a[k][0]))]
if c==2:
print("Yes")
print(len(ans1),len(ans2))
for x in ans1:
print(x,end='')
print()
for x in ans2:
print(x,end='')
print()
else:
print("No")
| 40 | 748 | 42,700,800 |
159295631
|
I = lambda:list(map(int, input().split()))
n,x1,x2 = I()
servers = I()
idx = list(range(n))
idx.sort(key = lambda i : servers[i])
def solve(a, b, rev):
j = len(idx)-1
while j >= 0 and (n-j)*servers[idx[j]] < b:
j -= 1
if j < 0:
return
i = j-1
while i >= 0 and (j-i)*servers[idx[i]] < a:
i -= 1
if i < 0:
return
print("Yes")
if not rev:
print(j-i, n-j)
print(" ".join(map(lambda x:str(x+1),idx[i:j])))
print(" ".join(map(lambda x:str(x+1), idx[j:])))
else:
print(n-j, j-i)
print(" ".join(map(lambda x:str(x+1), idx[j:])))
print(" ".join(map(lambda x:str(x+1),idx[i:j])))
exit()
solve(x1,x2,False)
solve(x2,x1,True)
print("No")
|
VK Cup 2018 - Round 3
|
CF
| 2,018 | 2 | 256 |
Resource Distribution
|
One department of some software company has $$$n$$$ servers of different specifications. Servers are indexed with consecutive integers from $$$1$$$ to $$$n$$$. Suppose that the specifications of the $$$j$$$-th server may be expressed with a single integer number $$$c_j$$$ of artificial resource units.
In order for production to work, it is needed to deploy two services $$$S_1$$$ and $$$S_2$$$ to process incoming requests using the servers of the department. Processing of incoming requests of service $$$S_i$$$ takes $$$x_i$$$ resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service $$$S_i$$$ is deployed using $$$k_i$$$ servers, then the load is divided equally between these servers and each server requires only $$$x_i / k_i$$$ (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services.
|
The first line contains three integers $$$n$$$, $$$x_1$$$, $$$x_2$$$ ($$$2 \leq n \leq 300\,000$$$, $$$1 \leq x_1, x_2 \leq 10^9$$$) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains $$$n$$$ space-separated integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \leq c_i \leq 10^9$$$) — the number of resource units provided by each of the servers.
|
If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers $$$k_1$$$ and $$$k_2$$$ ($$$1 \leq k_1, k_2 \leq n$$$) — the number of servers used for each of the services.
In the third line print $$$k_1$$$ integers, the indices of the servers that will be used for the first service.
In the fourth line print $$$k_2$$$ integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them.
| null |
In the first sample test each of the servers 1, 2 and 6 will will provide $$$8 / 3 = 2.(6)$$$ resource units and each of the servers 5, 4 will provide $$$16 / 2 = 8$$$ resource units.
In the second sample test the first server will provide $$$20$$$ resource units and each of the remaining servers will provide $$$32 / 3 = 10.(6)$$$ resource units.
|
[{"input": "6 8 16\n3 5 2 9 8 7", "output": "Yes\n3 2\n1 2 6\n5 4"}, {"input": "4 20 32\n21 11 11 12", "output": "Yes\n1 3\n1\n2 3 4"}, {"input": "4 11 32\n5 5 16 16", "output": "No"}, {"input": "5 12 20\n7 8 4 11 9", "output": "No"}]
| 1,700 |
["binary search", "implementation", "sortings"]
| 40 |
[{"input": "6 8 16\r\n3 5 2 9 8 7\r\n", "output": "Yes\r\n4 2\r\n3 1 2 6\r\n5 4\r\n"}, {"input": "4 20 32\r\n21 11 11 12\r\n", "output": "Yes\r\n1 3\r\n1\r\n2 3 4\r\n"}, {"input": "4 11 32\r\n5 5 16 16\r\n", "output": "No\r\n"}, {"input": "5 12 20\r\n7 8 4 11 9\r\n", "output": "No\r\n"}, {"input": "2 1 1\r\n1 1\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 1\r\n1 1000000\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 1\r\n1000000000 1000000000\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 2\r\n1 1\r\n", "output": "No\r\n"}, {"input": "15 250 200\r\n71 2 77 69 100 53 54 40 73 32 82 58 24 82 41\r\n", "output": "Yes\r\n11 3\r\n13 10 8 15 6 7 12 4 1 9 3\r\n11 14 5\r\n"}, {"input": "4 12 11\r\n4 4 6 11\r\n", "output": "Yes\r\n3 1\r\n1 2 3\r\n4\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, x1, x2 = map(int, f.readline().split())
c = list(map(int, f.readline().split()))
with open(output_path) as f:
ref_lines = [line.strip() for line in f]
with open(submission_path) as f:
sub_lines = [line.strip() for line in f]
non_empty_ref = [line for line in ref_lines if line]
non_empty_sub = [line for line in sub_lines if line]
if not non_empty_ref:
print(0)
return
ref_first = non_empty_ref[0]
if ref_first == 'No':
if len(non_empty_sub) == 1 and non_empty_sub[0] == 'No':
print(1)
else:
print(0)
return
else:
if len(non_empty_sub) != 4 or non_empty_sub[0] != 'Yes':
print(0)
return
try:
k1, k2 = map(int, non_empty_sub[1].split())
except:
print(0)
return
if k1 <= 0 or k2 <= 0 or k1 + k2 > n:
print(0)
return
try:
s1 = list(map(int, non_empty_sub[2].split()))
s2 = list(map(int, non_empty_sub[3].split()))
except:
print(0)
return
if len(s1) != k1 or len(s2) != k2:
print(0)
return
all_indices = set()
for s in s1:
if not (1 <= s <= n) or s in all_indices:
print(0)
return
all_indices.add(s)
for s in s2:
if not (1 <= s <= n) or s in all_indices:
print(0)
return
all_indices.add(s)
for s in s1:
if c[s-1] * k1 < x1:
print(0)
return
for s in s2:
if c[s-1] * k2 < x2:
print(0)
return
print(1)
return
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
908/E
|
908
|
E
|
PyPy 3
|
TESTS
| 4 | 93 | 23,756,800 |
33796723
|
from collections import defaultdict as di
MOD = int(1e9+7)
bells = di(int)
bells[0,0] = 1
K=60
for j in range(1,K):
bells[0,j] = bells[j-1,j-1]
for i in range(j):
bells[i+1,j] = (bells[i,j] + bells[i,j-1])%MOD
def bellman(n):
return bells[n-1,n-1]
m,n = [int(x) for x in input().split()]
Tlist = []
for _ in range(n):
Tlist.append(input())
numbs = []
for i in range(m):
numb = []
for j in range(n):
numb.append(Tlist[j][i])
numbs.append(int(''.join(numb),2))
eqsize = di(lambda:0)
for numb in numbs:
eqsize[numb]+=1
sets = []
for numb in eqsize:
sets.append(eqsize[numb])
parts = 1
for s in sets:
parts*=bellman(s)
parts%=MOD
print(parts)
| 37 | 77 | 6,451,200 |
33788749
|
import sys
#f = open('input', 'r')
f = sys.stdin
n,m = list(map(int, f.readline().split()))
s = [f.readline().strip() for _ in range(m)]
s = [list(x) for x in s]
d = {}
for k in zip(*s):
if k in d:
d[k] += 1
else:
d[k] = 1
dv = d.values()
got = [1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 382958538, 480142077, 864869230, 76801385, 742164233, 157873304, 812832668, 706900318, 546020311, 173093227, 759867260, 200033042, 40680577, 159122123, 665114805, 272358185, 365885605, 744733441, 692873095, 463056339, 828412002, 817756178, 366396447, 683685664, 681586780, 840750853, 683889724, 216039853, 954226396, 858087702, 540284076, 514254014, 647209774, 900185117, 348985796, 609459762, 781824096, 756600466, 654591160, 171792186, 748630189, 848074470, 75742990, 352494923, 278101098, 462072300, 334907097, 10474572, 495625635, 586051441, 159996073, 479379757, 707597945, 561063550, 974840072, 209152841, 906106015, 467465396, 82034048, 392794164, 700950185, 344807921, 475335490, 496881113, 358229039, 519104519, 784488542, 665151655, 307919717, 591199688, 692769253, 335414677, 884560880, 847374378, 791103220, 200350027, 485480275, 557337842, 434181960, 73976309, 792463021, 462067202, 677783523, 295755371, 435431099, 193120002, 513369106, 134597056, 143018012, 353529690, 591382993, 163160926, 287984994, 842145354, 703798750, 386436223, 618375990, 636477101, 536261496, 574800957, 34046224, 167415054, 961776342, 807141069, 218578541, 513967253, 460200768, 230725907, 239843627, 792763805, 368353031, 740982762, 126993201, 967654419, 588554507, 728057622, 239984996, 818342358, 882367644, 216705655, 267152940, 867213913, 330735015, 934583772, 59261085, 443816525, 568733052, 754405433, 244324432, 153903806, 292097031, 557968620, 311976469, 242994387, 773037141, 549999484, 243701468, 941251494, 7149216, 932327662, 456857477, 739044033, 645452229, 69273749, 304951367, 503353209, 243194926, 688663125, 239795364, 522687881, 121506491, 835250259, 159173149, 545801717, 19848500, 322507013, 106069527, 807985703, 290163328, 971751677, 238407093, 981758956, 301257197, 728003485, 817681690, 318332431, 864806329, 87958605, 929106232, 617996713, 519300437, 307911059, 137306007, 887695462, 633135243, 442387331, 730250437, 27723819, 80605394, 760335262, 822289356, 415861662, 558003999, 645049413, 347692428, 380668983, 897875109, 278111491, 106909073, 951914124, 374756177, 635211535, 286442394, 774619548, 756991257, 929298287, 923425488, 182439740, 266683608, 415378498, 728411148, 808161291, 436338820, 692451577, 228029692, 235546564, 895805974, 758052804, 700926159, 226442121, 579900323, 96916377, 243044550, 858703179, 30279679, 343764928, 100627558, 840734795, 291199760, 989808717, 370270411, 158336199, 393391701, 881731480, 507200370, 588418523, 340981140, 295449954, 683858381, 903859151, 866470542, 4959332, 237784075, 861373023, 950693473, 955867890, 400039807, 939877954, 124824910, 954530940, 204884446, 42218442, 234856311, 189836713, 179563650, 683193288, 929322036, 73574908, 943547254, 103031032, 141180580, 540183111, 680050153, 382916846, 948921599, 252835397, 199109508, 551172546, 700090782, 44999714, 970123610, 145637563, 33948107, 267648423, 504777414, 584919509, 212459491, 242880452, 351366578, 345323768, 285497541, 692868556, 706562675, 675626979, 620872182, 136458675, 971105139, 182064384, 948539342, 186406165, 706529481, 790490927, 888369436, 784409511, 835815713, 447895018, 17015606, 342727699, 321837918, 394134115, 563672582, 70390332, 61116103, 949269501, 833942074, 581389345, 570974405, 768179852, 765734098, 928340756, 541194960, 126833304, 427218334, 75800034, 100445725, 242810216, 330081440, 986329793, 298082322, 643160582, 505669854, 255287400, 403977567, 659185446, 596703087, 289443930, 478095124, 920175726, 205886838, 729278433, 535998256, 658801091, 606948240, 432632296, 552723022, 17794080, 234033713, 189986528, 444922724, 263196004, 846019724, 684703320, 895782046, 505050988, 44287113, 505335732, 436498414, 12098144, 714227851, 643983136, 647148160, 579243434, 951209063, 511291462, 426622734, 830870687, 949900312, 599926584, 633837711, 176405710, 913717356, 753535741, 874916804, 956692925, 220742732, 649500982, 584759931, 109573948, 937203173, 96881033, 305784835, 559854872, 894253854, 746366726, 951492991, 532579856, 822308583, 572042503, 397665465, 600979983, 914199453, 628402767, 594763006, 9791558, 451332658, 516069180, 651367831, 962708649, 687016963, 539878802, 107278296, 926059014, 371504543, 556987417, 447666745, 565595310, 778161513, 995461128, 121460302, 599892490, 242414970, 900391574, 362620950, 292857964, 495260535, 355054738, 176340034, 370047225, 509682533, 459314034, 40869728, 534741938, 788604648, 945028000, 701904601, 154924404, 695162652, 220536827, 615701976, 167761915, 833779942, 52430883, 368585637, 936409860, 654822736, 613850019, 941559844, 357840989, 218223326, 721900618, 171013438, 597980462, 193395922, 949112044, 859322955, 354602094, 807705992, 347609311, 451694117, 623122523, 252980054, 491785682, 13877214, 918727309, 750110421, 114981703, 174636266, 363160184, 781715298, 30575457, 862940854, 642129450, 34525888, 798422280, 792396821, 168367459, 344551406, 799847612, 626838494, 671596530, 167280197, 959000039, 614621296, 273560655, 8705247, 284372524, 940371542, 906010703, 582585495, 929449942, 308961449, 768816240, 674729787, 279648144, 286568146, 938661138, 536038536, 456529723, 18843013, 501518651, 457224675, 520694423, 938573228, 179014658, 169719825, 459657583, 302109678, 560375405, 556039265, 348713003, 957546568, 687116649, 3656313, 562760316, 716689588, 324677598, 570275686, 60738163, 996201577, 305457565, 38935942, 538451492, 228282207, 77975017, 389525459, 25000235, 152169430, 62331625, 618611219, 462328092, 106666757, 661839198, 177836427, 313546124, 392585017, 950280707, 551167559, 389204003, 77447456, 158414991, 766574847, 941433736, 363591676, 805565034, 312418363, 999641612, 122925536, 768845786, 608121932, 373163730, 783033644, 74564718, 894150080, 796617981, 274365270, 802488053, 947861187, 401960309, 143529635, 769621671, 249500752, 619408647, 849453216, 354838551, 69741157, 944781258, 135254314, 7413076, 416298064, 871313316, 343673168, 375656287, 868866230, 179060630, 399560227, 852555486, 987661859, 165863065, 12882359, 3688778, 380092596, 438366086, 720041886, 240796679, 588918084, 14802664, 17188673, 504951961, 842108931, 839289310, 256364811, 121095676, 164017670, 35340476, 875551801, 239615760, 262141182, 262741417, 456560451, 350350882, 777143297, 264469934, 807530935, 89546104, 246698645, 241166716, 125659016, 839103323, 418357064, 186866754, 179291319, 326054960, 172454707, 430532716, 558625738, 306201933, 61986384, 837357643, 575407529, 983555480, 13784333, 311989892, 153386582, 633092291, 722816631, 633510090, 551352594, 323601313, 248995449, 672011813, 612805937, 202743586, 215183002, 32688571, 38454892, 245790100, 451190956, 823199664, 12164578, 67389319, 584760532, 968838901, 307205626, 971537038, 836812364, 663878188, 468850566, 647599527, 839342879, 242347168, 169911213, 993779953, 251402771, 969281106, 416168275, 738337745, 8172262, 852101376, 879373674, 929752458, 452163141, 48347012, 500253327, 672444134, 406391337, 665852222, 499704706, 116418822, 67956495, 994761753, 808150613, 251453632, 543431315, 143101466, 381253760, 826943616, 763270983, 959511676, 323777679, 514214633, 669340157, 471626592, 557874503, 304789863, 116939617, 503636634, 660499296, 659726735, 273788323, 704107733, 718417780, 624033370, 355000823, 194537583, 760997582, 289828020, 778033293, 933152490, 910945024, 644565086, 434509630, 289427510, 502291783, 421699389, 159196930, 834667293, 313599675, 560298831, 812176354, 865521998, 126796474, 886921339, 937011401, 791993161, 583948688, 275408655, 665183437, 130342900, 699431744, 117550047, 460234251, 56770880, 306091228, 912949106, 626369877, 852501236, 241076503, 262988042, 737247015, 831044258, 475123008, 928949542, 332750699, 696284377, 689111142, 196862045, 570365577, 187156806, 451528865, 635110126, 385331290, 263486377, 200189955, 206842029, 457862597, 450522487, 818984909, 710634976, 461356455, 71985964, 781500456, 467334209, 46762760, 97663653, 870671823, 255977331, 79650379, 32876340, 636190780, 364339752, 597149326, 452604321, 748186407, 302032725, 779013981, 111971627, 175687535, 399961122, 451853028, 798326812, 902775588, 362836436, 498862780, 160000437, 629259604, 919729986, 5994845, 631476109, 371320167, 76164236, 448643023, 945220853, 111192011, 150577654, 827274836, 17668451, 938388515, 88566735, 27940328, 882026632, 712756966, 83642744, 873585716, 638987456, 405271094, 822216133, 345587303, 668558160, 314983205, 826678060, 563583341, 516998387, 77703032, 726563479, 155257773, 49705622, 891556456, 164127879, 842558039, 189099480, 956148983, 992226557, 671472701, 137476493, 871069222, 78241093, 728497057, 278888712, 332713952, 222597908, 235198692, 876216003, 167364357, 722341150, 519365334, 604855967, 834816722, 850786742, 416385106, 608404143, 311628039, 507077268, 571796589, 506398832, 305540948, 556971113, 444565912, 866477296, 411983920, 905854221, 901986568, 512703782, 684027511, 596294441, 916862272, 495347444, 802477106, 235968146, 257527513, 528476230, 969655767, 772044489, 682345813, 66418556, 603372280, 439233253, 278244332, 590581374, 353687769, 321352820, 245676729, 325255315, 91010070, 923699200, 837616604, 736177081, 528261400, 876353388, 339195128, 377206087, 769944080, 772953529, 123785293, 35984916, 461119619, 236140329, 884137913, 625494604, 791773064, 661436140, 308509072, 54134926, 279367618, 51918421, 149844467, 308119110, 948074070, 941738748, 890320056, 933243910, 430364344, 903312966, 574904506, 56353560, 861112413, 440392450, 937276559, 944662107, 599470900, 458887833, 962614595, 589151703, 997944986, 642961512, 63773929, 737273926, 110546606, 654813100, 374632916, 327432718, 307869727, 387738989, 133844439, 688886605, 989252194, 303514517, 79062408, 79381603, 941446109, 189307316, 728764788, 619946432, 359845738, 216171670, 690964059, 337106876, 762119224, 226624101, 401879891, 47069454, 41411521, 429556898, 188042667, 832342137, 770962364, 294422843, 991268380, 137519647, 903275202, 115040918, 521250780, 783585266, 98267582, 337193737, 717487549, 510794369, 206729333, 248526905, 412652544, 146948138, 103954760, 132289464, 938042429, 185735408, 640754677, 315573450, 956487685, 454822141, 783819416, 882547786, 976850791, 307258357, 929434429, 832158433, 334518103, 700273615, 734048238, 48618988, 693477108, 12561960, 598093056, 154072663, 174314067, 345548333, 479759833, 658594149, 282072153, 57970886, 905112877, 584117466, 472359245, 776860470, 324216896, 334199385, 321245477, 508188925, 521442872, 286692969, 245141864, 59342176, 896413224, 573301289, 869453643, 87399903, 60102262, 835514392, 493582549, 649986925, 576899388, 20454903, 271374500, 589229956, 505139242, 789538901, 243337905, 248443618, 39334644, 831631854, 541659849, 159802612, 524090232, 855135628, 542520502, 967119953, 597294058, 465231251]
MM = 10**9 + 7
ans = 1
for v in dv:
ans = ans*got[v-1]
ans = ans%MM
print(ans)
'''
t = [[0] * 1010 for _ in range(1010)]
t[1][1] = 1
for i in range(2,1001):
for j in range(1,i+1):
t[i][j] = t[i-1][j-1] + t[i-1][j]*j
t[i][j] = t[i][j] % MM
print([sum(t[i])%MM for i in range(1,1001)])
'''
|
Good Bye 2017
|
CF
| 2,017 | 2 | 256 |
New Year and Entity Enumeration
|
You are given an integer m.
Let M = 2m - 1.
You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.
A set of integers S is called "good" if the following hold.
1. If $$a \in S$$, then $$a \text{ XOR } M \in S$$.
2. If $$a,b\in S$$, then $$a \text{ AND } b \in S$$
3. $$T \subseteq S$$
4. All elements of S are less than or equal to M.
Here, $$\text{XOR}$$ and $$AND$$ refer to the bitwise XOR and bitwise AND operators, respectively.
Count the number of good sets S, modulo 109 + 7.
|
The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).
The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.
|
Print a single integer, the number of good sets modulo 109 + 7.
| null |
An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}.
|
[{"input": "5 3\n11010\n00101\n11000", "output": "4"}, {"input": "30 2\n010101010101010010101010101010\n110110110110110011011011011011", "output": "860616440"}]
| 2,500 |
["bitmasks", "combinatorics", "dp", "math"]
| 37 |
[{"input": "5 3\r\n11010\r\n00101\r\n11000\r\n", "output": "4\r\n"}, {"input": "30 2\r\n010101010101010010101010101010\r\n110110110110110011011011011011\r\n", "output": "860616440\r\n"}, {"input": "30 10\r\n001000000011000111000010010000\r\n000001100001010000000000000100\r\n000110100010100000000000101000\r\n110000010000000001000000000000\r\n100001000000000010010101000101\r\n001001000000000100000000110000\r\n000000010000100000001000000000\r\n001000010001000000001000000010\r\n000000110000000001001010000000\r\n000011001000000000010001000000\r\n", "output": "80\r\n"}]
| false |
stdio
| null | true |
158/C
|
158
|
C
|
Python 3
|
TESTS
| 0 | 92 | 0 |
230971727
|
vhod = input()
current_dir = '/'
last_input = ''
pwd_commands = []
counter = 0
for i in range(0, int(vhod)):
current_input = input()
if current_input == 'pwd':
#print(current_dir)
pwd_commands.append(current_dir)
elif 'cd' in current_input:
if '..' in current_input:
current_input = current_input.replace('cd ', '')
if current_input[0] == '/':
#print('zacne se z /')
if '/' not in current_input:
current_dir = current_dir.rsplit('/', 1)[0]
#print(current_dir)
else:
current_input_slashi = current_input.count('/')
#print(current_input_slashi)
razclenjen_ukaz = current_input.split('/')
del razclenjen_ukaz[0]
#print(razclenjen_ukaz)
counter = 0
for i in razclenjen_ukaz:
if razclenjen_ukaz[counter] == '..':
current_dir = current_dir.rsplit('/', 1)[0]
else:
current_dir = current_dir + '/' + razclenjen_ukaz[counter]
counter += 1
else:
current_input = '/' + current_input
if '/' not in current_input:
current_dir = current_dir.rsplit('/', 1)[0]
#print(current_dir)
else:
current_input_slashi = current_input.count('/')
#print(current_input_slashi)
razclenjen_ukaz = current_input.split('/')
del razclenjen_ukaz[0]
#print(razclenjen_ukaz)
counter = 0
for i in razclenjen_ukaz:
if razclenjen_ukaz[counter] == '..':
current_dir = current_dir.rsplit('/', 1)[0]
else:
current_dir = current_dir + '/' + razclenjen_ukaz[counter]
counter += 1
else:
last_input = current_input
current_input = current_input.replace('cd ', '')
current_dir += current_input
#print(current_input)
final_pwd_commands = [i[1:] if i.startswith('//') else i for i in pwd_commands]
for element in final_pwd_commands:
print(element)
| 29 | 124 | 0 |
18858108
|
class Shell:
def __init__(self):
self.wd = ['']
def cd(self, where: str):
glob = where.startswith('/')
directories = where.split('/')
if directories and not directories[0]:
directories.pop(0)
if glob:
self._reset()
for d in directories:
self._cd_one_dir(d)
def pwd(self):
if len(self.wd) == 1:
print('/')
else:
print('/', '/'.join(self.wd[1:]), '/', sep='')
def _reset(self):
self.wd = ['']
def _cd_one_dir(self, d):
if d == '..':
self.wd.pop()
else:
self.wd.append(d)
n = int(input())
sh = Shell()
for _ in range(n):
cmd = input()
if cmd.startswith('cd'):
sh.cd(cmd.split()[-1])
else:
sh.pwd()
|
VK Cup 2012 Qualification Round 1
|
CF
| 2,012 | 3 | 256 |
Cd and pwd commands
|
Vasya is writing an operating system shell, and it should have commands for working with directories. To begin with, he decided to go with just two commands: cd (change the current directory) and pwd (display the current directory).
Directories in Vasya's operating system form a traditional hierarchical tree structure. There is a single root directory, denoted by the slash character "/". Every other directory has a name — a non-empty string consisting of lowercase Latin letters. Each directory (except for the root) has a parent directory — the one that contains the given directory. It is denoted as "..".
The command cd takes a single parameter, which is a path in the file system. The command changes the current directory to the directory specified by the path. The path consists of the names of directories separated by slashes. The name of the directory can be "..", which means a step up to the parent directory. «..» can be used in any place of the path, maybe several times. If the path begins with a slash, it is considered to be an absolute path, that is, the directory changes to the specified one, starting from the root. If the parameter begins with a directory name (or ".."), it is considered to be a relative path, that is, the directory changes to the specified directory, starting from the current one.
The command pwd should display the absolute path to the current directory. This path must not contain "..".
Initially, the current directory is the root. All directories mentioned explicitly or passed indirectly within any command cd are considered to exist. It is guaranteed that there is no attempt of transition to the parent directory of the root directory.
|
The first line of the input data contains the single integer n (1 ≤ n ≤ 50) — the number of commands.
Then follow n lines, each contains one command. Each of these lines contains either command pwd, or command cd, followed by a space-separated non-empty parameter.
The command parameter cd only contains lower case Latin letters, slashes and dots, two slashes cannot go consecutively, dots occur only as the name of a parent pseudo-directory. The command parameter cd does not end with a slash, except when it is the only symbol that points to the root directory. The command parameter has a length from 1 to 200 characters, inclusive.
Directories in the file system can have the same names.
|
For each command pwd you should print the full absolute path of the given directory, ending with a slash. It should start with a slash and contain the list of slash-separated directories in the order of being nested from the root to the current folder. It should contain no dots.
| null | null |
[{"input": "7\npwd\ncd /home/vasya\npwd\ncd ..\npwd\ncd vasya/../petya\npwd", "output": "/\n/home/vasya/\n/home/\n/home/petya/"}, {"input": "4\ncd /a/b\npwd\ncd ../a/b\npwd", "output": "/a/b/\n/a/a/b/"}]
| 1,400 |
["*special", "data structures", "implementation"]
| 29 |
[{"input": "7\r\npwd\r\ncd /home/vasya\r\npwd\r\ncd ..\r\npwd\r\ncd vasya/../petya\r\npwd\r\n", "output": "/\r\n/home/vasya/\r\n/home/\r\n/home/petya/\r\n"}, {"input": "4\r\ncd /a/b\r\npwd\r\ncd ../a/b\r\npwd\r\n", "output": "/a/b/\r\n/a/a/b/\r\n"}, {"input": "1\r\npwd\r\n", "output": "/\r\n"}, {"input": "2\r\ncd /test/../test/../test/../test/../a/b/c/..\r\npwd\r\n", "output": "/a/b/\r\n"}, {"input": "9\r\ncd test\r\npwd\r\ncd ..\r\ncd /test\r\npwd\r\ncd ..\r\npwd\r\ncd test/test\r\npwd\r\n", "output": "/test/\r\n/test/\r\n/\r\n/test/test/\r\n"}, {"input": "6\r\ncd a/a/b/b\r\npwd\r\ncd ../..\r\npwd\r\ncd ..\r\npwd\r\n", "output": "/a/a/b/b/\r\n/a/a/\r\n/a/\r\n"}, {"input": "5\r\npwd\r\ncd /xgztbykka\r\npwd\r\ncd /gia/kxfls\r\npwd\r\n", "output": "/\r\n/xgztbykka/\r\n/gia/kxfls/\r\n"}, {"input": "17\r\npwd\r\ncd denwxe/../jhj/rxit/ie\r\npwd\r\ncd /tmyuylvul/qev/ezqit\r\npwd\r\ncd ../gxsfgyuspg/irleht\r\npwd\r\ncd wq/pqyz/tjotsmdzja\r\npwd\r\ncd ia/hs/../u/nemol/ffhf\r\npwd\r\ncd /lrdm/mvwxwb/llib\r\npwd\r\ncd /lmhu/wloover/rqd\r\npwd\r\ncd lkwabdw/../wrqn/x/../ien\r\npwd\r\n", "output": "/\r\n/jhj/rxit/ie/\r\n/tmyuylvul/qev/ezqit/\r\n/tmyuylvul/qev/gxsfgyuspg/irleht/\r\n/tmyuylvul/qev/gxsfgyuspg/irleht/wq/pqyz/tjotsmdzja/\r\n/tmyuylvul/qev/gxsfgyuspg/irleht/wq/pqyz/tjotsmdzja/ia/u/nemol/ffhf/\r\n/lrdm/mvwxwb/llib/\r\n/lmhu/wloover/rqd/\r\n/lmhu/wloover/rqd/wrqn/ien/\r\n"}, {"input": "5\r\ncd /xgztbykka\r\ncd /gia/kxfls\r\ncd /kiaxt/hcx\r\ncd /ufzoiv\r\npwd\r\n", "output": "/ufzoiv/\r\n"}, {"input": "17\r\ncd denwxe/../jhj/rxit/ie\r\ncd /tmyuylvul/qev/ezqit\r\ncd ../gxsfgyuspg/irleht\r\ncd wq/pqyz/tjotsmdzja\r\ncd ia/hs/../u/nemol/ffhf\r\ncd /lrdm/mvwxwb/llib\r\ncd /lmhu/wloover/rqd\r\ncd lkwabdw/../wrqn/x/../ien\r\ncd /rqljh/qyovqhiry/q\r\ncd /d/aargbeotxm/ovv\r\ncd /jaagwy/../xry/w/zdvx\r\ncd /nblqgcua/s/s/c/dgg\r\ncd /jktwitbkgj/ee/../../q\r\ncd wkx/jyphtd/h/../ygwc\r\ncd areyd/regf/ogvklan\r\ncd /wrbi/vbxefrd/jimis\r\npwd\r\n", "output": "/wrbi/vbxefrd/jimis/\r\n"}, {"input": "5\r\npwd\r\ncd ztb/kag\r\npwd\r\npwd\r\npwd\r\n", "output": "/\r\n/ztb/kag/\r\n/ztb/kag/\r\n/ztb/kag/\r\n"}, {"input": "17\r\ncd en/ebhjhjzlrx/pmieg\r\ncd uylvulohqe/wezq/oarx\r\npwd\r\ncd yus/fsi/ehtrs/../vjpq\r\ncd tjotsmdzja/diand/dqb\r\ncd emolqs/hff/rdmy/vw\r\ncd ../llibd/mhuos/oove\r\ncd /rqdqj/kwabd/nj/qng\r\npwd\r\ncd /yie/lrq/hmxq/vqhi\r\ncd qma/../aargbeotxm/ov\r\ncd /jaagwy/../xry/w/zdvx\r\npwd\r\ncd ../gcuagas/s/c/dggmz\r\npwd\r\npwd\r\ncd bkgjifee/../../../vfwkxjoj\r\n", "output": "/en/ebhjhjzlrx/pmieg/uylvulohqe/wezq/oarx/\r\n/rqdqj/kwabd/nj/qng/\r\n/xry/w/zdvx/\r\n/xry/w/gcuagas/s/c/dggmz/\r\n/xry/w/gcuagas/s/c/dggmz/\r\n"}, {"input": "50\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\n", "output": "/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n"}, {"input": "1\r\ncd /\r\n", "output": ""}, {"input": "11\r\npwd\r\ncd /home/vasya\r\npwd\r\ncd ..\r\npwd\r\ncd vasya/../../petya\r\npwd\r\ncd /a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a\r\npwd\r\ncd ..\r\npwd\r\n", "output": "/\r\n/home/vasya/\r\n/home/\r\n/petya/\r\n/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/\r\n/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/\r\n"}]
| false |
stdio
| null | true |
1006/E
|
1006
|
E
|
Python 3
|
TESTS
| 2 | 2,979 | 15,667,200 |
166707942
|
import sys
import threading
sys.setrecursionlimit(10 ** 6)
n , q = list(map(int,input().split()))
graph = [[] for i in range(n+1)]
a = list(map(int,input().split()))
a.insert(0 , 0)
for i in range(2,n+1):
graph[i].append(a[i-1])
graph[a[i-1]].append(i)
size = [0 for i in range(n+1)]
out = []
vis = [0 for i in range(n+1)]
def dfs(u , e):
size[u] = 1
for v in graph[u]:
if v == e:
continue
dfs(v , u)
size[u] += size[v]
def dfs2(u):
vis[u] = 1
out.append(u)
for v in graph[u]:
if vis[v] == 0:
dfs2(v)
def copy_ninja():
dfs(1 , 0)
dfs2(1)
for i in range(q):
u , k = list(map(int,input().split()))
if size[u] < k:
print(-1)
else:
print(out[out.index(u) + (k-1)])
threading.stack_size(10 ** 6)
t = threading.Thread(target = copy_ninja)
t.start()
t.join()
| 31 | 608 | 80,281,600 |
225580588
|
import random
import sys
# sys.setrecursionlimit(10**8)设置最大递归次数
class FastIO:
def __init__(self):
self.random_seed = random.randint(0, 10 ** 9 + 7)
return
@staticmethod
def read_int():
return int(sys.stdin.readline().strip())
@staticmethod
def read_float():
return float(sys.stdin.readline().strip())
@staticmethod
def read_list_ints():
return list(map(int, sys.stdin.readline().strip().split()))
@staticmethod
def read_list_floats():
return list(map(float, sys.stdin.readline().strip().split()))
@staticmethod
def read_list_ints_minus_one():
return list(map(lambda x: int(x) - 1, sys.stdin.readline().strip().split()))
@staticmethod
def read_str():
return sys.stdin.readline().strip()
@staticmethod
def read_list_strs():
return sys.stdin.readline().strip().split()
@staticmethod
def read_list_str():
return list(sys.stdin.readline().strip())
@staticmethod
def st(x):
return print(x)
@staticmethod
def lst(x):
return print(*x)
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def ceil(a, b):
return a // b + int(a % b != 0)
def hash_num(self, x):
return x ^ self.random_seed
@staticmethod
def accumulate(nums):
n = len(nums)
pre = [0] * (n + 1)
for i in range(n):
pre[i + 1] = pre[i] + nums[i]
return pre
def inter_ask(self, lst):
# CF交互题输出询问并读取结果
self.st(lst)
sys.stdout.flush()
res = self.read_int()
# 记得任何一个输出之后都要 sys.stdout.flush() 刷新
return res
def inter_out(self, lst):
# CF交互题输出最终答案
self.st(lst)
sys.stdout.flush()
return
class DfsEulerOrder:
def __init__(self, dct, root=0):
# 模板：dfs序与欧拉序，支持在线区间修改树上边，并且实时查询任意两点树上距离
n = len(dct)
self.start = [-1] * n # 每个原始节点的dfs序号开始点也是node_to_order
self.end = [-1] * n # 每个原始节点的dfs序号结束点
self.parent = [-1] * n # 每个原始节点的父节点
self.order_to_node = [-1] * n # 每个dfs序号对应的原始节点编号
self.build(dct, root)
return
def build(self, dct, root):
# 生成dfs序与欧拉序相关信息
order = 0
stack = [[root, -1]]
while stack:
i, fa = stack.pop()
if i >= 0:
self.start[i] = order
self.order_to_node[order] = i
self.end[i] = order
order += 1
stack.append([~i, fa])
for j in dct[i]:
if j != fa: # 注意访问顺序可以进行调整，比如字典序正序逆序
self.parent[j] = i
stack.append([j, i])
else:
i = ~i
if self.parent[i] != -1:
self.end[self.parent[i]] = self.end[i]
return
class Solution:
def __init__(self):
return
@staticmethod
def main(ac=FastIO()):
n, q = ac.read_list_ints()
dct = [[] for _ in range(n)]
p = ac.read_list_ints_minus_one()
for i in range(n-1):
dct[p[i]].append(i+1)
for i in range(n):
dct[i].reverse()
dfs = DfsEulerOrder(dct)
for _ in range(q):
u, k = ac.read_list_ints()
u -= 1
x = dfs.start[u]
if n-x < k or dfs.start[dfs.order_to_node[x+k-1]] > dfs.end[u]:
ac.st(-1)
else:
ac.st(dfs.order_to_node[x+k-1] + 1)
return
Solution().main()
|
Codeforces Round 498 (Div. 3)
|
ICPC
| 2,018 | 3 | 256 |
Military Problem
|
In this problem you will have to help Berland army with organizing their command delivery system.
There are $$$n$$$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $$$a$$$ is the direct superior of officer $$$b$$$, then we also can say that officer $$$b$$$ is a direct subordinate of officer $$$a$$$.
Officer $$$x$$$ is considered to be a subordinate (direct or indirect) of officer $$$y$$$ if one of the following conditions holds:
- officer $$$y$$$ is the direct superior of officer $$$x$$$;
- the direct superior of officer $$$x$$$ is a subordinate of officer $$$y$$$.
For example, on the picture below the subordinates of the officer $$$3$$$ are: $$$5, 6, 7, 8, 9$$$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $$$n$$$ vertices, in which vertex $$$u$$$ corresponds to officer $$$u$$$. The parent of vertex $$$u$$$ corresponds to the direct superior of officer $$$u$$$. The root (which has index $$$1$$$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $$$q$$$ queries, the $$$i$$$-th query is given as $$$(u_i, k_i)$$$, where $$$u_i$$$ is some officer, and $$$k_i$$$ is a positive integer.
To process the $$$i$$$-th query imagine how a command from $$$u_i$$$ spreads to the subordinates of $$$u_i$$$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $$$a$$$ and he spreads a command. Officer $$$a$$$ chooses $$$b$$$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $$$a$$$ chooses the one having minimal index. Officer $$$a$$$ gives a command to officer $$$b$$$. Afterwards, $$$b$$$ uses exactly the same algorithm to spread the command to its subtree. After $$$b$$$ finishes spreading the command, officer $$$a$$$ chooses the next direct subordinate again (using the same strategy). When officer $$$a$$$ cannot choose any direct subordinate who still hasn't received this command, officer $$$a$$$ finishes spreading the command.
Let's look at the following example:
If officer $$$1$$$ spreads a command, officers receive it in the following order: $$$[1, 2, 3, 5 ,6, 8, 7, 9, 4]$$$.
If officer $$$3$$$ spreads a command, officers receive it in the following order: $$$[3, 5, 6, 8, 7, 9]$$$.
If officer $$$7$$$ spreads a command, officers receive it in the following order: $$$[7, 9]$$$.
If officer $$$9$$$ spreads a command, officers receive it in the following order: $$$[9]$$$.
To answer the $$$i$$$-th query $$$(u_i, k_i)$$$, construct a sequence which describes the order in which officers will receive the command if the $$$u_i$$$-th officer spreads it. Return the $$$k_i$$$-th element of the constructed list or -1 if there are fewer than $$$k_i$$$ elements in it.
You should process queries independently. A query doesn't affect the following queries.
|
The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$$$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $$$n - 1$$$ integers $$$p_2, p_3, \dots, p_n$$$ ($$$1 \le p_i < i$$$), where $$$p_i$$$ is the index of the direct superior of the officer having the index $$$i$$$. The commander has index $$$1$$$ and doesn't have any superiors.
The next $$$q$$$ lines describe the queries. The $$$i$$$-th query is given as a pair ($$$u_i, k_i$$$) ($$$1 \le u_i, k_i \le n$$$), where $$$u_i$$$ is the index of the officer which starts spreading a command, and $$$k_i$$$ is the index of the required officer in the command spreading sequence.
|
Print $$$q$$$ numbers, where the $$$i$$$-th number is the officer at the position $$$k_i$$$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $$$u_i$$$. Print "-1" if the number of officers which receive the command is less than $$$k_i$$$.
You should process queries independently. They do not affect each other.
| null | null |
[{"input": "9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9", "output": "3\n6\n8\n-1\n9\n4"}]
| 1,600 |
["dfs and similar", "graphs", "trees"]
| 31 |
[{"input": "9 6\r\n1 1 1 3 5 3 5 7\r\n3 1\r\n1 5\r\n3 4\r\n7 3\r\n1 8\r\n1 9\r\n", "output": "3\r\n6\r\n8\r\n-1\r\n9\r\n4\r\n"}, {"input": "2 1\r\n1\r\n1 1\r\n", "output": "1\r\n"}, {"input": "13 12\r\n1 1 1 1 1 1 1 1 1 1 1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n"}]
| false |
stdio
| null | true |
898/C
|
898
|
C
|
PyPy 3
|
TESTS
| 2 | 93 | 0 |
120273289
|
n = int(input())
d = {}
for i in range(n):
l = list(map(str, input().split()))
if l[0] not in d:
d[l[0]] = list(set(l[2:]))
else:
d[l[0]] = list(d[l[0]] + list(set(l[2:])))
for i in d:
a = d[i]
ans = []
q = len(a)
for j in range(q):
c = 1
k = len(a[j])
for z in range(q):
if j != z and a[j] == a[z][-k: ]:
c = 0
if c:
ans.append(a[j])
d[i] = ans
print(len(d))
for i in d:
print(i, len(d[i]), *d[i])
| 59 | 62 | 5,529,600 |
33301284
|
n = int(input())
D = {}
for _ in range(n):
name, amount, *numbers = input().split()
if name in D:
D[name] += numbers
else:
D[name] = numbers
#for name, numbers in D.items():
# print(name, numbers)
for name, numbers in D.items():
ns = sorted(numbers, key=len)[::-1]
nn = [ns[0]]
for i in range(1, len(ns)):
if not any([j[-len(ns[i]):] == ns[i] for j in nn]):
nn += [ns[i]]
D[name] = nn
print(len(D))
for name, numbers in D.items():
print(name, len(numbers), ' '.join(numbers))
|
Codeforces Round 451 (Div. 2)
|
CF
| 2,017 | 2 | 256 |
Phone Numbers
|
Vasya has several phone books, in which he recorded the telephone numbers of his friends. Each of his friends can have one or several phone numbers.
Vasya decided to organize information about the phone numbers of friends. You will be given n strings — all entries from Vasya's phone books. Each entry starts with a friend's name. Then follows the number of phone numbers in the current entry, and then the phone numbers themselves. It is possible that several identical phones are recorded in the same record.
Vasya also believes that if the phone number a is a suffix of the phone number b (that is, the number b ends up with a), and both numbers are written by Vasya as the phone numbers of the same person, then a is recorded without the city code and it should not be taken into account.
The task is to print organized information about the phone numbers of Vasya's friends. It is possible that two different people have the same number. If one person has two numbers x and y, and x is a suffix of y (that is, y ends in x), then you shouldn't print number x. If the number of a friend in the Vasya's phone books is recorded several times in the same format, it is necessary to take it into account exactly once.
Read the examples to understand statement and format of the output better.
|
First line contains the integer n (1 ≤ n ≤ 20) — number of entries in Vasya's phone books.
The following n lines are followed by descriptions of the records in the format described in statement. Names of Vasya's friends are non-empty strings whose length does not exceed 10. They consists only of lowercase English letters. Number of phone numbers in one entry is not less than 1 is not more than 10. The telephone numbers consist of digits only. If you represent a phone number as a string, then its length will be in range from 1 to 10. Phone numbers can contain leading zeros.
|
Print out the ordered information about the phone numbers of Vasya's friends. First output m — number of friends that are found in Vasya's phone books.
The following m lines must contain entries in the following format "name number_of_phone_numbers phone_numbers". Phone numbers should be separated by a space. Each record must contain all the phone numbers of current friend.
Entries can be displayed in arbitrary order, phone numbers for one record can also be printed in arbitrary order.
| null | null |
[{"input": "2\nivan 1 00123\nmasha 1 00123", "output": "2\nmasha 1 00123\nivan 1 00123"}, {"input": "3\nkarl 2 612 12\npetr 1 12\nkatya 1 612", "output": "3\nkatya 1 612\npetr 1 12\nkarl 1 612"}, {"input": "4\nivan 3 123 123 456\nivan 2 456 456\nivan 8 789 3 23 6 56 9 89 2\ndasha 2 23 789", "output": "2\ndasha 2 23 789\nivan 4 789 123 2 456"}]
| 1,400 |
["implementation", "strings"]
| 59 |
[{"input": "2\r\nivan 1 00123\r\nmasha 1 00123\r\n", "output": "2\r\nmasha 1 00123 \r\nivan 1 00123 \r\n"}, {"input": "3\r\nkarl 2 612 12\r\npetr 1 12\r\nkatya 1 612\r\n", "output": "3\r\nkatya 1 612 \r\npetr 1 12 \r\nkarl 1 612 \r\n"}, {"input": "4\r\nivan 3 123 123 456\r\nivan 2 456 456\r\nivan 8 789 3 23 6 56 9 89 2\r\ndasha 2 23 789\r\n", "output": "2\r\ndasha 2 789 23 \r\nivan 4 789 123 456 2 \r\n"}, {"input": "20\r\nnxj 6 7 6 6 7 7 7\r\nnxj 10 8 5 1 7 6 1 0 7 0 6\r\nnxj 2 6 5\r\nnxj 10 6 7 6 6 5 8 3 6 6 8\r\nnxj 10 6 1 7 6 7 1 8 7 8 6\r\nnxj 10 8 5 8 6 5 6 1 9 6 3\r\nnxj 10 8 1 6 4 8 0 4 6 0 1\r\nnxj 9 2 6 6 8 1 1 3 6 6\r\nnxj 10 8 9 0 9 1 3 2 3 2 3\r\nnxj 6 6 7 0 8 1 2\r\nnxj 7 7 7 8 1 3 6 9\r\nnxj 10 2 7 0 1 5 1 9 1 2 6\r\nnxj 6 9 6 9 6 3 7\r\nnxj 9 0 1 7 8 2 6 6 5 6\r\nnxj 4 0 2 3 7\r\nnxj 10 0 4 0 6 1 1 8 8 4 7\r\nnxj 8 4 6 2 6 6 1 2 7\r\nnxj 10 5 3 4 2 1 0 7 0 7 6\r\nnxj 10 9 6 0 6 1 6 2 1 9 6\r\nnxj 4 2 9 0 1\r\n", "output": "1\r\nnxj 10 0 3 2 1 4 7 8 5 9 6 \r\n"}, {"input": "20\r\nl 6 02 02 2 02 02 2\r\nl 8 8 8 8 2 62 13 31 3\r\ne 9 0 91 0 0 60 91 60 2 44\r\ne 9 69 2 1 44 2 91 66 1 70\r\nl 9 7 27 27 3 1 3 7 80 81\r\nl 9 2 1 13 7 2 10 02 3 92\r\ne 9 0 15 3 5 5 15 91 09 44\r\nl 7 2 50 4 5 98 31 98\r\nl 3 26 7 3\r\ne 6 7 5 0 62 65 91\r\nl 8 80 0 4 0 2 2 0 13\r\nl 9 19 13 02 2 1 4 19 26 02\r\nl 10 7 39 7 9 22 22 26 2 90 4\r\ne 7 65 2 36 0 34 57 9\r\ne 8 13 02 09 91 73 5 36 62\r\nl 9 75 0 10 8 76 7 82 8 34\r\nl 7 34 0 19 80 6 4 7\r\ne 5 4 2 5 7 2\r\ne 7 4 02 69 7 07 20 2\r\nl 4 8 2 1 63\r\n", "output": "2\r\ne 18 15 62 07 70 91 57 02 66 65 69 09 13 20 44 73 34 60 36 \r\nl 21 27 50 22 63 75 19 26 90 02 92 62 31 10 76 82 80 98 81 39 34 13 \r\n"}, {"input": "20\r\no 10 6 6 97 45 6 6 6 6 5 6\r\nl 8 5 5 5 19 59 5 8 5\r\nj 9 2 30 58 2 2 1 0 30 4\r\nc 10 1 1 7 51 7 7 51 1 1 1\r\no 9 7 97 87 70 2 19 2 14 6\r\ne 6 26 6 6 6 26 5\r\ng 9 3 3 3 3 3 78 69 8 9\r\nl 8 8 01 1 5 8 41 72 3\r\nz 10 1 2 2 2 9 1 9 1 6 7\r\ng 8 7 78 05 36 7 3 67 9\r\no 5 6 9 9 7 7\r\ne 10 30 2 1 1 2 5 04 0 6 6\r\ne 9 30 30 2 2 0 26 30 79 8\r\nt 10 2 2 9 29 7 7 7 9 2 9\r\nc 7 7 51 1 31 2 7 4\r\nc 9 83 1 6 78 94 74 54 8 32\r\ng 8 4 1 01 9 39 28 6 6\r\nt 7 9 2 01 4 4 9 58\r\nj 5 0 1 58 02 4\r\nw 10 80 0 91 91 06 91 9 9 27 7\r\n", "output": "9\r\nw 5 91 27 06 9 80 \r\nt 6 7 58 4 29 2 01 \r\ne 8 8 79 5 30 2 26 04 1 \r\nl 8 72 3 19 59 41 01 5 8 \r\nj 5 4 30 58 1 02 \r\nz 5 6 1 9 2 7 \r\ng 10 05 39 4 3 36 01 67 69 28 78 \r\no 8 14 87 97 6 19 70 45 2 \r\nc 10 83 51 31 54 74 32 7 94 78 6 \r\n"}, {"input": "1\r\negew 5 3 123 23 1234 134\r\n", "output": "1\r\negew 3 123 134 1234 \r\n"}]
| false |
stdio
|
import sys
from collections import defaultdict
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input and build friends data
friends = defaultdict(set)
with open(input_path, 'r') as f:
n = int(f.readline())
for _ in range(n):
parts = f.readline().strip().split()
name = parts[0]
numbers = parts[2:]
friends[name].update(numbers) # Add all numbers, duplicates in same line are handled via set
# Read submission output
submission_friends = set()
try:
with open(submission_path, 'r') as f:
lines = f.read().splitlines()
except:
print(0)
return
if not lines:
print(0)
return
# Check m line
try:
m_line = lines[0]
m = int(m_line)
except:
print(0)
return
if m != len(friends):
print(0)
return
submission_lines = lines[1:]
if len(submission_lines) != m:
print(0)
return
for line in submission_lines:
parts = line.strip().split()
if len(parts) < 2:
print(0)
return
name = parts[0]
if name not in friends:
print(0)
return
submission_friends.add(name)
try:
k = int(parts[1])
except:
print(0)
return
numbers = parts[2:]
if len(numbers) != k:
print(0)
return
# Check for duplicates in submission's numbers
if len(set(numbers)) != k:
print(0)
return
# Check all numbers are present in friends' merged set
merged_numbers = friends[name]
for num in numbers:
if num not in merged_numbers:
print(0)
return
# Check no two numbers in submission are suffix of each other
for i in range(len(numbers)):
for j in range(i+1, len(numbers)):
a = numbers[i]
b = numbers[j]
if (len(a) >= len(b) and a.endswith(b)) or (len(b) >= len(a) and b.endswith(a)):
print(0)
return
# Check every merged number is either in submission or is a suffix of some submission number
for merged_num in merged_numbers:
if merged_num in numbers:
continue
found = False
for num in numbers:
if len(merged_num) <= len(num) and num.endswith(merged_num):
found = True
break
if not found:
print(0)
return
# Check all friends are present in submission
if submission_friends != set(friends.keys()):
print(0)
return
# All checks passed
print(100)
if __name__ == "__main__":
main()
| true |
400/A
|
400
|
A
|
PyPy 3
|
TESTS
| 1 | 92 | 0 |
216130444
|
t=int(input())
for i in range(t):
f=input()
if(f=="XXXXXXXXXXXX"):
print("6 1x12 2x6 3x4 4x3 6x2 12x1")
elif(f=="OOOOOOOOOOOO"):
print("0")
else:
k=list(f)
nb=0
st=""
if(k[11]=="X"):
st+="1x12 "
nb+=1
if(k[5]=="X" and k[11]=="X"):
st+="2x6 "
nb += 1
if(k[3]=="X" and k[7]=="X" and k[11]=="X"):
st+="3x4 "
nb += 1
if(k[2]=="X" and k[5]=="X" and k[8]=="X" and k[11]=="X"):
st+="4x3 "
nb += 1
if(k[1]=="X" and k[3]=="X" and k[5]=="X" and k[7]=="X" and k[9]=="X" and k[11]=="X"):
st+="6x2 "
nb += 1
print(nb,st)
| 44 | 46 | 0 |
156023101
|
import sys
input = sys.stdin.readline
for _ in range(int(input())):
s = input()[:-1]
d = []
l = [1,2,3,4,6,12]
for j in l:
k = 12//j
for i in range(k):
if s[i::k] == 'X' * j:
d += [str(j)+'x'+str(k)]
break
print(len(d), ' '.join(d))
|
Codeforces Round 234 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Inna and Choose Options
|
There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusual upgrade of this game. The rules of the game are given below:
There is one person playing the game. Before the beginning of the game he puts 12 cards in a row on the table. Each card contains a character: "X" or "O". Then the player chooses two positive integers a and b (a·b = 12), after that he makes a table of size a × b from the cards he put on the table as follows: the first b cards form the first row of the table, the second b cards form the second row of the table and so on, the last b cards form the last (number a) row of the table. The player wins if some column of the table contain characters "X" on all cards. Otherwise, the player loses.
Inna has already put 12 cards on the table in a row. But unfortunately, she doesn't know what numbers a and b to choose. Help her win the game: print to her all the possible ways of numbers a, b that she can choose and win.
|
The first line of the input contains integer t (1 ≤ t ≤ 100). This value shows the number of sets of test data in the input. Next follows the description of each of the t tests on a separate line.
The description of each test is a string consisting of 12 characters, each character is either "X", or "O". The i-th character of the string shows the character that is written on the i-th card from the start.
|
For each test, print the answer to the test on a single line. The first number in the line must represent the number of distinct ways to choose the pair a, b. Next, print on this line the pairs in the format axb. Print the pairs in the order of increasing first parameter (a). Separate the pairs in the line by whitespaces.
| null | null |
[{"input": "4\nOXXXOXOOXOOX\nOXOXOXOXOXOX\nXXXXXXXXXXXX\nOOOOOOOOOOOO", "output": "3 1x12 2x6 4x3\n4 1x12 2x6 3x4 6x2\n6 1x12 2x6 3x4 4x3 6x2 12x1\n0"}]
| 1,000 |
["implementation"]
| 44 |
[{"input": "4\r\nOXXXOXOOXOOX\r\nOXOXOXOXOXOX\r\nXXXXXXXXXXXX\r\nOOOOOOOOOOOO\r\n", "output": "3 1x12 2x6 4x3\r\n4 1x12 2x6 3x4 6x2\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n0\r\n"}, {"input": "2\r\nOOOOOOOOOOOO\r\nXXXXXXXXXXXX\r\n", "output": "0\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}, {"input": "13\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\n", "output": "6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}]
| false |
stdio
| null | true |
400/A
|
400
|
A
|
Python 3
|
TESTS
| 1 | 109 | 0 |
61170847
|
le=int(input())
for _ in range(le):
str1=input()
count=0
arr=[12,6,4,3,2,1]
arr1=[]
for num in arr:
j=0
flag=True
while(j<12):
j+=num
if(str1[j-1]!="X"):
flag=False
break
if(flag):
count+=1
arr1.append(str(str(12//num)+"x"+str(num)))
print(count,end=" ")
for j in arr1:
print(j,end=" ")
print("\t")
| 44 | 46 | 0 |
179040613
|
#!/usr/bin/env/python
# -*- coding: utf-8 -*-
tests = int(input())
for _ in range(tests):
m = input()
res = []
for i in [1, 2, 3, 4, 6, 12]:
j = 12 // i
flag = False
for kk in range(j):
stmp = ''
if flag:
break
for k in range(i):
stmp += m[k * j + kk]
# print(stmp)
if stmp.count('X') == len(stmp):
res.append(i)
flag = True
break
print(len(res), end=' ')
for r in res:
print('{}x{}'.format(r, 12//r), end=' ')
print()
|
Codeforces Round 234 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Inna and Choose Options
|
There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusual upgrade of this game. The rules of the game are given below:
There is one person playing the game. Before the beginning of the game he puts 12 cards in a row on the table. Each card contains a character: "X" or "O". Then the player chooses two positive integers a and b (a·b = 12), after that he makes a table of size a × b from the cards he put on the table as follows: the first b cards form the first row of the table, the second b cards form the second row of the table and so on, the last b cards form the last (number a) row of the table. The player wins if some column of the table contain characters "X" on all cards. Otherwise, the player loses.
Inna has already put 12 cards on the table in a row. But unfortunately, she doesn't know what numbers a and b to choose. Help her win the game: print to her all the possible ways of numbers a, b that she can choose and win.
|
The first line of the input contains integer t (1 ≤ t ≤ 100). This value shows the number of sets of test data in the input. Next follows the description of each of the t tests on a separate line.
The description of each test is a string consisting of 12 characters, each character is either "X", or "O". The i-th character of the string shows the character that is written on the i-th card from the start.
|
For each test, print the answer to the test on a single line. The first number in the line must represent the number of distinct ways to choose the pair a, b. Next, print on this line the pairs in the format axb. Print the pairs in the order of increasing first parameter (a). Separate the pairs in the line by whitespaces.
| null | null |
[{"input": "4\nOXXXOXOOXOOX\nOXOXOXOXOXOX\nXXXXXXXXXXXX\nOOOOOOOOOOOO", "output": "3 1x12 2x6 4x3\n4 1x12 2x6 3x4 6x2\n6 1x12 2x6 3x4 4x3 6x2 12x1\n0"}]
| 1,000 |
["implementation"]
| 44 |
[{"input": "4\r\nOXXXOXOOXOOX\r\nOXOXOXOXOXOX\r\nXXXXXXXXXXXX\r\nOOOOOOOOOOOO\r\n", "output": "3 1x12 2x6 4x3\r\n4 1x12 2x6 3x4 6x2\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n0\r\n"}, {"input": "2\r\nOOOOOOOOOOOO\r\nXXXXXXXXXXXX\r\n", "output": "0\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}, {"input": "13\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\n", "output": "6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}]
| false |
stdio
| null | true |
898/C
|
898
|
C
|
PyPy 3-64
|
TESTS
| 2 | 46 | 512,000 |
153721776
|
n = int(input())
records = {}
for i in range(n):
record = input().split()
records[record[0]] = list(set(record[2:]))
res = {k: [] for k in records}
for k in records:
for number in records[k]:
isGood = True
for p in records[k]:
isGood &= not (p.endswith(number) and p != number)
if isGood:
res[k].append(number)
print(len(res))
for k in records:
print(k, len(res[k]), end=' ')
for x in res[k]:
print(x, end=' ')
print()
| 59 | 62 | 5,529,600 |
33302491
|
k = int(input())
d = dict()
for _ in range(k):
name, *nums = input().split()
nums = set(nums[1:])
if name in d:
d[name] = d[name].union(nums)
else:
d[name] = nums
print(len(d))
for name in d:
arr = sorted(list(map(lambda x: x[::-1], d[name])))
i = 0
while i < len(arr)-1:
if (arr[i+1].startswith(arr[i])):
arr.pop(i)
i -= 1
i += 1
print(name, len(list(arr)), ' '.join([k[::-1] for k in arr]))
|
Codeforces Round 451 (Div. 2)
|
CF
| 2,017 | 2 | 256 |
Phone Numbers
|
Vasya has several phone books, in which he recorded the telephone numbers of his friends. Each of his friends can have one or several phone numbers.
Vasya decided to organize information about the phone numbers of friends. You will be given n strings — all entries from Vasya's phone books. Each entry starts with a friend's name. Then follows the number of phone numbers in the current entry, and then the phone numbers themselves. It is possible that several identical phones are recorded in the same record.
Vasya also believes that if the phone number a is a suffix of the phone number b (that is, the number b ends up with a), and both numbers are written by Vasya as the phone numbers of the same person, then a is recorded without the city code and it should not be taken into account.
The task is to print organized information about the phone numbers of Vasya's friends. It is possible that two different people have the same number. If one person has two numbers x and y, and x is a suffix of y (that is, y ends in x), then you shouldn't print number x. If the number of a friend in the Vasya's phone books is recorded several times in the same format, it is necessary to take it into account exactly once.
Read the examples to understand statement and format of the output better.
|
First line contains the integer n (1 ≤ n ≤ 20) — number of entries in Vasya's phone books.
The following n lines are followed by descriptions of the records in the format described in statement. Names of Vasya's friends are non-empty strings whose length does not exceed 10. They consists only of lowercase English letters. Number of phone numbers in one entry is not less than 1 is not more than 10. The telephone numbers consist of digits only. If you represent a phone number as a string, then its length will be in range from 1 to 10. Phone numbers can contain leading zeros.
|
Print out the ordered information about the phone numbers of Vasya's friends. First output m — number of friends that are found in Vasya's phone books.
The following m lines must contain entries in the following format "name number_of_phone_numbers phone_numbers". Phone numbers should be separated by a space. Each record must contain all the phone numbers of current friend.
Entries can be displayed in arbitrary order, phone numbers for one record can also be printed in arbitrary order.
| null | null |
[{"input": "2\nivan 1 00123\nmasha 1 00123", "output": "2\nmasha 1 00123\nivan 1 00123"}, {"input": "3\nkarl 2 612 12\npetr 1 12\nkatya 1 612", "output": "3\nkatya 1 612\npetr 1 12\nkarl 1 612"}, {"input": "4\nivan 3 123 123 456\nivan 2 456 456\nivan 8 789 3 23 6 56 9 89 2\ndasha 2 23 789", "output": "2\ndasha 2 23 789\nivan 4 789 123 2 456"}]
| 1,400 |
["implementation", "strings"]
| 59 |
[{"input": "2\r\nivan 1 00123\r\nmasha 1 00123\r\n", "output": "2\r\nmasha 1 00123 \r\nivan 1 00123 \r\n"}, {"input": "3\r\nkarl 2 612 12\r\npetr 1 12\r\nkatya 1 612\r\n", "output": "3\r\nkatya 1 612 \r\npetr 1 12 \r\nkarl 1 612 \r\n"}, {"input": "4\r\nivan 3 123 123 456\r\nivan 2 456 456\r\nivan 8 789 3 23 6 56 9 89 2\r\ndasha 2 23 789\r\n", "output": "2\r\ndasha 2 789 23 \r\nivan 4 789 123 456 2 \r\n"}, {"input": "20\r\nnxj 6 7 6 6 7 7 7\r\nnxj 10 8 5 1 7 6 1 0 7 0 6\r\nnxj 2 6 5\r\nnxj 10 6 7 6 6 5 8 3 6 6 8\r\nnxj 10 6 1 7 6 7 1 8 7 8 6\r\nnxj 10 8 5 8 6 5 6 1 9 6 3\r\nnxj 10 8 1 6 4 8 0 4 6 0 1\r\nnxj 9 2 6 6 8 1 1 3 6 6\r\nnxj 10 8 9 0 9 1 3 2 3 2 3\r\nnxj 6 6 7 0 8 1 2\r\nnxj 7 7 7 8 1 3 6 9\r\nnxj 10 2 7 0 1 5 1 9 1 2 6\r\nnxj 6 9 6 9 6 3 7\r\nnxj 9 0 1 7 8 2 6 6 5 6\r\nnxj 4 0 2 3 7\r\nnxj 10 0 4 0 6 1 1 8 8 4 7\r\nnxj 8 4 6 2 6 6 1 2 7\r\nnxj 10 5 3 4 2 1 0 7 0 7 6\r\nnxj 10 9 6 0 6 1 6 2 1 9 6\r\nnxj 4 2 9 0 1\r\n", "output": "1\r\nnxj 10 0 3 2 1 4 7 8 5 9 6 \r\n"}, {"input": "20\r\nl 6 02 02 2 02 02 2\r\nl 8 8 8 8 2 62 13 31 3\r\ne 9 0 91 0 0 60 91 60 2 44\r\ne 9 69 2 1 44 2 91 66 1 70\r\nl 9 7 27 27 3 1 3 7 80 81\r\nl 9 2 1 13 7 2 10 02 3 92\r\ne 9 0 15 3 5 5 15 91 09 44\r\nl 7 2 50 4 5 98 31 98\r\nl 3 26 7 3\r\ne 6 7 5 0 62 65 91\r\nl 8 80 0 4 0 2 2 0 13\r\nl 9 19 13 02 2 1 4 19 26 02\r\nl 10 7 39 7 9 22 22 26 2 90 4\r\ne 7 65 2 36 0 34 57 9\r\ne 8 13 02 09 91 73 5 36 62\r\nl 9 75 0 10 8 76 7 82 8 34\r\nl 7 34 0 19 80 6 4 7\r\ne 5 4 2 5 7 2\r\ne 7 4 02 69 7 07 20 2\r\nl 4 8 2 1 63\r\n", "output": "2\r\ne 18 15 62 07 70 91 57 02 66 65 69 09 13 20 44 73 34 60 36 \r\nl 21 27 50 22 63 75 19 26 90 02 92 62 31 10 76 82 80 98 81 39 34 13 \r\n"}, {"input": "20\r\no 10 6 6 97 45 6 6 6 6 5 6\r\nl 8 5 5 5 19 59 5 8 5\r\nj 9 2 30 58 2 2 1 0 30 4\r\nc 10 1 1 7 51 7 7 51 1 1 1\r\no 9 7 97 87 70 2 19 2 14 6\r\ne 6 26 6 6 6 26 5\r\ng 9 3 3 3 3 3 78 69 8 9\r\nl 8 8 01 1 5 8 41 72 3\r\nz 10 1 2 2 2 9 1 9 1 6 7\r\ng 8 7 78 05 36 7 3 67 9\r\no 5 6 9 9 7 7\r\ne 10 30 2 1 1 2 5 04 0 6 6\r\ne 9 30 30 2 2 0 26 30 79 8\r\nt 10 2 2 9 29 7 7 7 9 2 9\r\nc 7 7 51 1 31 2 7 4\r\nc 9 83 1 6 78 94 74 54 8 32\r\ng 8 4 1 01 9 39 28 6 6\r\nt 7 9 2 01 4 4 9 58\r\nj 5 0 1 58 02 4\r\nw 10 80 0 91 91 06 91 9 9 27 7\r\n", "output": "9\r\nw 5 91 27 06 9 80 \r\nt 6 7 58 4 29 2 01 \r\ne 8 8 79 5 30 2 26 04 1 \r\nl 8 72 3 19 59 41 01 5 8 \r\nj 5 4 30 58 1 02 \r\nz 5 6 1 9 2 7 \r\ng 10 05 39 4 3 36 01 67 69 28 78 \r\no 8 14 87 97 6 19 70 45 2 \r\nc 10 83 51 31 54 74 32 7 94 78 6 \r\n"}, {"input": "1\r\negew 5 3 123 23 1234 134\r\n", "output": "1\r\negew 3 123 134 1234 \r\n"}]
| false |
stdio
|
import sys
from collections import defaultdict
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input and build friends data
friends = defaultdict(set)
with open(input_path, 'r') as f:
n = int(f.readline())
for _ in range(n):
parts = f.readline().strip().split()
name = parts[0]
numbers = parts[2:]
friends[name].update(numbers) # Add all numbers, duplicates in same line are handled via set
# Read submission output
submission_friends = set()
try:
with open(submission_path, 'r') as f:
lines = f.read().splitlines()
except:
print(0)
return
if not lines:
print(0)
return
# Check m line
try:
m_line = lines[0]
m = int(m_line)
except:
print(0)
return
if m != len(friends):
print(0)
return
submission_lines = lines[1:]
if len(submission_lines) != m:
print(0)
return
for line in submission_lines:
parts = line.strip().split()
if len(parts) < 2:
print(0)
return
name = parts[0]
if name not in friends:
print(0)
return
submission_friends.add(name)
try:
k = int(parts[1])
except:
print(0)
return
numbers = parts[2:]
if len(numbers) != k:
print(0)
return
# Check for duplicates in submission's numbers
if len(set(numbers)) != k:
print(0)
return
# Check all numbers are present in friends' merged set
merged_numbers = friends[name]
for num in numbers:
if num not in merged_numbers:
print(0)
return
# Check no two numbers in submission are suffix of each other
for i in range(len(numbers)):
for j in range(i+1, len(numbers)):
a = numbers[i]
b = numbers[j]
if (len(a) >= len(b) and a.endswith(b)) or (len(b) >= len(a) and b.endswith(a)):
print(0)
return
# Check every merged number is either in submission or is a suffix of some submission number
for merged_num in merged_numbers:
if merged_num in numbers:
continue
found = False
for num in numbers:
if len(merged_num) <= len(num) and num.endswith(merged_num):
found = True
break
if not found:
print(0)
return
# Check all friends are present in submission
if submission_friends != set(friends.keys()):
print(0)
return
# All checks passed
print(100)
if __name__ == "__main__":
main()
| true |
852/E
|
852
|
E
|
PyPy 3
|
TESTS
| 3 | 904 | 6,553,600 |
30146944
|
n=int(input())
a=[0]*n
for i in range(n-1):
for j in input().split():
a[int(j)-1]+=1
l=a.count(1)
print(l*2**(n-l+1)+(n-l)*2**(n-l))
| 12 | 264 | 102,400 |
203394745
|
n=int(input())
a = [0]*(n+1)
for i in range(n-1):
for i in input().split():
a[int(i)]+=1
l = a.count(1)
print ((l*2**(n-l+1)+(n-l)*(2**(n-l)))%(10**9+7))
|
Bubble Cup X - Finals [Online Mirror]
|
ICPC
| 2,017 | 1 | 256 |
Casinos and travel
|
John has just bought a new car and is planning a journey around the country. Country has N cities, some of which are connected by bidirectional roads. There are N - 1 roads and every city is reachable from any other city. Cities are labeled from 1 to N.
John first has to select from which city he will start his journey. After that, he spends one day in a city and then travels to a randomly choosen city which is directly connected to his current one and which he has not yet visited. He does this until he can't continue obeying these rules.
To select the starting city, he calls his friend Jack for advice. Jack is also starting a big casino business and wants to open casinos in some of the cities (max 1 per city, maybe nowhere). Jack knows John well and he knows that if he visits a city with a casino, he will gamble exactly once before continuing his journey.
He also knows that if John enters a casino in a good mood, he will leave it in a bad mood and vice versa. Since he is John's friend, he wants him to be in a good mood at the moment when he finishes his journey. John is in a good mood before starting the journey.
In how many ways can Jack select a starting city for John and cities where he will build casinos such that no matter how John travels, he will be in a good mood at the end? Print answer modulo 109 + 7.
|
In the first line, a positive integer N (1 ≤ N ≤ 100000), the number of cities.
In the next N - 1 lines, two numbers a, b (1 ≤ a, b ≤ N) separated by a single space meaning that cities a and b are connected by a bidirectional road.
|
Output one number, the answer to the problem modulo 109 + 7.
| null |
Example 1: If Jack selects city 1 as John's starting city, he can either build 0 casinos, so John will be happy all the time, or build a casino in both cities, so John would visit a casino in city 1, become unhappy, then go to city 2, visit a casino there and become happy and his journey ends there because he can't go back to city 1. If Jack selects city 2 for start, everything is symmetrical, so the answer is 4.
Example 2: If Jack tells John to start from city 1, he can either build casinos in 0 or 2 cities (total 4 possibilities). If he tells him to start from city 2, then John's journey will either contain cities 2 and 1 or 2 and 3. Therefore, Jack will either have to build no casinos, or build them in all three cities. With other options, he risks John ending his journey unhappy. Starting from 3 is symmetric to starting from 1, so in total we have 4 + 2 + 4 = 10 options.
|
[{"input": "2\n1 2", "output": "4"}, {"input": "3\n1 2\n2 3", "output": "10"}]
| 2,100 |
["dp"]
| 12 |
[{"input": "2\r\n1 2\r\n", "output": "4\r\n"}, {"input": "3\r\n1 2\r\n2 3\r\n", "output": "10\r\n"}, {"input": "4\r\n1 2\r\n2 3\r\n3 4\r\n", "output": "24\r\n"}]
| false |
stdio
| null | true |
21/D
|
21
|
D
|
PyPy 3
|
TESTS
| 0 | 122 | 0 |
152774562
|
from heapq import heappop,heappush
n,m=map(int,input().split())
E=[[] for i in range(n)]
for i in range(m):
x,y,c=map(int,input().split())
x-=1
y-=1
E[x].append((y,c))
E[y].append((x,c))
DP=[[1<<30]*(1<<n) for i in range(n)]
DP[0][1<<0]=0
Q=[]
Q.append((0,0,1<<0))
while Q:
dis,fr,passed=heappop(Q)
if DP[fr][passed]!=dis:
continue
for to,cost in E[fr]:
if DP[to][passed|(1<<to)]>dis+cost:
DP[to][passed|(1<<to)]=dis+cost
heappush(Q,(dis+cost,to,passed|(1<<to)))
#ANS=1<<30
#for i in range(n):
# ANS=min(ANS,DP[i][(1<<n)-1])
#print(ANS)
print(DP[0][(1<<n)-1])
| 60 | 280 | 5,017,600 |
153611194
|
from itertools import repeat
inf = 10000000000
ans = inf
mat = 0
def solve(odd, graph_matrix):
global ans
global mat
if (len(odd) == 0):
ans = min(ans, mat)
return
e = odd.pop()
i = 0
while i<len(odd):
cur = odd.pop(i)
mat+=graph_matrix[cur][e]
solve(odd, graph_matrix)
mat-=graph_matrix[cur][e]
odd.insert(i, cur)
i+=1
odd.append(e)
def main():
v,e= map(int, input().split())
graph = [[inf]*v for i in repeat(None, v)]
deg=[0]*v
sum = 0
for i in range(v):
graph[i][i]=0
for i in range(e):
in1, in2, in3= map(int,input().split())
in1-=1
in2-=1
deg[in1]+=1
deg[in2]+=1
graph[in1][in2]= min(graph[in1][in2], in3)
graph[in2][in1]= min(graph[in2][in1], in3)
sum+=in3
for i in range(v):
for j in range(v):
for k in range(v):
graph[j][k] = min(graph[j][k], graph[j][i]+graph[i][k])
for i in range(v):
if graph[0][i] == inf and deg[i]>0:
print(-1)
return
odd = []
for i in range(v):
if deg[i] % 2 != 0:
odd.append(i)
solve(odd, graph_matrix=graph)
print(sum + ans)
main()
|
Codeforces Alpha Round 21 (Codeforces format)
|
CF
| 2,010 | 0.5 | 64 |
Traveling Graph
|
You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).
|
The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and w is the edge length.
|
Output minimal cycle length or -1 if it doesn't exists.
| null | null |
[{"input": "3 3\n1 2 1\n2 3 1\n3 1 1", "output": "3"}, {"input": "3 2\n1 2 3\n2 3 4", "output": "14"}]
| 2,400 |
["bitmasks", "graph matchings", "graphs"]
| 60 |
[{"input": "4 6\r\n1 2 10\r\n2 3 1000\r\n3 4 10\r\n4 1 1000\r\n4 2 5000\r\n1 3 2\r\n", "output": "7042\r\n"}, {"input": "2 9\r\n1 2 9\r\n1 2 9\r\n2 1 9\r\n1 2 8\r\n2 1 9\r\n1 2 9\r\n1 2 9\r\n1 2 11\r\n1 2 9\r\n", "output": "90\r\n"}, {"input": "2 10\r\n1 2 9\r\n1 2 9\r\n2 1 9\r\n1 2 8\r\n2 1 9\r\n1 2 9\r\n1 2 9\r\n1 2 11\r\n1 2 9\r\n1 2 9\r\n", "output": "91\r\n"}, {"input": "15 14\r\n1 2 1\r\n2 3 1\r\n2 4 1\r\n3 5 1\r\n3 6 1\r\n4 7 1\r\n4 8 1\r\n5 9 1\r\n5 10 1\r\n6 11 1\r\n6 12 1\r\n7 13 1\r\n7 14 1\r\n8 15 1\r\n", "output": "28\r\n"}, {"input": "4 5\r\n1 2 3\r\n2 3 4\r\n3 4 5\r\n1 4 10\r\n1 3 12\r\n", "output": "41\r\n"}, {"input": "4 5\r\n1 2 3\r\n2 3 4\r\n3 4 5\r\n1 4 10\r\n1 3 12\r\n", "output": "41\r\n"}, {"input": "5 0\r\n", "output": "0\r\n"}, {"input": "1 0\r\n", "output": "0\r\n"}, {"input": "1 1\r\n1 1 44\r\n", "output": "44\r\n"}, {"input": "1 2\r\n1 1 5\r\n1 1 3\r\n", "output": "8\r\n"}, {"input": "2 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\n2 1 3\r\n", "output": "6\r\n"}, {"input": "2 1\r\n1 1 3\r\n", "output": "3\r\n"}, {"input": "2 1\r\n2 2 44\r\n", "output": "-1\r\n"}, {"input": "2 2\r\n1 1 44\r\n2 2 44\r\n", "output": "-1\r\n"}, {"input": "2 3\r\n1 1 1\r\n2 2 2\r\n2 1 3\r\n", "output": "9\r\n"}, {"input": "7 3\r\n4 4 1\r\n7 7 1\r\n2 2 1\r\n", "output": "-1\r\n"}, {"input": "15 0\r\n", "output": "0\r\n"}, {"input": "4 2\r\n1 2 1\r\n3 4 1\r\n", "output": "-1\r\n"}, {"input": "7 1\r\n3 4 4\r\n", "output": "-1\r\n"}, {"input": "2 1\r\n2 2 5741\r\n", "output": "-1\r\n"}, {"input": "2 2\r\n2 1 4903\r\n1 1 4658\r\n", "output": "14464\r\n"}, {"input": "2 4\r\n1 2 7813\r\n2 1 6903\r\n1 2 6587\r\n2 2 7372\r\n", "output": "35262\r\n"}, {"input": "2 8\r\n1 2 4618\r\n1 1 6418\r\n2 2 2815\r\n1 1 4077\r\n2 1 4239\r\n1 2 5359\r\n1 2 3971\r\n1 2 7842\r\n", "output": "43310\r\n"}, {"input": "3 1\r\n3 2 6145\r\n", "output": "-1\r\n"}, {"input": "3 2\r\n1 1 1169\r\n1 2 1250\r\n", "output": "3669\r\n"}, {"input": "3 4\r\n1 1 5574\r\n3 1 5602\r\n3 2 5406\r\n2 1 5437\r\n", "output": "22019\r\n"}, {"input": "3 8\r\n3 3 9507\r\n2 1 9560\r\n3 3 9328\r\n2 2 9671\r\n2 2 9641\r\n1 2 9717\r\n1 3 9535\r\n1 2 9334\r\n", "output": "95162\r\n"}, {"input": "4 1\r\n1 3 3111\r\n", "output": "6222\r\n"}, {"input": "4 2\r\n3 2 6816\r\n1 3 7161\r\n", "output": "27954\r\n"}, {"input": "4 4\r\n1 3 1953\r\n3 2 2844\r\n1 3 2377\r\n3 2 2037\r\n", "output": "9211\r\n"}, {"input": "4 8\r\n1 2 4824\r\n3 1 436\r\n2 2 3087\r\n2 4 2955\r\n2 4 2676\r\n4 3 2971\r\n3 4 3185\r\n3 1 3671\r\n", "output": "28629\r\n"}, {"input": "5 1\r\n5 5 1229\r\n", "output": "-1\r\n"}, {"input": "5 2\r\n2 2 2515\r\n2 4 3120\r\n", "output": "-1\r\n"}, {"input": "5 4\r\n5 1 404\r\n3 1 551\r\n1 1 847\r\n5 1 706\r\n", "output": "3059\r\n"}, {"input": "5 8\r\n1 5 1016\r\n4 5 918\r\n1 4 926\r\n2 3 928\r\n5 4 994\r\n2 3 1007\r\n1 4 946\r\n3 4 966\r\n", "output": "9683\r\n"}, {"input": "6 1\r\n3 6 2494\r\n", "output": "-1\r\n"}, {"input": "6 2\r\n5 3 5039\r\n2 3 4246\r\n", "output": "-1\r\n"}, {"input": "6 4\r\n5 4 6847\r\n3 6 7391\r\n1 6 7279\r\n2 5 7250\r\n", "output": "-1\r\n"}, {"input": "6 8\r\n2 4 8044\r\n6 4 7952\r\n2 5 6723\r\n6 4 8105\r\n1 5 6648\r\n1 6 6816\r\n1 3 7454\r\n5 3 6857\r\n", "output": "73199\r\n"}, {"input": "15 1\r\n7 5 7838\r\n", "output": "-1\r\n"}, {"input": "15 2\r\n5 13 9193\r\n14 5 9909\r\n", "output": "-1\r\n"}, {"input": "15 4\r\n1 5 5531\r\n9 15 3860\r\n8 4 6664\r\n13 3 4320\r\n", "output": "-1\r\n"}, {"input": "15 8\r\n14 6 9084\r\n1 12 8967\r\n11 12 8866\r\n12 2 8795\r\n7 10 9102\r\n10 12 9071\r\n12 10 9289\r\n4 11 8890\r\n", "output": "-1\r\n"}, {"input": "15 16\r\n3 3 2551\r\n6 11 2587\r\n2 4 2563\r\n3 6 2569\r\n3 1 2563\r\n4 11 2487\r\n7 15 2580\r\n7 14 2534\r\n10 7 2530\r\n3 5 2587\r\n5 14 2596\r\n14 14 2556\r\n15 9 2547\r\n12 4 2586\r\n6 8 2514\r\n2 12 2590\r\n", "output": "69034\r\n"}]
| false |
stdio
| null | true |
383/C
|
383
|
C
|
Python 3
|
TESTS
| 5 | 623 | 127,692,800 |
119294347
|
from collections import defaultdict
import threading
def dfs(node,parent,g,sons):
for e in g[node]:
if e==parent:
continue
else:
sons[node].append(e)
dfs(e,node,g,sons)
return
def prop(parent,val,sons,par,ar):
if par==0:
ar[parent]+=val
else:
ar[parent]+=-(val)
par^=1
for i in sons[parent]:
prop(i,val,sons,par,ar)
def main():
n,m=map(int,input().strip().split())
ar=[*map(int,input().strip().split())]
ar=[0]+ar
g=defaultdict(list)
sons=defaultdict(list)
for i in range(n-1):
x,y=map(int,input().strip().split())
g[x].append(y)
g[y].append(x)
dfs(1,-1,g,sons)
q=[]
for _ in range(m):
l=list(map(int,input().strip().split()))
if l[0]==1:
prop(l[1],l[2],sons,0,ar)
else:
print(ar[l[1]])
threading.stack_size(10 ** 8)
t = threading.Thread(target=main)
t.start()
t.join()
| 54 | 1,107 | 130,252,800 |
192697070
|
class BIT():
"""区間加算、一点取得クエリをそれぞれO(logN)で応えるデータ構造を構築する
add: 区間[begin, end)にvalを加える
get_val: i番目(0-indexed)の値を求める
"""
def __init__(self, n):
self.n = n
self.bit = [0] * (n + 1)
def get_val(self, i):
i = i + 1
s = 0
while i <= self.n:
s += self.bit[i]
i += i & -i
return s
def _add(self, i, val):
while i > 0:
self.bit[i] += val
i -= i & -i
def add(self, i, j, val):
self._add(j, val)
self._add(i, -val)
from collections import deque
import sys
input = sys.stdin.readline
def eular_tour(tree: list, root: int):
n = len(tree)
res = []
begin = [-1] * n
end = [-1] * n
visited = [False] * n
visited[root] = True
q = deque([root])
while q:
pos = q.pop()
res.append(pos)
end[pos] = len(res)
if begin[pos] == -1:
begin[pos] = len(res) - 1
for next_pos in tree[pos]:
if visited[next_pos]:
continue
else:
visited[next_pos] = True
q.append(pos)
q.append(next_pos)
return res, begin, end
n, q = map(int, input().split())
init_cost = list(map(int, input().split()))
info = [list(map(int, input().split())) for i in range(n-1)]
query = [list(map(int, input().split())) for i in range(q)]
tree = [[] for i in range(n)]
for i in range(n-1):
a, b = info[i]
a -= 1
b -= 1
tree[a].append(b)
tree[b].append(a)
res, begin, end = eular_tour(tree, 0)
even_res = []
odd_res = []
for i in range(len(res)):
if i % 2 == 0:
even_res.append(res[i])
else:
odd_res.append(res[i])
even_bit = BIT(len(even_res))
odd_bit = BIT(len(odd_res))
for i in range(q):
if query[i][0] == 1:
_, pos, cost = query[i]
pos -= 1
if begin[pos] % 2 == 0:
even_bit.add(begin[pos] // 2, (end[pos] + 1) // 2, cost)
odd_bit.add(begin[pos] // 2, end[pos] // 2, -cost)
else:
odd_bit.add(begin[pos] // 2, end[pos] // 2, cost)
even_bit.add((begin[pos] + 1) // 2, end[pos] // 2, -cost)
else:
_, pos = query[i]
pos -= 1
if begin[pos] % 2 == 0:
ans = even_bit.get_val(begin[pos] // 2)
else:
ans = odd_bit.get_val(begin[pos] // 2)
print(ans + init_cost[pos])
|
Codeforces Round 225 (Div. 1)
|
CF
| 2,014 | 2 | 256 |
Propagating tree
|
Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.
This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.
This tree supports two types of queries:
- "1 x val" — val is added to the value of node x;
- "2 x" — print the current value of node x.
In order to help Iahub understand the tree better, you must answer m queries of the preceding type.
|
The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.
Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.
|
For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.
| null |
The values of the nodes are [1, 2, 1, 1, 2] at the beginning.
Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].
Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].
You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
|
[{"input": "5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4", "output": "3\n3\n0"}]
| 2,000 |
["data structures", "dfs and similar", "trees"]
| 54 |
[{"input": "5 5\r\n1 2 1 1 2\r\n1 2\r\n1 3\r\n2 4\r\n2 5\r\n1 2 3\r\n1 1 2\r\n2 1\r\n2 2\r\n2 4\r\n", "output": "3\r\n3\r\n0\r\n"}, {"input": "10 10\r\n137 197 856 768 825 894 86 174 218 326\r\n7 8\r\n4 7\r\n8 9\r\n7 10\r\n1 2\r\n2 4\r\n3 6\r\n3 5\r\n2 3\r\n1 9 624\r\n2 1\r\n2 4\r\n1 6 505\r\n1 8 467\r\n1 3 643\r\n2 1\r\n1 8 631\r\n2 4\r\n1 7 244\r\n", "output": "137\r\n768\r\n137\r\n768\r\n"}, {"input": "10 10\r\n418 45 865 869 745 901 177 773 854 462\r\n4 8\r\n1 4\r\n3 6\r\n1 5\r\n1 10\r\n5 9\r\n1 2\r\n4 7\r\n1 3\r\n2 2\r\n1 6 246\r\n1 4 296\r\n1 2 378\r\n1 8 648\r\n2 6\r\n1 5 288\r\n1 6 981\r\n1 2 868\r\n2 7\r\n", "output": "45\r\n1147\r\n-119\r\n"}]
| false |
stdio
| null | true |
474/D
|
474
|
D
|
PyPy 3-64
|
TESTS
| 2 | 78 | 7,884,800 |
188602238
|
import sys
import math
import heapq as hp
#hp.heapify hp.heappush hp.heappop
from collections import deque
#appendleft append pop popleft
input=sys.stdin.readline
m=1000000007
def inp():
return int(input())
def minp():
return map(int,input().split())
def strinp():
S=input().rstrip()
return S
def lst():
return list(map(int,input().split()))
#------------------------------------#
t,k=minp()
queries=[]
mxm=0
for i in range(t):
a,b=minp()
mxm=max(mxm,b)
queries.append((a,b))
sum=[0]*(mxm+1)
dp=[[0]*(mxm+1) for _ in range(2)]
#0 for R 1 for W
dp[0][0]=1
for i in range(1,mxm+1):
dp[0][i]=(dp[1][i-1]+dp[0][i-1])%m
if i>=k:
dp[1][i]=(dp[0][i-k]+dp[1][i-k])%m
sum[i]=(sum[i-1]+dp[0][i]+dp[1][i])%m
for i in range(len(queries)):
print(sum[queries[i][1]]-sum[queries[i][0]-1])
| 43 | 186 | 18,944,000 |
211154609
|
import sys
input=sys.stdin.readline
from itertools import permutations, accumulate
from collections import defaultdict as dd
from collections import deque
from bisect import bisect_left,bisect_right
from math import lcm,gcd,sqrt,ceil,comb
from heapq import heappop,heappush,heapify
toBin=lambda x:bin(x).replace("0b","")
inf=float("inf");DIRS=[[1,0],[-1,0],[0,1],[0,-1]];CHARS="abcdefghijklmnopqrstuvwxyz"
MOD=10**9+7;N=300010;fac=[1]*N;invfac=[1]*N #Fill out N to calculate combinations
for i in range(2,N):fac[i]=fac[i-1]*i%MOD
invfac[N-1]=pow(fac[N-1],MOD-2,MOD)
for i in range(N-1)[::-1]:invfac[i]=invfac[i+1]*(i+1)%MOD
def c(i,j):return 0 if i<j else fac[i]*invfac[j]*invfac[i-j]
def solve():
t,k=map(int,input().split());dp=[1]+[0]*(10**5);dp0=[0]
for i in range(1,10**5+1):
dp[i]=(dp[i]+dp[i-1]+(dp[i-k] if i-k>=0 else 0))%MOD
dp0.append((dp0[-1]+dp[i])%MOD)
for _ in range(t):
a,b=map(int,input().split())
print((dp0[b]-dp0[a-1])%MOD)
if __name__=="__main__":
solve()
|
Codeforces Round 271 (Div. 2)
|
CF
| 2,014 | 1.5 | 256 |
Flowers
|
We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.
But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.
Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (109 + 7).
|
Input contains several test cases.
The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.
The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.
|
Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).
| null |
- For K = 2 and length 1 Marmot can eat (R).
- For K = 2 and length 2 Marmot can eat (RR) and (WW).
- For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).
- For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).
|
[{"input": "3 2\n1 3\n2 3\n4 4", "output": "6\n5\n5"}]
| 1,700 |
["dp"]
| 43 |
[{"input": "3 2\r\n1 3\r\n2 3\r\n4 4\r\n", "output": "6\r\n5\r\n5\r\n"}, {"input": "1 1\r\n1 3\r\n", "output": "14\r\n"}, {"input": "1 2\r\n64329 79425\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
995/D
|
995
|
D
|
Python 3
|
TESTS
| 3 | 951 | 32,153,600 |
39730476
|
from sys import stdin
from math import fsum
def main():
n, m = map(int, input().split())
ff = list(map(float, input().split()))
scale, r = .5 ** n, fsum(ff)
res = [r * scale]
for i, f in map(str.split, stdin.read().splitlines()):
r += float(f) - ff[int(i)]
res.append(r * scale)
print('\n'.join(map(str, res)))
if __name__ == '__main__':
main()
| 53 | 1,154 | 32,051,200 |
39730546
|
from sys import stdin
from math import fsum
def main():
n, m = map(int, input().split())
ff = list(map(float, input().split()))
scale, r = .5 ** n, fsum(ff)
res = [r * scale]
for si, sf in map(str.split, stdin.read().splitlines()):
i, f = int(si), float(sf)
r += f - ff[i]
ff[i] = f
res.append(r * scale)
print('\n'.join(map(str, res)))
if __name__ == '__main__':
main()
|
Codeforces Round 492 (Div. 1) [Thanks, uDebug!]
|
CF
| 2,018 | 3 | 256 |
Game
|
Allen and Bessie are playing a simple number game. They both know a function $$$f: \{0, 1\}^n \to \mathbb{R}$$$, i. e. the function takes $$$n$$$ binary arguments and returns a real value. At the start of the game, the variables $$$x_1, x_2, \dots, x_n$$$ are all set to $$$-1$$$. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an $$$i$$$ such that $$$x_i = -1$$$ and either setting $$$x_i \to 0$$$ or $$$x_i \to 1$$$.
After $$$n$$$ rounds all variables are set, and the game value resolves to $$$f(x_1, x_2, \dots, x_n)$$$. Allen wants to maximize the game value, and Bessie wants to minimize it.
Your goal is to help Allen and Bessie find the expected game value! They will play $$$r+1$$$ times though, so between each game, exactly one value of $$$f$$$ changes. In other words, between rounds $$$i$$$ and $$$i+1$$$ for $$$1 \le i \le r$$$, $$$f(z_1, \dots, z_n) \to g_i$$$ for some $$$(z_1, \dots, z_n) \in \{0, 1\}^n$$$. You are to find the expected game value in the beginning and after each change.
|
The first line contains two integers $$$n$$$ and $$$r$$$ ($$$1 \le n \le 18$$$, $$$0 \le r \le 2^{18}$$$).
The next line contains $$$2^n$$$ integers $$$c_0, c_1, \dots, c_{2^n-1}$$$ ($$$0 \le c_i \le 10^9$$$), denoting the initial values of $$$f$$$. More specifically, $$$f(x_0, x_1, \dots, x_{n-1}) = c_x$$$, if $$$x = \overline{x_{n-1} \ldots x_0}$$$ in binary.
Each of the next $$$r$$$ lines contains two integers $$$z$$$ and $$$g$$$ ($$$0 \le z \le 2^n - 1$$$, $$$0 \le g \le 10^9$$$). If $$$z = \overline{z_{n-1} \dots z_0}$$$ in binary, then this means to set $$$f(z_0, \dots, z_{n-1}) \to g$$$.
|
Print $$$r+1$$$ lines, the $$$i$$$-th of which denotes the value of the game $$$f$$$ during the $$$i$$$-th round. Your answer must have absolute or relative error within $$$10^{-6}$$$.
Formally, let your answer be $$$a$$$, and the jury's answer be $$$b$$$. Your answer is considered correct if $$$\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$$$.
| null |
Consider the second test case. If Allen goes first, he will set $$$x_1 \to 1$$$, so the final value will be $$$3$$$. If Bessie goes first, then she will set $$$x_1 \to 0$$$ so the final value will be $$$2$$$. Thus the answer is $$$2.5$$$.
In the third test case, the game value will always be $$$1$$$ regardless of Allen and Bessie's play.
|
[{"input": "2 2\n0 1 2 3\n2 5\n0 4", "output": "1.500000\n2.250000\n3.250000"}, {"input": "1 0\n2 3", "output": "2.500000"}, {"input": "2 0\n1 1 1 1", "output": "1.000000"}]
| 2,500 |
["math"]
| 53 |
[{"input": "2 2\r\n0 1 2 3\r\n2 5\r\n0 4\r\n", "output": "1.500000\r\n2.250000\r\n3.250000\r\n"}, {"input": "1 0\r\n2 3\r\n", "output": "2.500000\r\n"}, {"input": "2 0\r\n1 1 1 1\r\n", "output": "1.000000\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n, r = map(int, f.readline().split())
expected_lines = r + 1
with open(output_path) as f:
correct = [float(line.strip()) for line in f]
if len(correct) != expected_lines:
print(0)
return
with open(submission_path) as f:
submission = [line.strip() for line in f]
if len(submission) != expected_lines:
print(0)
return
for a_str, b in zip(submission, correct):
try:
a = float(a_str)
except:
print(0)
return
max_denominator = max(1.0, abs(b))
if abs(a - b) > 1e-6 * max_denominator:
print(0)
return
print(1)
if __name__ == "__main__":
main()
| true |
32/A
|
32
|
A
|
Python 3
|
TESTS
| 2 | 62 | 0 |
185880775
|
n,d=map(int,input().split())
ls=list(map(int,input().split()))
ls.sort()
cntr=0
for x in range(n):
if abs(ls[x]-ls[x-1])<=d:
cntr+=1
print(2*cntr)
| 32 | 122 | 2,355,200 |
151388817
|
from sys import stdin
input = stdin.readline
nums, diff = map(int,input().split())
heights = list(map(int,input().split()))
heights.sort()
matched = 0
for i,shorter in enumerate(heights):
for higher in heights[i+1:]:
if higher - shorter > diff:
break
matched += 1
print(matched*2)
|
Codeforces Beta Round 32 (Div. 2, Codeforces format)
|
CF
| 2,010 | 2 | 256 |
Reconnaissance
|
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most d centimeters. Captain Bob has n soldiers in his detachment. Their heights are a1, a2, ..., an centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1, 2) and (2, 1) should be regarded as different.
|
The first line contains two integers n and d (1 ≤ n ≤ 1000, 1 ≤ d ≤ 109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains n space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
|
Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed d.
| null | null |
[{"input": "5 10\n10 20 50 60 65", "output": "6"}, {"input": "5 1\n55 30 29 31 55", "output": "6"}]
| 800 |
["brute force"]
| 32 |
[{"input": "5 10\r\n10 20 50 60 65\r\n", "output": "6\r\n"}, {"input": "5 1\r\n55 30 29 31 55\r\n", "output": "6\r\n"}, {"input": "6 10\r\n4 6 4 1 9 3\r\n", "output": "30\r\n"}, {"input": "7 100\r\n19 1694 261 162 1 234 513\r\n", "output": "8\r\n"}, {"input": "8 42\r\n37 53 74 187 568 22 5 65\r\n", "output": "20\r\n"}, {"input": "10 4\r\n11 6 76 49 28 20 57 152 5 32\r\n", "output": "4\r\n"}, {"input": "100 100\r\n51 93 101 960 2 477 213 129 663 925 254 78 1486 274 160 481 132 156 412 372 5 57 152 298 1771 7 359 468 254 406 202 929 221 366 552 97 555 29 822 118 539 140 992 854 7 163 134 103 940 30 409 1003 398 43 555 79 107 40 23 103 643 171 310 382 770 337 18 189 570 177 29 54 855 171 205 291 299 935 620 180 114 358 88 292 118 400 218 537 369 60 683 192 13 537 59 824 264 191 3 300\r\n", "output": "2404\r\n"}]
| false |
stdio
| null | true |
756/F
|
756
|
F
|
Python 3
|
TESTS
| 0 | 31 | 0 |
203818251
|
MOD = int(1e9) + 7
def eval_expression(p):
w = 0
result = 0
while w < len(p):
if p[w].isdigit():
num = int(p[w])
w += 1
while w < len(p) and p[w].isdigit():
num = num * 10 + int(p[w])
w += 1
result = (result * 10 + num) % MOD
elif p[w] == '(':
cnt = 1
j = w + 1
while j < len(p) and cnt > 0:
if p[j] == '(':
cnt += 1
elif p[j] == ')':
cnt -= 1
j += 1
num = int(p[w-1])
subexpr = p[w+1:j-1]
val = eval_expression(subexpr)
result = (result + num * val) % MOD
w = j
elif p[w] == '-':
l = int(p[w-1])
r = int(p[w+1])
j = w + 2
while j < len(p) and p[j].isdigit():
r = r * 10 + int(p[j])
j += 1
num = int(''.join(str(k) for k in range(l, r+1)))
result = (result + num) % MOD
w = j
elif p[w] == '+':
subexpr = p[w+1:]
val = eval_expression(subexpr)
result = (result + val) % MOD
break
return result
p = input().strip()
print(eval_expression(p))
| 41 | 842 | 145,920,000 |
164916281
|
# a ^ (P-1) ≡ 1 (mod P)
# a ^ k ≡ (a % P) ^ (k % (P-1)) (mod P)
import sys
sys.setrecursionlimit(10**5)
MOD = 10**9 + 7
def inv(x):
return pow(x, MOD - 2, MOD)
class Number:
def __init__(self, value, length):
self.value = value % MOD
self.length = length % (MOD - 1)
def __add__(self, other):
value = self.value * pow(10, other.length, MOD) + other.value
length = self.length + other.length
return Number(value, length)
def __mul__(self, other):
base = pow(10, self.length, MOD)
if base == 1:
value = self.value * other
else:
value = self.value * (pow(base, other, MOD) - 1) * inv(base - 1)
return Number(value, self.length * other)
@classmethod
def _range(cls, a, r, rpow, length):
base = pow(10, length, MOD)
u = pow(base, rpow, MOD) - 1
v = inv(base - 1)
return cls((a * u + u * v - r) * v, length * rpow)
@classmethod
def range(cls, a, b):
a, b, la, lb = int(a), int(b) + 1, len(a), len(b)
if la == lb:
return cls._range(a, b - a, b - a, la)
term = cls._range(a, p[la] - a, q[la] - a, la)
for i in range(la + 1, lb):
term += cls._range(p[i - 1], p[i] - p[i - 1], q[i] - q[i - 1], i)
term += cls._range(p[lb - 1], b - p[lb - 1], b - q[lb - 1], lb)
return term
def parse_term(start, stop):
if s[stop - 1] == ")":
left = s.index("(", start)
return parse_expression(left + 1, stop - 1) * int(s[start:left])
try:
sep = s.index("-", start, stop)
except ValueError:
return Number(int(s[start:stop]), stop - start)
return Number.range(s[start:sep], s[sep + 1 : stop])
def parse_expression(start, stop):
expression = Number(0, 0)
i = start
while i < stop:
if s[i] == "+":
expression += parse_term(start, i)
start = i + 1
i = right[i] if s[i] == "(" else i + 1
expression += parse_term(start, i)
return expression
s = input()
n = len(s)
p = [1]
for _ in range(n):
p.append(p[-1] * 10 % MOD)
q = [1]
for _ in range(n):
q.append(q[-1] * 10 % (MOD - 1))
right = [None] * n
stack = []
for i, x in enumerate(s):
if x == "(":
stack.append(i)
elif x == ")":
right[stack.pop()] = i
print(parse_expression(0, n).value)
|
8VC Venture Cup 2017 - Final Round
|
CF
| 2,017 | 2 | 512 |
Long number
|
Consider the following grammar:
- <expression> ::= <term> | <expression> '+' <term>
- <term> ::= <number> | <number> '-' <number> | <number> '(' <expression> ')'
- <number> ::= <pos_digit> | <number> <digit>
- <digit> ::= '0' | <pos_digit>
- <pos_digit> ::= '1' | '2' | '3' | '4' | '5' | '6' | '7' | '8' | '9'
This grammar describes a number in decimal system using the following rules:
- <number> describes itself,
- <number>-<number> (l-r, l ≤ r) describes integer which is concatenation of all integers from l to r, written without leading zeros. For example, 8-11 describes 891011,
- <number>(<expression>) describes integer which is concatenation of <number> copies of integer described by <expression>,
- <expression>+<term> describes integer which is concatenation of integers described by <expression> and <term>.
For example, 2(2-4+1)+2(2(17)) describes the integer 2341234117171717.
You are given an expression in the given grammar. Print the integer described by it modulo 109 + 7.
|
The only line contains a non-empty string at most 105 characters long which is valid according to the given grammar. In particular, it means that in terms l-r l ≤ r holds.
|
Print single integer — the number described by the expression modulo 109 + 7.
| null | null |
[{"input": "8-11", "output": "891011"}, {"input": "2(2-4+1)+2(2(17))", "output": "100783079"}, {"input": "1234-5678", "output": "745428774"}, {"input": "1+2+3+4-5+6+7-9", "output": "123456789"}]
| 3,400 |
["expression parsing", "math", "number theory"]
| 41 |
[{"input": "8-11\r\n", "output": "891011\r\n"}, {"input": "2(2-4+1)+2(2(17))\r\n", "output": "100783079\r\n"}, {"input": "1234-5678\r\n", "output": "745428774\r\n"}, {"input": "1+2+3+4-5+6+7-9\r\n", "output": "123456789\r\n"}, {"input": "598777\r\n", "output": "598777\r\n"}, {"input": "49603501749575096738857\r\n", "output": "586922407\r\n"}, {"input": "11-57\r\n", "output": "486296559\r\n"}, {"input": "7177-57797\r\n", "output": "110843609\r\n"}, {"input": "4510812433666-7741104736713\r\n", "output": "665706430\r\n"}, {"input": "4778066503(27032-80044+16+51+58(9)+5668114736297420472+2336+6-117+8495(7265))\r\n", "output": "865860620\r\n"}, {"input": "9678329648012859556810940272201-70370609657815505164700664074744+28870231062776633852(2098997519485)\r\n", "output": "219564419\r\n"}, {"input": "52(6(1)+100000000(2-10))+777(40000000(7-20)+876543)\r\n", "output": "133241273\r\n"}, {"input": "1(1000000005(5)+4)\r\n", "output": "1000000006\r\n"}, {"input": "1-1\r\n", "output": "1\r\n"}, {"input": "10-10\r\n", "output": "10\r\n"}, {"input": "1+2\r\n", "output": "12\r\n"}, {"input": "9\r\n", "output": "9\r\n"}, {"input": "1(1)+1(1)\r\n", "output": "11\r\n"}]
| false |
stdio
| null | true |
69/C
|
69
|
C
|
PyPy 3
|
TESTS
| 2 | 248 | 0 |
81754120
|
import sys
inp = sys.stdin.readlines()
all_lines = []
first_line = inp[0].split(' ')
allies = int(first_line[0])
basic = int(first_line[1])
composite= int(first_line[2])
friend_purchases = int(first_line[3])
friend = {}
for count in range(allies+1)[1:]:
friend[count] = []
for owned in inp[basic+1+composite:basic+1+composite+friend_purchases]:
owned = owned.strip()
friend_count = int(owned.split(' ')[0])
friend_item = str(owned.split(' ')[1])
friend[friend_count].append(friend_item)
new = {}
for line in inp[basic+1:basic+1+composite]:
line = line.strip()
item = line.split(': ')
item_name = item[0]
item_info = item[1].split(', ')
item_info = [item.split(' ') for item in item_info]
for friend_count, attribute in friend.items():
temp = attribute.copy()
not_enough = False
item_length = 0
for name in item_info:
for num_item in range(int(name[1])):
item_length +=1
if name[0] in temp:
temp.remove(name[0])
if len(attribute)-len(temp) != item_length:
pass
else:
temp.append(item_name)
friend[friend_count] = temp
for key, item in friend.items():
print(key)
item.sort()
counted = {i:item.count(i) for i in item}
for k, count in counted.items():
print(k, str(count))
| 47 | 374 | 2,969,600 |
95919675
|
import sys
from array import array # noqa: F401
from collections import defaultdict
def input():
return sys.stdin.buffer.readline().decode('utf-8')
k, n, m, q = map(int, input().split())
basic = [input().rstrip() for _ in range(n)]
composite = defaultdict(list)
for _ in range(m):
name, desc = input().split(':')
for item in desc.split(','):
pair = item.split()
pair[1] = int(pair[1])
composite[name].append(pair)
backpack = [defaultdict(int) for _ in range(k + 1)]
for _ in range(q):
f, name = input().split()
f = int(f)
backpack[f][name] += 1
for key, recipe in composite.items():
if all(backpack[f][base] >= cnt for base, cnt in recipe):
for base, cnt in recipe:
backpack[f][base] -= cnt
backpack[f][key] += 1
for i in range(1, k + 1):
ans = [f'{name} {cnt}' for name, cnt in backpack[i].items() if cnt > 0]
ans.sort()
print(len(ans))
if ans:
print('\n'.join(ans))
|
Codeforces Beta Round 63 (Div. 2)
|
CF
| 2,011 | 2 | 256 |
Game
|
In one school with Vasya there is a student Kostya. Kostya does not like physics, he likes different online games. Every day, having come home, Kostya throws his bag in the farthest corner and sits down at his beloved computer. Kostya even eats glued to the game. A few days ago Kostya bought a new RPG game "HaresButtle", which differs from all other games in this genre. It has a huge number of artifacts. As we know, artifacts are divided into basic and composite ones. Only the basic artifacts are available on sale. More powerful composite artifacts are collected from some number of basic artifacts.
After the composing composite artifact, all the components disappear.
Kostya is the head of the alliance, so he has to remember, what artifacts has not only himself, but also his allies. You must identify by sequence of artifacts purchased by Kostya and his allies, how many and which artifacts has been collected by each of them. It is believed that initially no one has any artifacts.
|
The first line has 4 natural numbers: k (1 ≤ k ≤ 100) — the number of Kostya's allies, n (1 ≤ n ≤ 50) — the number of basic artifacts, m (0 ≤ m ≤ 50) — the number of composite artifacts, q (1 ≤ q ≤ 500) — the number of his friends' purchases. The following n lines contain the names of basic artifacts. After them m lines contain the descriptions of composite artifacts in the following format:
<Art. Name>: <Art. №1> <Art. №1 Number>, <Art. №2> <Art. №2 Number>, ... <Art. №X> <Art. №Х Number>
All the numbers are natural numbers not exceeding 100 (1 ≤ X ≤ n).
The names of all artifacts are different, they are composed of lowercase Latin letters, and the length of each name is from 1 to 100 characters inclusive. All the words in the format of the description of a composite artifact are separated by exactly one space. It is guaranteed that all components of the new artifact are different and have already been met in the input data as the names of basic artifacts.
Next, each of the following q lines is characterized by the number ai, the number of a friend who has bought the artifact (1 ≤ ai ≤ k), and the name of the purchased basic artifact. Let's assume that the backpacks of the heroes are infinitely large and any artifact bought later can fit in there.
It is guaranteed that after the i-th purchase no more than one opportunity to collect the composite artifact appears. If such an opportunity arose, the hero must take advantage of it.
|
The output file should consist of k blocks. The first line should contain number bi — the number of different artifacts the i-th ally has. Then the block should contain bi lines with the names of these artifacts and the number of these artifacts. At that the lines should be printed in accordance with the lexicographical order of the names of the artifacts. In each block all the artifacts must be different, and all the numbers except the bi should be positive.
| null | null |
[{"input": "2 3 2 5\ndesolator\nrefresher\nperseverance\nvanguard: desolator 1, refresher 1\nmaelstorm: perseverance 2\n1 desolator\n2 perseverance\n1 refresher\n2 desolator\n2 perseverance", "output": "1\nvanguard 1\n2\ndesolator 1\nmaelstorm 1"}]
| 2,000 |
["implementation"]
| 47 |
[{"input": "2 3 2 5\r\ndesolator\r\nrefresher\r\nperseverance\r\nvanguard: desolator 1, refresher 1\r\nmaelstorm: perseverance 2\r\n1 desolator\r\n2 perseverance\r\n1 refresher\r\n2 desolator\r\n2 perseverance\r\n", "output": "1\r\nvanguard 1\r\n2\r\ndesolator 1\r\nmaelstorm 1\r\n"}, {"input": "2 3 2 5\r\na\r\nb\r\nc\r\nd: a 1, b 1\r\ne: c 2\r\n1 a\r\n2 c\r\n1 b\r\n2 c\r\n2 c\r\n", "output": "1\r\nd 1\r\n2\r\nc 1\r\ne 1\r\n"}, {"input": "10 1 1 3\r\na\r\nb: a 2\r\n1 a\r\n1 a\r\n1 a\r\n", "output": "2\r\na 1\r\nb 1\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "2 3 2 5\r\ndesolator\r\nrefresher\r\nperseverance\r\nvanguard: desolator 1, refresher 1\r\nmaelstorm: refresher 2\r\n1 desolator\r\n2 perseverance\r\n1 refresher\r\n1 refresher\r\n2 perseverance\r\n", "output": "2\r\nrefresher 1\r\nvanguard 1\r\n1\r\nperseverance 2\r\n"}, {"input": "1 1 1 1\r\na\r\nb: a 1\r\n1 a\r\n", "output": "1\r\nb 1\r\n"}, {"input": "2 3 2 5\r\ndesolator\r\nrefresher\r\nperseverance\r\nvanguard: desolator 1, refresher 3\r\nmaelstorm: refresher 2\r\n1 desolator\r\n2 perseverance\r\n1 refresher\r\n1 refresher\r\n2 perseverance\r\n", "output": "2\r\ndesolator 1\r\nmaelstorm 1\r\n1\r\nperseverance 2\r\n"}, {"input": "1 2 0 2\r\nb\r\na\r\n1 a\r\n1 b\r\n", "output": "2\r\na 1\r\nb 1\r\n"}, {"input": "2 3 2 5\r\na\r\nb\r\nc\r\nd: a 1, b 21\r\ne: c 2\r\n1 a\r\n2 c\r\n1 b\r\n2 c\r\n2 c\r\n", "output": "2\r\na 1\r\nb 1\r\n2\r\nc 1\r\ne 1\r\n"}, {"input": "1 1 2 2\r\na\r\nb: a 2\r\nc: a 1\r\n1 a\r\n1 a\r\n", "output": "1\r\nc 2\r\n"}]
| false |
stdio
| null | true |
393/B
|
393
|
B
|
PyPy 3
|
TESTS
| 0 | 124 | 0 |
93462429
|
n=int(input())
w=[list(map(int,input().split()))for i in range(n)]
a=[[0for i in range(n)]for i in range(n)]
b=[[0for i in range(n)]for i in range(n)]
for i in range(n):
for j in range(n):
if i==j:
a[i][j]=w[i][j]
else:
a[i][j]=w[i][j]//2
b[i][j]=w[i][j]-a[i][j]
for i in a:
print(*i)
for i in b:
print(*i)
| 40 | 155 | 9,113,600 |
127862133
|
n=int(input())
lis=[]
for x in range(n):
lis.append(list(map(int,input().split())))
A=[]
B=[]
for x in range(n):
A.append([ (lis[x][i]+lis[i][x])/2 for i in range(n) ])
B.append([lis[x][i]-A[x][i] for i in range(n)])
for x in A:
for y in x:
print(y,end=' ')
print()
for x in B:
for y in x:
print(y,end=' ')
print()
|
Codeforces Round 230 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Three matrices
|
Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold:
- Aij = Aji, for all i, j (1 ≤ i, j ≤ n);
- Bij = - Bji, for all i, j (1 ≤ i, j ≤ n);
- Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n).
Can you solve the problem?
|
The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij (0 ≤ |Wij| < 1717).
|
The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.
| null | null |
[{"input": "2\n1 4\n3 2", "output": "1.00000000 3.50000000\n3.50000000 2.00000000\n0.00000000 0.50000000\n-0.50000000 0.00000000"}, {"input": "3\n1 2 3\n4 5 6\n7 8 9", "output": "1.00000000 3.00000000 5.00000000\n3.00000000 5.00000000 7.00000000\n5.00000000 7.00000000 9.00000000\n0.00000000 -1.00000000 -2.00000000\n1.00000000 0.00000000 -1.00000000\n2.00000000 1.00000000 0.00000000"}]
| null |
[]
| 40 |
[{"input": "2\r\n1 4\r\n3 2\r\n", "output": "1.00000000 3.50000000\r\n3.50000000 2.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}, {"input": "3\r\n1 2 3\r\n4 5 6\r\n7 8 9\r\n", "output": "1.00000000 3.00000000 5.00000000\r\n3.00000000 5.00000000 7.00000000\r\n5.00000000 7.00000000 9.00000000\r\n0.00000000 -1.00000000 -2.00000000\r\n1.00000000 0.00000000 -1.00000000\r\n2.00000000 1.00000000 0.00000000\r\n"}, {"input": "8\r\n62 567 1382 1279 728 1267 1262 568\r\n77 827 717 1696 774 248 822 1266\r\n563 612 995 424 1643 1197 338 1141\r\n1579 806 1254 468 184 1571 716 772\r\n1087 182 1312 772 605 1674 720 1349\r\n1393 988 873 157 403 301 1519 1192\r\n1085 625 1395 1087 847 1360 1004 594\r\n1368 1056 916 839 472 840 53 1238\r\n", "output": "62.000000000 322.000000000 972.500000000 1429.000000000 907.500000000 1330.000000000 1173.500000000 968.000000000\n322.000000000 827.000000000 664.500000000 1251.000000000 478.000000000 618.000000000 723.500000000 1161.000000000\n972.500000000 664.500000000 995.000000000 839.000000000 1477.500000000 1035.000000000 866.500000000 1028.500000000\n1429.000000000 1251.000000000 839.000000000 468.000000000 478.000000000 864.000000000 901.500000000 805.500000000\n907.500000000 478.000000000 1477.500000000 478.000000000 605.000000000 1038.500000000 783.500000000 910.500000000\n1330.000000000 618.000000000 1035.000000000 864.000000000 1038.500000000 301.000000000 1439.500000000 1016.000000000\n1173.500000000 723.500000000 866.500000000 901.500000000 783.500000000 1439.500000000 1004.000000000 323.500000000\n968.000000000 1161.000000000 1028.500000000 805.500000000 910.500000000 1016.000000000 323.500000000 1238.000000000\n0.000000000 245.000000000 409.500000000 -150.000000000 -179.500000000 -63.000000000 88.500000000 -400.000000000\n-245.000000000 0.000000000 52.500000000 445.000000000 296.000000000 -370.000000000 98.500000000 105.000000000\n-409.500000000 -52.500000000 0.000000000 -415.000000000 165.500000000 162.000000000 -528.500000000 112.500000000\n150.000000000 -445.000000000 415.000000000 0.000000000 -294.000000000 707.000000000 -185.500000000 -33.500000000\n179.500000000 -296.000000000 -165.500000000 294.000000000 0.000000000 635.500000000 -63.500000000 438.500000000\n63.000000000 370.000000000 -162.000000000 -707.000000000 -635.500000000 0.000000000 79.500000000 176.000000000\n-88.500000000 -98.500000000 528.500000000 185.500000000 63.500000000 -79.500000000 0.000000000 270.500000000\n400.000000000 -105.000000000 -112.500000000 33.500000000 -438.500000000 -176.000000000 -270.500000000 0.000000000\n"}, {"input": "7\r\n926 41 1489 72 749 375 940\r\n464 1148 858 1010 285 1469 1506\r\n1112 1087 225 917 480 511 1090\r\n759 945 627 230 220 1456 529\r\n318 83 203 134 1192 1167 6\r\n440 1158 1614 683 1358 1140 1196\r\n1175 900 126 1562 1220 813 148\r\n", "output": "926.000000000 252.500000000 1300.500000000 415.500000000 533.500000000 407.500000000 1057.500000000\n252.500000000 1148.000000000 972.500000000 977.500000000 184.000000000 1313.500000000 1203.000000000\n1300.500000000 972.500000000 225.000000000 772.000000000 341.500000000 1062.500000000 608.000000000\n415.500000000 977.500000000 772.000000000 230.000000000 177.000000000 1069.500000000 1045.500000000\n533.500000000 184.000000000 341.500000000 177.000000000 1192.000000000 1262.500000000 613.000000000\n407.500000000 1313.500000000 1062.500000000 1069.500000000 1262.500000000 1140.000000000 1004.500000000\n1057.500000000 1203.000000000 608.000000000 1045.500000000 613.000000000 1004.500000000 148.000000000\n0.000000000 -211.500000000 188.500000000 -343.500000000 215.500000000 -32.500000000 -117.500000000\n211.500000000 0.000000000 -114.500000000 32.500000000 101.000000000 155.500000000 303.000000000\n-188.500000000 114.500000000 0.000000000 145.000000000 138.500000000 -551.500000000 482.000000000\n343.500000000 -32.500000000 -145.000000000 0.000000000 43.000000000 386.500000000 -516.500000000\n-215.500000000 -101.000000000 -138.500000000 -43.000000000 0.000000000 -95.500000000 -607.000000000\n32.500000000 -155.500000000 551.500000000 -386.500000000 95.500000000 0.000000000 191.500000000\n117.500000000 -303.000000000 -482.000000000 516.500000000 607.000000000 -191.500000000 0.000000000\n"}, {"input": "1\r\n1\r\n", "output": "1.00000000\r\n0.00000000\r\n"}, {"input": "1\r\n0\r\n", "output": "0.00000000\r\n0.00000000\r\n"}, {"input": "2\r\n0 0\r\n0 0\r\n", "output": "0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n"}, {"input": "2\r\n0 1\r\n0 1\r\n", "output": "0.00000000 0.50000000\r\n0.50000000 1.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}]
| false |
stdio
|
import sys
def is_close(a, b):
abs_tol = 1e-4
rel_tol = 1e-4
diff = abs(a - b)
if diff <= abs_tol:
return True
max_val = max(abs(a), abs(b))
return diff <= rel_tol * max_val
def main(input_path, output_path, submission_path):
with open(input_path) as f:
lines = f.read().splitlines()
n = int(lines[0])
w = []
for line in lines[1:1+n]:
row = list(map(int, line.split()))
w.append(row)
with open(submission_path) as f:
submission_lines = f.read().splitlines()
if len(submission_lines) != 2 * n:
print(0)
return
A = []
B = []
for i in range(n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
A.append(row)
for i in range(n, 2 * n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
B.append(row)
for i in range(n):
for j in range(n):
if not is_close(A[i][j], A[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
if not is_close(B[i][j], -B[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
sum_ab = A[i][j] + B[i][j]
expected = w[i][j]
if not is_close(sum_ab, expected):
print(0)
return
print(1)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
393/B
|
393
|
B
|
Python 3
|
TESTS
| 0 | 93 | 0 |
61959882
|
n = int(input())
mat = []
for k in range(0, n):
mat.append(list(map(int, input().rstrip().split())))
A = mat
B = [[0] * n for _ in range(n)]
for i in range(1, n):
for j in range(0, i):
a = (mat[i][j] + mat[j][i]) / 2
b = mat[j][i] - a
A[i][j] = a
B[i][j] = b
A[j][i] = a
B[j][i] = -b
for i in range(n):
print(*A[i])
for i in range(n):
print(*B[i])
| 40 | 171 | 1,945,600 |
114797070
|
R = lambda: list(map(int, input().split()))
n = int(input())
w = [R() for _ in ' '*n]
wd = [[0]*n for _ in ' '*n]
for i in range(n):
for j in range(n):
wd[i][j] = w[j][i]
a = [[0]*n for _ in ' '*n]
b = [[0]*n for _ in ' '*n]
for i in range(n):
for j in range(n):
a[i][j] = (w[i][j] + wd[i][j])/2
b[i][j] = (w[i][j] - wd[i][j])/2
for i in a: print(*i)
for i in b: print(*i)
|
Codeforces Round 230 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Three matrices
|
Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold:
- Aij = Aji, for all i, j (1 ≤ i, j ≤ n);
- Bij = - Bji, for all i, j (1 ≤ i, j ≤ n);
- Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n).
Can you solve the problem?
|
The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij (0 ≤ |Wij| < 1717).
|
The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.
| null | null |
[{"input": "2\n1 4\n3 2", "output": "1.00000000 3.50000000\n3.50000000 2.00000000\n0.00000000 0.50000000\n-0.50000000 0.00000000"}, {"input": "3\n1 2 3\n4 5 6\n7 8 9", "output": "1.00000000 3.00000000 5.00000000\n3.00000000 5.00000000 7.00000000\n5.00000000 7.00000000 9.00000000\n0.00000000 -1.00000000 -2.00000000\n1.00000000 0.00000000 -1.00000000\n2.00000000 1.00000000 0.00000000"}]
| null |
[]
| 40 |
[{"input": "2\r\n1 4\r\n3 2\r\n", "output": "1.00000000 3.50000000\r\n3.50000000 2.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}, {"input": "3\r\n1 2 3\r\n4 5 6\r\n7 8 9\r\n", "output": "1.00000000 3.00000000 5.00000000\r\n3.00000000 5.00000000 7.00000000\r\n5.00000000 7.00000000 9.00000000\r\n0.00000000 -1.00000000 -2.00000000\r\n1.00000000 0.00000000 -1.00000000\r\n2.00000000 1.00000000 0.00000000\r\n"}, {"input": "8\r\n62 567 1382 1279 728 1267 1262 568\r\n77 827 717 1696 774 248 822 1266\r\n563 612 995 424 1643 1197 338 1141\r\n1579 806 1254 468 184 1571 716 772\r\n1087 182 1312 772 605 1674 720 1349\r\n1393 988 873 157 403 301 1519 1192\r\n1085 625 1395 1087 847 1360 1004 594\r\n1368 1056 916 839 472 840 53 1238\r\n", "output": "62.000000000 322.000000000 972.500000000 1429.000000000 907.500000000 1330.000000000 1173.500000000 968.000000000\n322.000000000 827.000000000 664.500000000 1251.000000000 478.000000000 618.000000000 723.500000000 1161.000000000\n972.500000000 664.500000000 995.000000000 839.000000000 1477.500000000 1035.000000000 866.500000000 1028.500000000\n1429.000000000 1251.000000000 839.000000000 468.000000000 478.000000000 864.000000000 901.500000000 805.500000000\n907.500000000 478.000000000 1477.500000000 478.000000000 605.000000000 1038.500000000 783.500000000 910.500000000\n1330.000000000 618.000000000 1035.000000000 864.000000000 1038.500000000 301.000000000 1439.500000000 1016.000000000\n1173.500000000 723.500000000 866.500000000 901.500000000 783.500000000 1439.500000000 1004.000000000 323.500000000\n968.000000000 1161.000000000 1028.500000000 805.500000000 910.500000000 1016.000000000 323.500000000 1238.000000000\n0.000000000 245.000000000 409.500000000 -150.000000000 -179.500000000 -63.000000000 88.500000000 -400.000000000\n-245.000000000 0.000000000 52.500000000 445.000000000 296.000000000 -370.000000000 98.500000000 105.000000000\n-409.500000000 -52.500000000 0.000000000 -415.000000000 165.500000000 162.000000000 -528.500000000 112.500000000\n150.000000000 -445.000000000 415.000000000 0.000000000 -294.000000000 707.000000000 -185.500000000 -33.500000000\n179.500000000 -296.000000000 -165.500000000 294.000000000 0.000000000 635.500000000 -63.500000000 438.500000000\n63.000000000 370.000000000 -162.000000000 -707.000000000 -635.500000000 0.000000000 79.500000000 176.000000000\n-88.500000000 -98.500000000 528.500000000 185.500000000 63.500000000 -79.500000000 0.000000000 270.500000000\n400.000000000 -105.000000000 -112.500000000 33.500000000 -438.500000000 -176.000000000 -270.500000000 0.000000000\n"}, {"input": "7\r\n926 41 1489 72 749 375 940\r\n464 1148 858 1010 285 1469 1506\r\n1112 1087 225 917 480 511 1090\r\n759 945 627 230 220 1456 529\r\n318 83 203 134 1192 1167 6\r\n440 1158 1614 683 1358 1140 1196\r\n1175 900 126 1562 1220 813 148\r\n", "output": "926.000000000 252.500000000 1300.500000000 415.500000000 533.500000000 407.500000000 1057.500000000\n252.500000000 1148.000000000 972.500000000 977.500000000 184.000000000 1313.500000000 1203.000000000\n1300.500000000 972.500000000 225.000000000 772.000000000 341.500000000 1062.500000000 608.000000000\n415.500000000 977.500000000 772.000000000 230.000000000 177.000000000 1069.500000000 1045.500000000\n533.500000000 184.000000000 341.500000000 177.000000000 1192.000000000 1262.500000000 613.000000000\n407.500000000 1313.500000000 1062.500000000 1069.500000000 1262.500000000 1140.000000000 1004.500000000\n1057.500000000 1203.000000000 608.000000000 1045.500000000 613.000000000 1004.500000000 148.000000000\n0.000000000 -211.500000000 188.500000000 -343.500000000 215.500000000 -32.500000000 -117.500000000\n211.500000000 0.000000000 -114.500000000 32.500000000 101.000000000 155.500000000 303.000000000\n-188.500000000 114.500000000 0.000000000 145.000000000 138.500000000 -551.500000000 482.000000000\n343.500000000 -32.500000000 -145.000000000 0.000000000 43.000000000 386.500000000 -516.500000000\n-215.500000000 -101.000000000 -138.500000000 -43.000000000 0.000000000 -95.500000000 -607.000000000\n32.500000000 -155.500000000 551.500000000 -386.500000000 95.500000000 0.000000000 191.500000000\n117.500000000 -303.000000000 -482.000000000 516.500000000 607.000000000 -191.500000000 0.000000000\n"}, {"input": "1\r\n1\r\n", "output": "1.00000000\r\n0.00000000\r\n"}, {"input": "1\r\n0\r\n", "output": "0.00000000\r\n0.00000000\r\n"}, {"input": "2\r\n0 0\r\n0 0\r\n", "output": "0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n"}, {"input": "2\r\n0 1\r\n0 1\r\n", "output": "0.00000000 0.50000000\r\n0.50000000 1.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}]
| false |
stdio
|
import sys
def is_close(a, b):
abs_tol = 1e-4
rel_tol = 1e-4
diff = abs(a - b)
if diff <= abs_tol:
return True
max_val = max(abs(a), abs(b))
return diff <= rel_tol * max_val
def main(input_path, output_path, submission_path):
with open(input_path) as f:
lines = f.read().splitlines()
n = int(lines[0])
w = []
for line in lines[1:1+n]:
row = list(map(int, line.split()))
w.append(row)
with open(submission_path) as f:
submission_lines = f.read().splitlines()
if len(submission_lines) != 2 * n:
print(0)
return
A = []
B = []
for i in range(n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
A.append(row)
for i in range(n, 2 * n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
B.append(row)
for i in range(n):
for j in range(n):
if not is_close(A[i][j], A[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
if not is_close(B[i][j], -B[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
sum_ab = A[i][j] + B[i][j]
expected = w[i][j]
if not is_close(sum_ab, expected):
print(0)
return
print(1)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
393/B
|
393
|
B
|
Python 3
|
TESTS
| 0 | 77 | 307,200 |
69578715
|
# Design_by_JOKER
import math
import cmath
n = int(input())
w = []
a = []
b = []
for _ in range(n):
w.append(list(map(int, input().split())))
a.append([0.0]*n)
b.append([0.0]*n)
for x in range(n):
a[x][x] = 1.0*w[x][x]
for y in range(x):
a[x][y] = a[y][x] = (w[x][y] + w[y][x])/2
b[x][y] = a[x][y] - w[x][y]
b[y][x] = -b[x][y]
for x in range(n):
for y in range(n):
print(a[x][y], end=' ')
print("")
for x in range(n):
for y in range(n):
print(b[x][y], end=' ')
print("")
| 40 | 171 | 8,806,400 |
130600943
|
# import numpy as np
n = int(input())
arr = []
for i in range(n):
arr1 = [int(x) for x in input().split()]
arr.append(arr1)
dasharr = [[None for j in range(n)] for i in range(n)]
for i in range(n):
for j in range(n):
dasharr[i][j] = arr[j][i]
a = [[None for i in range(n)] for i in range(n)]
b = [[None for i in range(n)] for i in range(n)]
for i in range(n):
for j in range(n):
a[i][j] = arr[i][j]/2 + dasharr[i][j]/2
b[i][j] = arr[i][j]/2 - dasharr[i][j]/2
# print(dasharr)
for sub in a :
print(*sub)
for sub in b:
print(*sub)
|
Codeforces Round 230 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Three matrices
|
Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold:
- Aij = Aji, for all i, j (1 ≤ i, j ≤ n);
- Bij = - Bji, for all i, j (1 ≤ i, j ≤ n);
- Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n).
Can you solve the problem?
|
The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij (0 ≤ |Wij| < 1717).
|
The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.
| null | null |
[{"input": "2\n1 4\n3 2", "output": "1.00000000 3.50000000\n3.50000000 2.00000000\n0.00000000 0.50000000\n-0.50000000 0.00000000"}, {"input": "3\n1 2 3\n4 5 6\n7 8 9", "output": "1.00000000 3.00000000 5.00000000\n3.00000000 5.00000000 7.00000000\n5.00000000 7.00000000 9.00000000\n0.00000000 -1.00000000 -2.00000000\n1.00000000 0.00000000 -1.00000000\n2.00000000 1.00000000 0.00000000"}]
| null |
[]
| 40 |
[{"input": "2\r\n1 4\r\n3 2\r\n", "output": "1.00000000 3.50000000\r\n3.50000000 2.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}, {"input": "3\r\n1 2 3\r\n4 5 6\r\n7 8 9\r\n", "output": "1.00000000 3.00000000 5.00000000\r\n3.00000000 5.00000000 7.00000000\r\n5.00000000 7.00000000 9.00000000\r\n0.00000000 -1.00000000 -2.00000000\r\n1.00000000 0.00000000 -1.00000000\r\n2.00000000 1.00000000 0.00000000\r\n"}, {"input": "8\r\n62 567 1382 1279 728 1267 1262 568\r\n77 827 717 1696 774 248 822 1266\r\n563 612 995 424 1643 1197 338 1141\r\n1579 806 1254 468 184 1571 716 772\r\n1087 182 1312 772 605 1674 720 1349\r\n1393 988 873 157 403 301 1519 1192\r\n1085 625 1395 1087 847 1360 1004 594\r\n1368 1056 916 839 472 840 53 1238\r\n", "output": "62.000000000 322.000000000 972.500000000 1429.000000000 907.500000000 1330.000000000 1173.500000000 968.000000000\n322.000000000 827.000000000 664.500000000 1251.000000000 478.000000000 618.000000000 723.500000000 1161.000000000\n972.500000000 664.500000000 995.000000000 839.000000000 1477.500000000 1035.000000000 866.500000000 1028.500000000\n1429.000000000 1251.000000000 839.000000000 468.000000000 478.000000000 864.000000000 901.500000000 805.500000000\n907.500000000 478.000000000 1477.500000000 478.000000000 605.000000000 1038.500000000 783.500000000 910.500000000\n1330.000000000 618.000000000 1035.000000000 864.000000000 1038.500000000 301.000000000 1439.500000000 1016.000000000\n1173.500000000 723.500000000 866.500000000 901.500000000 783.500000000 1439.500000000 1004.000000000 323.500000000\n968.000000000 1161.000000000 1028.500000000 805.500000000 910.500000000 1016.000000000 323.500000000 1238.000000000\n0.000000000 245.000000000 409.500000000 -150.000000000 -179.500000000 -63.000000000 88.500000000 -400.000000000\n-245.000000000 0.000000000 52.500000000 445.000000000 296.000000000 -370.000000000 98.500000000 105.000000000\n-409.500000000 -52.500000000 0.000000000 -415.000000000 165.500000000 162.000000000 -528.500000000 112.500000000\n150.000000000 -445.000000000 415.000000000 0.000000000 -294.000000000 707.000000000 -185.500000000 -33.500000000\n179.500000000 -296.000000000 -165.500000000 294.000000000 0.000000000 635.500000000 -63.500000000 438.500000000\n63.000000000 370.000000000 -162.000000000 -707.000000000 -635.500000000 0.000000000 79.500000000 176.000000000\n-88.500000000 -98.500000000 528.500000000 185.500000000 63.500000000 -79.500000000 0.000000000 270.500000000\n400.000000000 -105.000000000 -112.500000000 33.500000000 -438.500000000 -176.000000000 -270.500000000 0.000000000\n"}, {"input": "7\r\n926 41 1489 72 749 375 940\r\n464 1148 858 1010 285 1469 1506\r\n1112 1087 225 917 480 511 1090\r\n759 945 627 230 220 1456 529\r\n318 83 203 134 1192 1167 6\r\n440 1158 1614 683 1358 1140 1196\r\n1175 900 126 1562 1220 813 148\r\n", "output": "926.000000000 252.500000000 1300.500000000 415.500000000 533.500000000 407.500000000 1057.500000000\n252.500000000 1148.000000000 972.500000000 977.500000000 184.000000000 1313.500000000 1203.000000000\n1300.500000000 972.500000000 225.000000000 772.000000000 341.500000000 1062.500000000 608.000000000\n415.500000000 977.500000000 772.000000000 230.000000000 177.000000000 1069.500000000 1045.500000000\n533.500000000 184.000000000 341.500000000 177.000000000 1192.000000000 1262.500000000 613.000000000\n407.500000000 1313.500000000 1062.500000000 1069.500000000 1262.500000000 1140.000000000 1004.500000000\n1057.500000000 1203.000000000 608.000000000 1045.500000000 613.000000000 1004.500000000 148.000000000\n0.000000000 -211.500000000 188.500000000 -343.500000000 215.500000000 -32.500000000 -117.500000000\n211.500000000 0.000000000 -114.500000000 32.500000000 101.000000000 155.500000000 303.000000000\n-188.500000000 114.500000000 0.000000000 145.000000000 138.500000000 -551.500000000 482.000000000\n343.500000000 -32.500000000 -145.000000000 0.000000000 43.000000000 386.500000000 -516.500000000\n-215.500000000 -101.000000000 -138.500000000 -43.000000000 0.000000000 -95.500000000 -607.000000000\n32.500000000 -155.500000000 551.500000000 -386.500000000 95.500000000 0.000000000 191.500000000\n117.500000000 -303.000000000 -482.000000000 516.500000000 607.000000000 -191.500000000 0.000000000\n"}, {"input": "1\r\n1\r\n", "output": "1.00000000\r\n0.00000000\r\n"}, {"input": "1\r\n0\r\n", "output": "0.00000000\r\n0.00000000\r\n"}, {"input": "2\r\n0 0\r\n0 0\r\n", "output": "0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n"}, {"input": "2\r\n0 1\r\n0 1\r\n", "output": "0.00000000 0.50000000\r\n0.50000000 1.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}]
| false |
stdio
|
import sys
def is_close(a, b):
abs_tol = 1e-4
rel_tol = 1e-4
diff = abs(a - b)
if diff <= abs_tol:
return True
max_val = max(abs(a), abs(b))
return diff <= rel_tol * max_val
def main(input_path, output_path, submission_path):
with open(input_path) as f:
lines = f.read().splitlines()
n = int(lines[0])
w = []
for line in lines[1:1+n]:
row = list(map(int, line.split()))
w.append(row)
with open(submission_path) as f:
submission_lines = f.read().splitlines()
if len(submission_lines) != 2 * n:
print(0)
return
A = []
B = []
for i in range(n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
A.append(row)
for i in range(n, 2 * n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
B.append(row)
for i in range(n):
for j in range(n):
if not is_close(A[i][j], A[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
if not is_close(B[i][j], -B[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
sum_ab = A[i][j] + B[i][j]
expected = w[i][j]
if not is_close(sum_ab, expected):
print(0)
return
print(1)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
886/F
|
886
|
F
|
Python 3
|
PRETESTS
| 2 | 61 | 0 |
32267865
|
n=int(input())
a=[]
ma=0
for i in range(n):
s=input()
z=s.split()
k1=int(z[0])-1
k2=int(z[1])-1
ma=max(k1,k2,ma)
a.append(s)
b=[[0]*(ma+1) for i in range(ma+1)]
for i in a:
z=i.split()
k1=int(z[0])-1
k2=int(z[1])-1
b[k1][k2]=1
a=[]
k=0
fl=True
for i in range(ma+1):
if b[i][i]==1 and fl:
k+=1
fl=False
for j in range(i+1,ma+1):
if b[i][j]==b[j][i] and b[j][i]!=0:
k+=2
if k==0:
print('-1')
else:
print(k)
| 51 | 794 | 19,046,400 |
211944348
|
from fractions import Fraction
import time
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def to_tuple(self):
return (self.x, self.y)
def __repr__(self):
return "Point({}, {})".format(self.x, self.y)
def __eq__(self, other):
return self.to_tuple() == other.to_tuple()
def __hash__(self):
return hash(self.to_tuple())
def __neg__(self):
return Point(-self.x, -self.y)
def __add__(self, other):
return Point(self.x+other.x, self.y+other.y)
def __sub__(self, other):
return self+(-other)
def scalar_mul(self, mu):
return Point(mu*self.x, mu*self.y)
def int_divide(self, den):
return Point(self.x//den, self.y//den)
class Line:
def __init__(self, a, b, c):
# ax+by+c=0
self.a = a
self.b = b
self.c = c
def __repr__(self):
return "{}*x + {}*y + {} = 0".format(self.a, self.b, self.c)
@classmethod
def between_two_points(cls, P, Q):
return cls(P.y-Q.y, Q.x-P.x, P.x*Q.y-P.y*Q.x)
def evaluate(self, P):
return self.a*P.x+self.b*P.y+self.c
def direction(self):
if self.a == 0:
return (0, 1)
return (1, Fraction(self.b, self.a))
true_start = time.time()
n = int(input())
points = set()
center = Point(0, 0)
for i in range(n):
row = input().split(" ")
cur = Point(int(row[0]), int(row[1])).scalar_mul(2*n)
center += cur
points.add(cur)
center = center.int_divide(n)
dcenter = center+center
sym_points_set = set()
for p in points:
sym_points_set.add(dcenter-p)
nosym = list(points - sym_points_set)
if len(nosym) == 0:
print(-1)
exit(0)
cnt = 0
p0 = nosym[0]
good_lines = set()
for p in nosym:
m = (p+p0).int_divide(2)
line = Line.between_two_points(m, center)
distances = list(map(line.evaluate, nosym))
ok = True
mydict = {}
for dd in distances:
dda = abs(dd)
if dda not in mydict:
mydict[dda] = 1
else:
mydict[dda] += 1
for k in mydict:
if mydict[k] % 2 == 1 and k != 0:
ok = False
break
if ok:
good_lines.add(line.direction())
print(len(good_lines))
|
Технокубок 2018 - Отборочный Раунд 3
|
CF
| 2,017 | 2 | 256 |
Symmetric Projections
|
You are given a set of n points on the plane. A line containing the origin is called good, if projection of the given set to this line forms a symmetric multiset of points. Find the total number of good lines.
Multiset is a set where equal elements are allowed.
Multiset is called symmetric, if there is a point P on the plane such that the multiset is centrally symmetric in respect of point P.
|
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of points in the set.
Each of the next n lines contains two integers xi and yi ( - 106 ≤ xi, yi ≤ 106) — the coordinates of the points. It is guaranteed that no two points coincide.
|
If there are infinitely many good lines, print -1.
Otherwise, print single integer — the number of good lines.
| null |
Picture to the first sample test:
In the second sample, any line containing the origin is good.
|
[{"input": "3\n1 2\n2 1\n3 3", "output": "3"}, {"input": "2\n4 3\n1 2", "output": "-1"}]
| 2,900 |
["geometry"]
| 51 |
[{"input": "3\r\n1 2\r\n2 1\r\n3 3\r\n", "output": "3\r\n"}, {"input": "2\r\n4 3\r\n1 2\r\n", "output": "-1\r\n"}, {"input": "6\r\n0 4\r\n1 5\r\n2 1\r\n3 2\r\n4 3\r\n5 0\r\n", "output": "5\r\n"}, {"input": "1\r\n5 2\r\n", "output": "-1\r\n"}, {"input": "4\r\n2 4\r\n1 2\r\n0 0\r\n-2 -4\r\n", "output": "1\r\n"}, {"input": "10\r\n0 5\r\n1 0\r\n2 3\r\n3 2\r\n4 6\r\n5 9\r\n6 1\r\n7 8\r\n8 4\r\n9 7\r\n", "output": "5\r\n"}, {"input": "9\r\n-1000000 -500000\r\n-750000 250000\r\n-500000 1000000\r\n-250000 -250000\r\n0 -1000000\r\n250000 750000\r\n500000 0\r\n750000 -750000\r\n1000000 500000\r\n", "output": "5\r\n"}, {"input": "10\r\n-84 -60\r\n-41 -100\r\n8 -8\r\n-52 -62\r\n-61 -76\r\n-52 -52\r\n14 -11\r\n-2 -54\r\n46 8\r\n26 -17\r\n", "output": "0\r\n"}, {"input": "5\r\n-1000000 -500000\r\n-500000 0\r\n0 500000\r\n500000 1000000\r\n1000000 -1000000\r\n", "output": "3\r\n"}, {"input": "6\r\n-100000 100000\r\n-60000 -20000\r\n-20000 20000\r\n20000 60000\r\n60000 -100000\r\n100000 -60000\r\n", "output": "5\r\n"}, {"input": "8\r\n-10000 4285\r\n-7143 -10000\r\n-4286 -1429\r\n-1429 -7143\r\n1428 1428\r\n4285 9999\r\n7142 7142\r\n9999 -4286\r\n", "output": "5\r\n"}, {"input": "10\r\n-1000000 -777778\r\n-777778 555554\r\n-555556 333332\r\n-333334 111110\r\n-111112 999998\r\n111110 -333334\r\n333332 -1000000\r\n555554 -555556\r\n777776 -111112\r\n999998 777776\r\n", "output": "3\r\n"}, {"input": "7\r\n14 -3\r\n2 -13\r\n12 -1\r\n10 -7\r\n8 -11\r\n4 -9\r\n6 -5\r\n", "output": "5\r\n"}, {"input": "24\r\n-1 -7\r\n-37 -45\r\n-1 -97\r\n-37 -25\r\n9 -107\r\n-47 -85\r\n-73 -43\r\n-73 -63\r\n9 -87\r\n-63 -3\r\n-47 -35\r\n-47 -15\r\n15 -39\r\n-11 -87\r\n-63 -73\r\n-17 -65\r\n-1 -77\r\n9 -17\r\n-53 -63\r\n-1 -27\r\n-63 -53\r\n-57 -25\r\n-11 3\r\n-11 -17\r\n", "output": "2\r\n"}, {"input": "8\r\n11 -3\r\n12 -5\r\n10 -6\r\n9 -4\r\n8 -8\r\n6 -7\r\n7 -10\r\n5 -9\r\n", "output": "4\r\n"}, {"input": "32\r\n16 37\r\n-26 41\r\n5 -6\r\n12 -5\r\n17 -30\r\n-31 -14\r\n-35 4\r\n-23 -20\r\n17 -20\r\n-25 34\r\n-33 40\r\n-32 33\r\n15 24\r\n22 -25\r\n-30 -21\r\n13 -12\r\n6 -13\r\n6 37\r\n-40 -1\r\n22 25\r\n16 17\r\n-16 21\r\n11 42\r\n11 32\r\n-26 21\r\n-35 -6\r\n12 -25\r\n23 18\r\n-21 16\r\n-24 -13\r\n-21 26\r\n-30 -1\r\n", "output": "3\r\n"}, {"input": "10\r\n-7 6\r\n-16 11\r\n-9 -5\r\n3 4\r\n-8 12\r\n-17 6\r\n2 -1\r\n-5 15\r\n-7 4\r\n-6 -2\r\n", "output": "-1\r\n"}, {"input": "10\r\n-8 11\r\n1 10\r\n2 10\r\n2 11\r\n0 9\r\n3 12\r\n-7 16\r\n11 4\r\n4 8\r\n12 9\r\n", "output": "3\r\n"}, {"input": "10\r\n9 6\r\n8 1\r\n-10 13\r\n-11 8\r\n-1 6\r\n0 8\r\n-2 7\r\n-1 7\r\n1 17\r\n-3 -3\r\n", "output": "3\r\n"}, {"input": "20\r\n12 -3\r\n-18 -24\r\n-13 7\r\n17 -23\r\n15 11\r\n-17 5\r\n0 -26\r\n18 10\r\n12 -18\r\n-14 -26\r\n-20 -24\r\n16 4\r\n-19 -21\r\n-14 -11\r\n-15 -19\r\n-18 12\r\n16 10\r\n-2 12\r\n11 9\r\n13 -25\r\n", "output": "5\r\n"}, {"input": "50\r\n-38 -107\r\n-34 -75\r\n-200 -143\r\n-222 -139\r\n-34 55\r\n-102 -79\r\n48 -99\r\n2 -237\r\n-118 -167\r\n-56 -41\r\n10 17\r\n68 -89\r\n-32 41\r\n-100 -93\r\n84 -1\r\n86 -15\r\n46 -145\r\n-58 -117\r\n8 31\r\n-36 -61\r\n-12 21\r\n-116 79\r\n88 -205\r\n70 -103\r\n-78 -37\r\n106 -5\r\n-96 -201\r\n-60 -103\r\n-54 45\r\n-138 -177\r\n-178 -47\r\n-154 -5\r\n-138 83\r\n44 -131\r\n-76 -191\r\n-176 -61\r\n-14 -65\r\n-210 53\r\n-116 -181\r\n-74 -205\r\n-174 -15\r\n0 -223\r\n-136 69\r\n-198 -57\r\n-76 -51\r\n-152 -19\r\n-80 -83\r\n22 -227\r\n24 -141\r\n-220 -153\r\n", "output": "7\r\n"}, {"input": "6\r\n-10 -10\r\n-10 10\r\n10 10\r\n10 -10\r\n10 11\r\n10 -11\r\n", "output": "1\r\n"}]
| false |
stdio
| null | true |
400/A
|
400
|
A
|
Python 3
|
TESTS
| 1 | 46 | 0 |
18541074
|
from sys import stdin, stdout
n = int(stdin.readline())
a = [1,2,3,4,6,12]
for i in range(n):
s = stdin.readline()
ans = list()
for x in a:
p = 12//x
flag = True
for j in range(p,13,p):
if s[j-1] != 'X':
flag = False
if flag:
ans.append(" "+str(x)+"x"+str(p))
stdout.write(str(len(ans)))
for q in ans:
stdout.write(q)
stdout.write("\n")
| 44 | 46 | 0 |
193117487
|
import math
loops = int(input())
for x in range(loops):
line = input()
line = line.lower()
out = ""
sum = 0
for i in range(12):
if line[i] == "x":
out = out + "1x12 "
sum+=1
break
for i in range(6):
if line[i] == "x" and line[i + 6] == "x":
out = out + "2x6 "
sum+=1
break
for i in range(4):
if line[i] == "x" and line[i + 4] == "x" and line[i + 8] == "x":
out = out + "3x4 "
sum+=1
break
for i in range(3):
if line[i] == "x" and line[i + 3] == "x" and line[i + 6] == "x" and line[i + 9] == "x":
out = out + "4x3 "
sum += 1
break
for i in range(2):
if line[i] == "x" and line[i + 2] == "x" and line[i + 4] == "x" and line[i + 6] == "x" and line[i + 8] == "x" and line[i + 10] == "x":
out = out + "6x2 "
sum += 1
break
a = True
for i in range(12):
if line[i] != "x":
a = False
break
if a == True:
out = out + "12x1 "
sum+= 1
print(str(sum) + " " + out[:-1])
|
Codeforces Round 234 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Inna and Choose Options
|
There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusual upgrade of this game. The rules of the game are given below:
There is one person playing the game. Before the beginning of the game he puts 12 cards in a row on the table. Each card contains a character: "X" or "O". Then the player chooses two positive integers a and b (a·b = 12), after that he makes a table of size a × b from the cards he put on the table as follows: the first b cards form the first row of the table, the second b cards form the second row of the table and so on, the last b cards form the last (number a) row of the table. The player wins if some column of the table contain characters "X" on all cards. Otherwise, the player loses.
Inna has already put 12 cards on the table in a row. But unfortunately, she doesn't know what numbers a and b to choose. Help her win the game: print to her all the possible ways of numbers a, b that she can choose and win.
|
The first line of the input contains integer t (1 ≤ t ≤ 100). This value shows the number of sets of test data in the input. Next follows the description of each of the t tests on a separate line.
The description of each test is a string consisting of 12 characters, each character is either "X", or "O". The i-th character of the string shows the character that is written on the i-th card from the start.
|
For each test, print the answer to the test on a single line. The first number in the line must represent the number of distinct ways to choose the pair a, b. Next, print on this line the pairs in the format axb. Print the pairs in the order of increasing first parameter (a). Separate the pairs in the line by whitespaces.
| null | null |
[{"input": "4\nOXXXOXOOXOOX\nOXOXOXOXOXOX\nXXXXXXXXXXXX\nOOOOOOOOOOOO", "output": "3 1x12 2x6 4x3\n4 1x12 2x6 3x4 6x2\n6 1x12 2x6 3x4 4x3 6x2 12x1\n0"}]
| 1,000 |
["implementation"]
| 44 |
[{"input": "4\r\nOXXXOXOOXOOX\r\nOXOXOXOXOXOX\r\nXXXXXXXXXXXX\r\nOOOOOOOOOOOO\r\n", "output": "3 1x12 2x6 4x3\r\n4 1x12 2x6 3x4 6x2\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n0\r\n"}, {"input": "2\r\nOOOOOOOOOOOO\r\nXXXXXXXXXXXX\r\n", "output": "0\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}, {"input": "13\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\n", "output": "6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}]
| false |
stdio
| null | true |
194/B
|
194
|
B
|
PyPy 3
|
TESTS
| 3 | 93 | 3,891,200 |
150667371
|
t=int(input())
arr=[int(i) for i in input().split()]
for i in range(t):
if(arr[i]%4==0):
arr[i]=(4*arr[i])+1
elif(arr[i]%4==1):
arr[i]=(2*arr[i])+1
elif(arr[i]%4==2):
arr[i]=(4*arr[i])
elif(arr[i]%4==3):
arr[i]=arr[i]+1
for i in range(t):
print(arr[i])
| 8 | 46 | 614,400 |
147587674
|
n = int(input())
a=[]
b = input()
b = b.split()
for i in range(len(b)):
a.append(int(b[i]))
for i in range(n):
if a[i]%4 == 1 :
print(1 + a[i]*2)
elif a[i]%2 == 0 :
print(1+a[i]*4)
else:
print(int(4*((a[i]+1)/4)))
|
Codeforces Round 122 (Div. 2)
|
CF
| 2,012 | 2 | 256 |
Square
|
There is a square painted on a piece of paper, the square's side equals n meters. John Doe draws crosses on the square's perimeter. John paints the first cross in the lower left corner of the square. Then John moves along the square's perimeter in the clockwise direction (first upwards, then to the right, then downwards, then to the left and so on). Every time he walks (n + 1) meters, he draws a cross (see picture for clarifications).
John Doe stops only when the lower left corner of the square has two crosses. How many crosses will John draw?
The figure shows the order in which John draws crosses for a square with side 4. The lower left square has two crosses. Overall John paints 17 crosses.
|
The first line contains integer t (1 ≤ t ≤ 104) — the number of test cases.
The second line contains t space-separated integers ni (1 ≤ ni ≤ 109) — the sides of the square for each test sample.
|
For each test sample print on a single line the answer to it, that is, the number of crosses John will draw as he will move along the square of the corresponding size. Print the answers to the samples in the order in which the samples are given in the input.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
| null | null |
[{"input": "3\n4 8 100", "output": "17\n33\n401"}]
| 1,200 |
["math"]
| 8 |
[{"input": "3\r\n4 8 100\r\n", "output": "17\r\n33\r\n401\r\n"}, {"input": "8\r\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 13\r\n", "output": "4000000001\r\n4000000001\r\n4000000001\r\n4000000001\r\n4000000001\r\n4000000001\r\n4000000001\r\n27\r\n"}, {"input": "3\r\n13 17 21\r\n", "output": "27\r\n35\r\n43\r\n"}]
| false |
stdio
| null | true |
175/B
|
175
|
B
|
Python 3
|
TESTS
| 1 | 62 | 0 |
214310458
|
# LUOGU_RID: 116207345
n=int(input())
d={}
c=[]
for _ in range(n):
a,b=input().split()
if a not in c:
c.append(a)
if a not in d:
d[a]=b
else:
if d[a]<b:
d[a]=b
print(len(d))
m=len(c)
x=[0]*m
for i in range(m):
for j in range(m):
if d[c[i]]<d[c[j]]:
x[i]+=1
for i in range(m):
d=x[i]/m
if d>0.5:
x[i]='noob'
elif d>0.2:
x[i]='random'
elif d>0.1:
x[i]='average'
else:
x[i]='pro'
for i in range(len(c)):
print(c[i],x[i])
| 46 | 436 | 204,800 |
214355001
|
# LUOGU_RID: 116281298
n=int(input())
d={}
c=[]
for _ in range(n):
a,b=input().split()
b=int(b)
if a not in c:
c.append(a)
if a not in d:
d[a]=b
else:
if d[a]<b:
d[a]=b
print(len(d))
m=len(c)
x=[0]*m
for i in range(m):
for j in range(m):
if d[c[i]]<d[c[j]]:
x[i]+=1
for i in range(m):
print(c[i],end=' ')
if x[i]<=0.01*m:
print('pro')
elif x[i]<=0.10*m:
print('hardcore')
elif x[i]<=0.20*m:
print('average')
elif x[i]<=0.50*m:
print('random')
else:
print('noob')
|
Codeforces Round 115
|
CF
| 2,012 | 2 | 256 |
Plane of Tanks: Pro
|
Vasya has been playing Plane of Tanks with his friends the whole year. Now it is time to divide the participants into several categories depending on their results.
A player is given a non-negative integer number of points in each round of the Plane of Tanks. Vasya wrote results for each round of the last year. He has n records in total.
In order to determine a player's category consider the best result obtained by the player and the best results of other players. The player belongs to category:
- "noob" — if more than 50% of players have better results;
- "random" — if his result is not worse than the result that 50% of players have, but more than 20% of players have better results;
- "average" — if his result is not worse than the result that 80% of players have, but more than 10% of players have better results;
- "hardcore" — if his result is not worse than the result that 90% of players have, but more than 1% of players have better results;
- "pro" — if his result is not worse than the result that 99% of players have.
When the percentage is calculated the player himself is taken into account. That means that if two players played the game and the first one gained 100 points and the second one 1000 points, then the first player's result is not worse than the result that 50% of players have, and the second one is not worse than the result that 100% of players have.
Vasya gave you the last year Plane of Tanks results. Help Vasya determine each player's category.
|
The first line contains the only integer number n (1 ≤ n ≤ 1000) — a number of records with the players' results.
Each of the next n lines contains a player's name and the amount of points, obtained by the player for the round, separated with a space. The name contains not less than 1 and no more than 10 characters. The name consists of lowercase Latin letters only. It is guaranteed that any two different players have different names. The amount of points, obtained by the player for the round, is a non-negative integer number and does not exceed 1000.
|
Print on the first line the number m — the number of players, who participated in one round at least.
Each one of the next m lines should contain a player name and a category he belongs to, separated with space. Category can be one of the following: "noob", "random", "average", "hardcore" or "pro" (without quotes). The name of each player should be printed only once. Player names with respective categories can be printed in an arbitrary order.
| null |
In the first example the best result, obtained by artem is not worse than the result that 25% of players have (his own result), so he belongs to category "noob". vasya and kolya have best results not worse than the results that 75% players have (both of them and artem), so they belong to category "random". igor has best result not worse than the result that 100% of players have (all other players and himself), so he belongs to category "pro".
In the second example both players have the same amount of points, so they have results not worse than 100% players have, so they belong to category "pro".
|
[{"input": "5\nvasya 100\nvasya 200\nartem 100\nkolya 200\nigor 250", "output": "4\nartem noob\nigor pro\nkolya random\nvasya random"}, {"input": "3\nvasya 200\nkolya 1000\nvasya 1000", "output": "2\nkolya pro\nvasya pro"}]
| 1,400 |
["implementation"]
| 78 |
[{"input": "5\r\nvasya 100\r\nvasya 200\r\nartem 100\r\nkolya 200\r\nigor 250\r\n", "output": "4\r\nartem noob\r\nigor pro\r\nkolya random\r\nvasya random\r\n"}, {"input": "3\r\nvasya 200\r\nkolya 1000\r\nvasya 1000\r\n", "output": "2\r\nkolya pro\r\nvasya pro\r\n"}, {"input": "1\r\nvasya 1000\r\n", "output": "1\r\nvasya pro\r\n"}, {"input": "5\r\nvasya 1000\r\nvasya 100\r\nkolya 200\r\npetya 300\r\noleg 400\r\n", "output": "4\r\nkolya noob\r\noleg random\r\npetya random\r\nvasya pro\r\n"}, {"input": "10\r\na 1\r\nb 2\r\nc 3\r\nd 4\r\ne 5\r\nf 6\r\ng 7\r\nh 8\r\ni 9\r\nj 10\r\n", "output": "10\r\na noob\r\nb noob\r\nc noob\r\nd noob\r\ne random\r\nf random\r\ng random\r\nh average\r\ni hardcore\r\nj pro\r\n"}, {"input": "10\r\nj 10\r\ni 9\r\nh 8\r\ng 7\r\nf 6\r\ne 5\r\nd 4\r\nc 3\r\nb 2\r\na 1\r\n", "output": "10\r\na noob\r\nb noob\r\nc noob\r\nd noob\r\ne random\r\nf random\r\ng random\r\nh average\r\ni hardcore\r\nj pro\r\n"}, {"input": "1\r\ntest 0\r\n", "output": "1\r\ntest pro\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_path):
# Read input data
with open(input_path) as f:
n = int(f.readline())
players = {}
for _ in range(n):
name, points = f.readline().strip().split()
points = int(points)
if name not in players or points > players[name]:
players[name] = points
m = len(players)
best_scores = list(players.values())
# Compute correct categories
correct = {}
for name in players:
score = players[name]
cnt = sum(1 for s in best_scores if s > score)
percentage = (cnt * 100.0) / m
if percentage > 50:
cat = 'noob'
elif percentage > 20:
cat = 'random'
elif percentage > 10:
cat = 'average'
elif percentage > 1:
cat = 'hardcore'
else:
cat = 'pro'
correct[name] = cat
# Read submission output
with open(submission_path) as f:
lines = f.read().splitlines()
if not lines:
print(0)
return
try:
m_sub = int(lines[0])
except:
print(0)
return
if m_sub != m:
print(0)
return
submission = {}
for line in lines[1:]:
parts = line.strip().split()
if len(parts) != 2:
print(0)
return
name, cat = parts
if name in submission:
print(0)
return
submission[name] = cat
# Check all players are present and correct
if submission.keys() != correct.keys():
print(0)
return
for name, cat in submission.items():
if correct.get(name) != cat:
print(0)
return
print(1)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
400/A
|
400
|
A
|
PyPy 3-64
|
TESTS
| 1 | 31 | 1,433,600 |
213973055
|
n = int(input())
otv = []
result = ""
counter = 0
flag = True
for _ in range(n):
temp = input()
for i in range(12):
if 12 % (i+1) == 0 and temp[i] == "X":
for j in range((i*2)+1, 12, i+1):
if temp[j] != temp[i]:
flag = False
break
if flag:
counter += 1
otv.append(f"{round(12/(i+1))} x {i+1} ")
flag = True
otv = sorted(otv, key=lambda x: int(x.split()[0].strip('x')))
result += str(counter) + " " + "".join(otv) + "\n"
result = result.replace(" x ", "x")
counter = 0
otv = []
else:
print(result)# 1689471988.4487488
| 44 | 46 | 0 |
195790246
|
t = int(input())
N = 12
divisors = [1,2,3,4,6,12]
L = 6
for _ in range(t):
cards = input()
goodDivs = []
for k in range(L):
numCols = divisors[k]
state = [True] * numCols
for m in range(N):
if cards[m] == 'O':
state[m % numCols] = False
if any(state):
goodDivs.append(numCols)
print(len(goodDivs), end=' ')
for k in range(len(goodDivs)-1, -1, -1):
print(f"{N//goodDivs[k]}x{goodDivs[k]}", end=' ')
print()
|
Codeforces Round 234 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Inna and Choose Options
|
There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusual upgrade of this game. The rules of the game are given below:
There is one person playing the game. Before the beginning of the game he puts 12 cards in a row on the table. Each card contains a character: "X" or "O". Then the player chooses two positive integers a and b (a·b = 12), after that he makes a table of size a × b from the cards he put on the table as follows: the first b cards form the first row of the table, the second b cards form the second row of the table and so on, the last b cards form the last (number a) row of the table. The player wins if some column of the table contain characters "X" on all cards. Otherwise, the player loses.
Inna has already put 12 cards on the table in a row. But unfortunately, she doesn't know what numbers a and b to choose. Help her win the game: print to her all the possible ways of numbers a, b that she can choose and win.
|
The first line of the input contains integer t (1 ≤ t ≤ 100). This value shows the number of sets of test data in the input. Next follows the description of each of the t tests on a separate line.
The description of each test is a string consisting of 12 characters, each character is either "X", or "O". The i-th character of the string shows the character that is written on the i-th card from the start.
|
For each test, print the answer to the test on a single line. The first number in the line must represent the number of distinct ways to choose the pair a, b. Next, print on this line the pairs in the format axb. Print the pairs in the order of increasing first parameter (a). Separate the pairs in the line by whitespaces.
| null | null |
[{"input": "4\nOXXXOXOOXOOX\nOXOXOXOXOXOX\nXXXXXXXXXXXX\nOOOOOOOOOOOO", "output": "3 1x12 2x6 4x3\n4 1x12 2x6 3x4 6x2\n6 1x12 2x6 3x4 4x3 6x2 12x1\n0"}]
| 1,000 |
["implementation"]
| 44 |
[{"input": "4\r\nOXXXOXOOXOOX\r\nOXOXOXOXOXOX\r\nXXXXXXXXXXXX\r\nOOOOOOOOOOOO\r\n", "output": "3 1x12 2x6 4x3\r\n4 1x12 2x6 3x4 6x2\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n0\r\n"}, {"input": "2\r\nOOOOOOOOOOOO\r\nXXXXXXXXXXXX\r\n", "output": "0\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}, {"input": "13\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\nXXXXXXXXXXXX\r\n", "output": "6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n6 1x12 2x6 3x4 4x3 6x2 12x1\r\n"}]
| false |
stdio
| null | true |
863/D
|
863
|
D
|
PyPy 3
|
TESTS
| 6 | 1,777 | 16,486,400 |
93021857
|
n,q,m=map(int,input().split())
l=list(map(int,input().split()))
w=[]
for i in range(q):
a,b,c=map(int,input().split())
w.append((a,b,c))
w.reverse()
b=list(map(int,input().split()))
for j in range(m):
for i in range(q):
a,b1,c=w[i]
if b1<=b[j]<=c:
if a==1:
b[j]-=1
if b[j]==0:
b[j]=c
else:
b[j]=b1+c-b[j]
#print(b,l)
for j in range(m):
b[j]=l[b[j]-1]
print(*b)
| 23 | 499 | 19,660,800 |
113065724
|
import sys
n, q, m = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
a = list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
query = [list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
for _ in range(q)]
b = list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
for qt, l, r in reversed(query):
if qt == 1:
for i in range(m):
if l <= b[i] <= r:
b[i] = r if b[i] == l else b[i]-1
else:
for i in range(m):
if l <= b[i] <= r:
b[i] = r - (b[i] - l)
print(*(a[i-1] for i in b))
|
Educational Codeforces Round 29
|
ICPC
| 2,017 | 2 | 256 |
Yet Another Array Queries Problem
|
You are given an array a of size n, and q queries to it. There are queries of two types:
- 1 li ri — perform a cyclic shift of the segment [li, ri] to the right. That is, for every x such that li ≤ x < ri new value of ax + 1 becomes equal to old value of ax, and new value of ali becomes equal to old value of ari;
- 2 li ri — reverse the segment [li, ri].
There are m important indices in the array b1, b2, ..., bm. For each i such that 1 ≤ i ≤ m you have to output the number that will have index bi in the array after all queries are performed.
|
The first line contains three integer numbers n, q and m (1 ≤ n, q ≤ 2·105, 1 ≤ m ≤ 100).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109).
Then q lines follow. i-th of them contains three integer numbers ti, li, ri, where ti is the type of i-th query, and [li, ri] is the segment where this query is performed (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n).
The last line contains m integer numbers b1, b2, ..., bm (1 ≤ bi ≤ n) — important indices of the array.
|
Print m numbers, i-th of which is equal to the number at index bi after all queries are done.
| null | null |
[{"input": "6 3 5\n1 2 3 4 5 6\n2 1 3\n2 3 6\n1 1 6\n2 2 1 5 3", "output": "3 3 1 5 2"}]
| 1,800 |
["data structures", "implementation"]
| 23 |
[{"input": "6 3 5\r\n1 2 3 4 5 6\r\n2 1 3\r\n2 3 6\r\n1 1 6\r\n2 2 1 5 3\r\n", "output": "3 3 1 5 2 \r\n"}, {"input": "5 2 5\r\n64 3 4 665 2\r\n1 1 3\r\n2 1 5\r\n1 2 3 4 5\r\n", "output": "2 665 3 64 4 \r\n"}, {"input": "1 1 1\r\n474812122\r\n2 1 1\r\n1\r\n", "output": "474812122 \r\n"}]
| false |
stdio
| null | true |
393/B
|
393
|
B
|
PyPy 3
|
TESTS
| 0 | 124 | 0 |
70848947
|
def nine(n,queries):
A=[[0 for _ in range(n)] for _ in range(n)]
B=[[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
for j in range(i+1,n):
a=queries[i][j]
b=queries[j][i]
x=(a+b)/2
y=a-x
A[i][j]=x
A[j][i]=x
B[i][j]=-y
B[j][i]=y
for k in range(n):
A[k][k]=queries[k][k]
for i in range(n):
print(' '.join(map(str,A[i])))
for i in range(n):
print(' '.join(map(str,B[i])))
n=int(input())
queries=[]
for _ in range(n):
queries.append(list(map(int,input().split())))
nine(n,queries)
| 40 | 171 | 9,318,400 |
115451908
|
import sys
from os import path
from collections import Counter
if(path.exists("inp.txt")):
sys.stdin = open("inp.txt",'r')
sys.stdout = open("out.txt",'w')
import math
n=int(input())
w=[]
li=[0]*((2*n)-1)
for i in range(n):
temp=list(map(int,input().split()))
w.append(temp)
a=[[0 for i in range(n)] for j in range(n)]
b=[[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
a[i][j]=(w[i][j]+w[j][i])/2
b[i][j]=(w[i][j]-w[j][i])/2
for i in a:
print(*i)
for i in b:
print(*i)
|
Codeforces Round 230 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
Three matrices
|
Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold:
- Aij = Aji, for all i, j (1 ≤ i, j ≤ n);
- Bij = - Bji, for all i, j (1 ≤ i, j ≤ n);
- Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n).
Can you solve the problem?
|
The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij (0 ≤ |Wij| < 1717).
|
The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.
| null | null |
[{"input": "2\n1 4\n3 2", "output": "1.00000000 3.50000000\n3.50000000 2.00000000\n0.00000000 0.50000000\n-0.50000000 0.00000000"}, {"input": "3\n1 2 3\n4 5 6\n7 8 9", "output": "1.00000000 3.00000000 5.00000000\n3.00000000 5.00000000 7.00000000\n5.00000000 7.00000000 9.00000000\n0.00000000 -1.00000000 -2.00000000\n1.00000000 0.00000000 -1.00000000\n2.00000000 1.00000000 0.00000000"}]
| null |
[]
| 40 |
[{"input": "2\r\n1 4\r\n3 2\r\n", "output": "1.00000000 3.50000000\r\n3.50000000 2.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}, {"input": "3\r\n1 2 3\r\n4 5 6\r\n7 8 9\r\n", "output": "1.00000000 3.00000000 5.00000000\r\n3.00000000 5.00000000 7.00000000\r\n5.00000000 7.00000000 9.00000000\r\n0.00000000 -1.00000000 -2.00000000\r\n1.00000000 0.00000000 -1.00000000\r\n2.00000000 1.00000000 0.00000000\r\n"}, {"input": "8\r\n62 567 1382 1279 728 1267 1262 568\r\n77 827 717 1696 774 248 822 1266\r\n563 612 995 424 1643 1197 338 1141\r\n1579 806 1254 468 184 1571 716 772\r\n1087 182 1312 772 605 1674 720 1349\r\n1393 988 873 157 403 301 1519 1192\r\n1085 625 1395 1087 847 1360 1004 594\r\n1368 1056 916 839 472 840 53 1238\r\n", "output": "62.000000000 322.000000000 972.500000000 1429.000000000 907.500000000 1330.000000000 1173.500000000 968.000000000\n322.000000000 827.000000000 664.500000000 1251.000000000 478.000000000 618.000000000 723.500000000 1161.000000000\n972.500000000 664.500000000 995.000000000 839.000000000 1477.500000000 1035.000000000 866.500000000 1028.500000000\n1429.000000000 1251.000000000 839.000000000 468.000000000 478.000000000 864.000000000 901.500000000 805.500000000\n907.500000000 478.000000000 1477.500000000 478.000000000 605.000000000 1038.500000000 783.500000000 910.500000000\n1330.000000000 618.000000000 1035.000000000 864.000000000 1038.500000000 301.000000000 1439.500000000 1016.000000000\n1173.500000000 723.500000000 866.500000000 901.500000000 783.500000000 1439.500000000 1004.000000000 323.500000000\n968.000000000 1161.000000000 1028.500000000 805.500000000 910.500000000 1016.000000000 323.500000000 1238.000000000\n0.000000000 245.000000000 409.500000000 -150.000000000 -179.500000000 -63.000000000 88.500000000 -400.000000000\n-245.000000000 0.000000000 52.500000000 445.000000000 296.000000000 -370.000000000 98.500000000 105.000000000\n-409.500000000 -52.500000000 0.000000000 -415.000000000 165.500000000 162.000000000 -528.500000000 112.500000000\n150.000000000 -445.000000000 415.000000000 0.000000000 -294.000000000 707.000000000 -185.500000000 -33.500000000\n179.500000000 -296.000000000 -165.500000000 294.000000000 0.000000000 635.500000000 -63.500000000 438.500000000\n63.000000000 370.000000000 -162.000000000 -707.000000000 -635.500000000 0.000000000 79.500000000 176.000000000\n-88.500000000 -98.500000000 528.500000000 185.500000000 63.500000000 -79.500000000 0.000000000 270.500000000\n400.000000000 -105.000000000 -112.500000000 33.500000000 -438.500000000 -176.000000000 -270.500000000 0.000000000\n"}, {"input": "7\r\n926 41 1489 72 749 375 940\r\n464 1148 858 1010 285 1469 1506\r\n1112 1087 225 917 480 511 1090\r\n759 945 627 230 220 1456 529\r\n318 83 203 134 1192 1167 6\r\n440 1158 1614 683 1358 1140 1196\r\n1175 900 126 1562 1220 813 148\r\n", "output": "926.000000000 252.500000000 1300.500000000 415.500000000 533.500000000 407.500000000 1057.500000000\n252.500000000 1148.000000000 972.500000000 977.500000000 184.000000000 1313.500000000 1203.000000000\n1300.500000000 972.500000000 225.000000000 772.000000000 341.500000000 1062.500000000 608.000000000\n415.500000000 977.500000000 772.000000000 230.000000000 177.000000000 1069.500000000 1045.500000000\n533.500000000 184.000000000 341.500000000 177.000000000 1192.000000000 1262.500000000 613.000000000\n407.500000000 1313.500000000 1062.500000000 1069.500000000 1262.500000000 1140.000000000 1004.500000000\n1057.500000000 1203.000000000 608.000000000 1045.500000000 613.000000000 1004.500000000 148.000000000\n0.000000000 -211.500000000 188.500000000 -343.500000000 215.500000000 -32.500000000 -117.500000000\n211.500000000 0.000000000 -114.500000000 32.500000000 101.000000000 155.500000000 303.000000000\n-188.500000000 114.500000000 0.000000000 145.000000000 138.500000000 -551.500000000 482.000000000\n343.500000000 -32.500000000 -145.000000000 0.000000000 43.000000000 386.500000000 -516.500000000\n-215.500000000 -101.000000000 -138.500000000 -43.000000000 0.000000000 -95.500000000 -607.000000000\n32.500000000 -155.500000000 551.500000000 -386.500000000 95.500000000 0.000000000 191.500000000\n117.500000000 -303.000000000 -482.000000000 516.500000000 607.000000000 -191.500000000 0.000000000\n"}, {"input": "1\r\n1\r\n", "output": "1.00000000\r\n0.00000000\r\n"}, {"input": "1\r\n0\r\n", "output": "0.00000000\r\n0.00000000\r\n"}, {"input": "2\r\n0 0\r\n0 0\r\n", "output": "0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n0.00000000 0.00000000\r\n"}, {"input": "2\r\n0 1\r\n0 1\r\n", "output": "0.00000000 0.50000000\r\n0.50000000 1.00000000\r\n0.00000000 0.50000000\r\n-0.50000000 0.00000000\r\n"}]
| false |
stdio
|
import sys
def is_close(a, b):
abs_tol = 1e-4
rel_tol = 1e-4
diff = abs(a - b)
if diff <= abs_tol:
return True
max_val = max(abs(a), abs(b))
return diff <= rel_tol * max_val
def main(input_path, output_path, submission_path):
with open(input_path) as f:
lines = f.read().splitlines()
n = int(lines[0])
w = []
for line in lines[1:1+n]:
row = list(map(int, line.split()))
w.append(row)
with open(submission_path) as f:
submission_lines = f.read().splitlines()
if len(submission_lines) != 2 * n:
print(0)
return
A = []
B = []
for i in range(n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
A.append(row)
for i in range(n, 2 * n):
if i >= len(submission_lines):
print(0)
return
parts = submission_lines[i].split()
if len(parts) != n:
print(0)
return
try:
row = list(map(float, parts))
except:
print(0)
return
B.append(row)
for i in range(n):
for j in range(n):
if not is_close(A[i][j], A[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
if not is_close(B[i][j], -B[j][i]):
print(0)
return
for i in range(n):
for j in range(n):
sum_ab = A[i][j] + B[i][j]
expected = w[i][j]
if not is_close(sum_ab, expected):
print(0)
return
print(1)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path)
| true |
522/C
|
522
|
C
|
Python 3
|
PRETESTS
| 3 | 46 | 0 |
10199729
|
t = int(input())
for i in range(t):
input()
m,k = map(int,input().split())
ak = list(map(int,input().split()))
ak2 = [0]*k
tjrj = [list(map(int,input().split())) for j in range(m-1)]
num = 0
num2 = 0
for j in range(m-1):
if num2 == 1 or tjrj[j][1] == 0:
if tjrj[j][0] != 0:
ak[tjrj[j][0]-1] -= 1
else:
num += 1
else:
for z in range(k):
if ak[z] - num < 1:
ak2[z] = 1
num2 = 1
if tjrj[j][0] != 0:
ak[tjrj[j][0]-1] -= 1
else:
num += 1
for f in range(j,m-1):
if tjrj[f][0] != 0:
ak2[tjrj[f][0]-1] = 0
for z in range(k):
if ak[z] - num < 1 or ak2[z] == 1:
print("Y",end="")
else:
print("N",end="")
print()
| 48 | 654 | 10,342,400 |
10207272
|
import sys
test_count = int(sys.stdin.readline())
for test in range(test_count):
sys.stdin.readline()
m, k = map(int, sys.stdin.readline().split())
counts = list(map(int, sys.stdin.readline().split()))
took = []
unhappy = []
for i in range(m - 1):
t, r = map(int, sys.stdin.readline().split())
t -= 1 # -1 means unknown
took.append(t)
unhappy.append(r)
took_in_total = [0 for dish_count in counts]
for i in range(m - 1):
if took[i] != -1:
took_in_total[took[i]] += 1
took_already = [0 for dish_count in counts]
left = [dish_count for dish_count in counts]
answer = [False for dish_count in counts]
unknown = 0
all_present = True
for i in range(m - 1):
if unhappy[i] == 1 and all_present:
could_exhaust = []
for j in range(k):
if took_already[j] < took_in_total[j]:
continue
if left[j] > unknown:
continue
could_exhaust.append(j)
if len(could_exhaust) == 0:
raise AssertionError
for j in could_exhaust:
answer[j] = True
unknown -= min(map(lambda j: left[j], could_exhaust))
all_present = False
if took[i] != -1:
left[took[i]] -= 1
took_already[took[i]] += 1
if left[took[i]] == 0:
all_present = False
else:
unknown += 1
for j in range(k):
if left[j] <= unknown:
answer[j] = True
sys.stdout.write(
''.join(map(lambda x: 'Y' if x else 'N', answer)) + '\n'
)
|
VK Cup 2015 - Qualification Round 1
|
CF
| 2,015 | 1 | 256 |
Chicken or Fish?
|
Polycarp is flying in the airplane. Finally, it is his favorite time — the lunchtime. The BerAvia company stewardess is giving food consecutively to all the passengers from the 1-th one to the last one. Polycarp is sitting on seat m, that means, he will be the m-th person to get food.
The flight menu has k dishes in total and when Polycarp boarded the flight, he had time to count the number of portions of each dish on board. Thus, he knows values a1, a2, ..., ak, where ai is the number of portions of the i-th dish.
The stewardess has already given food to m - 1 passengers, gave Polycarp a polite smile and asked him what he would prefer. That's when Polycarp realized that they might have run out of some dishes by that moment. For some of the m - 1 passengers ahead of him, he noticed what dishes they were given. Besides, he's heard some strange mumbling from some of the m - 1 passengers ahead of him, similar to phrase 'I'm disappointed'. That happened when a passenger asked for some dish but the stewardess gave him a polite smile and said that they had run out of that dish. In that case the passenger needed to choose some other dish that was available. If Polycarp heard no more sounds from a passenger, that meant that the passenger chose his dish at the first try.
Help Polycarp to find out for each dish: whether they could have run out of the dish by the moment Polyarp was served or that dish was definitely available.
|
Each test in this problem consists of one or more input sets. First goes a string that contains a single integer t (1 ≤ t ≤ 100 000) — the number of input data sets in the test. Then the sets follow, each set is preceded by an empty line.
The first line of each set of the input contains integers m, k (2 ≤ m ≤ 100 000, 1 ≤ k ≤ 100 000) — the number of Polycarp's seat and the number of dishes, respectively.
The second line contains a sequence of k integers a1, a2, ..., ak (1 ≤ ai ≤ 100 000), where ai is the initial number of portions of the i-th dish.
Then m - 1 lines follow, each line contains the description of Polycarp's observations about giving food to a passenger sitting in front of him: the j-th line contains a pair of integers tj, rj (0 ≤ tj ≤ k, 0 ≤ rj ≤ 1), where tj is the number of the dish that was given to the j-th passenger (or 0, if Polycarp didn't notice what dish was given to the passenger), and rj — a 1 or a 0, depending on whether the j-th passenger was or wasn't disappointed, respectively.
We know that sum ai equals at least m, that is,Polycarp will definitely get some dish, even if it is the last thing he wanted. It is guaranteed that the data is consistent.
Sum m for all input sets doesn't exceed 100 000. Sum k for all input sets doesn't exceed 100 000.
|
For each input set print the answer as a single line. Print a string of k letters "Y" or "N". Letter "Y" in position i should be printed if they could have run out of the i-th dish by the time the stewardess started serving Polycarp.
| null |
In the first input set depending on the choice of the second passenger the situation could develop in different ways:
- If he chose the first dish, then by the moment the stewardess reaches Polycarp, they will have run out of the first dish;
- If he chose the fourth dish, then by the moment the stewardess reaches Polycarp, they will have run out of the fourth dish;
- Otherwise, Polycarp will be able to choose from any of the four dishes.
Thus, the answer is "YNNY".
In the second input set there is, for example, the following possible scenario. First, the first passenger takes the only third dish, then the second passenger takes the second dish. Then, the third passenger asks for the third dish, but it is not available, so he makes disappointed muttering and ends up with the second dish. Then the fourth passenger takes the fourth dish, and Polycarp ends up with the choice between the first, fourth and fifth dish.
Likewise, another possible scenario is when by the time the stewardess comes to Polycarp, they will have run out of either the first or the fifth dish (this can happen if one of these dishes is taken by the second passenger). It is easy to see that there is more than enough of the fourth dish, so Polycarp can always count on it. Thus, the answer is "YYYNY".
|
[{"input": "2\n3 4\n2 3 2 1\n1 0\n0 0\n5 5\n1 2 1 3 1\n3 0\n0 0\n2 1\n4 0", "output": "YNNY\nYYYNY"}]
| 2,100 |
["greedy"]
| 48 |
[{"input": "2\r\n\r\n3 4\r\n2 3 2 1\r\n1 0\r\n0 0\r\n\r\n5 5\r\n1 2 1 3 1\r\n3 0\r\n0 0\r\n2 1\r\n4 0\r\n", "output": "YNNY\r\nYYYNY\r\n"}, {"input": "4\r\n\r\n2 1\r\n42\r\n0 0\r\n\r\n2 1\r\n2\r\n0 0\r\n\r\n2 1\r\n42\r\n1 0\r\n\r\n2 1\r\n2\r\n1 0\r\n", "output": "N\r\nN\r\nN\r\nN\r\n"}, {"input": "5\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n2 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 0\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 1\r\n", "output": "YYY\r\nYYN\r\nYYY\r\nYYY\r\nYYY\r\n"}, {"input": "1\r\n\r\n4 2\r\n2 2\r\n0 0\r\n0 0\r\n1 1\r\n", "output": "NY\r\n"}]
| false |
stdio
| null | true |
522/C
|
522
|
C
|
Python 3
|
PRETESTS
| 3 | 46 | 0 |
10202926
|
for ts in range(int(input())):
input()
m, k = tuple(map(int, input().split()))
a = list(map(int, input().split()))
at = []
o = 'N'*k
for i in range(m-1):
tj, rj = tuple(map(int, input().split()))
if tj > 0:
a[tj-1] -= 1
if a[tj-1] == 0:
o = o[:tj-1]+'Y'+o[tj:]
else:
if at == []:
at += a
for j in range(k):
a[j] -= 1
if a[j] == 0:
o = o[:j]+'Y'+o[j+1:]
if rj == 1:
if at != []:
for j in range(k):
if o[j] == 'N':
a[j] = at[j]
at = []
print(o)
| 48 | 967 | 7,475,200 |
10211431
|
t = int(input())
for j in range(t):
e = input()
m, k = map(int, input().split())
arr = [int(i) for i in input().split()]
sum, bfail = [0] * k, [0] * k
ffail, undef = -1, 0
used = [False] * k
ubfail = 0
for i in range(m - 1):
c, ns = map(int, input().split())
if c == 0:
undef += 1
if ns == 0 and ffail == -1:
ubfail += 1
else:
sum[c - 1] += 1
if ns == 0 and ffail == -1:
bfail[c - 1] += 1
if ns and ffail == -1:
ffail = i
if ffail != -1 and c > 0:
used[c - 1] = True
if ffail == -1:
for i in range(k):
if sum[i] + undef >= arr[i]:
print('Y', end = '')
else:
print('N', end = '')
print()
continue
minu = 10 ** 6
for i in range(k):
if not used[i] and arr[i] - bfail[i] < minu:
minu = arr[i] - bfail[i]
best = i
for i in range(k):
if i == best or undef - minu + sum[i] >= arr[i]:
print('Y', end = '')
elif bfail[i] + ubfail >= arr[i] and not used[i]:
print('Y', end = '')
else:
print('N', end = '')
print()
|
VK Cup 2015 - Qualification Round 1
|
CF
| 2,015 | 1 | 256 |
Chicken or Fish?
|
Polycarp is flying in the airplane. Finally, it is his favorite time — the lunchtime. The BerAvia company stewardess is giving food consecutively to all the passengers from the 1-th one to the last one. Polycarp is sitting on seat m, that means, he will be the m-th person to get food.
The flight menu has k dishes in total and when Polycarp boarded the flight, he had time to count the number of portions of each dish on board. Thus, he knows values a1, a2, ..., ak, where ai is the number of portions of the i-th dish.
The stewardess has already given food to m - 1 passengers, gave Polycarp a polite smile and asked him what he would prefer. That's when Polycarp realized that they might have run out of some dishes by that moment. For some of the m - 1 passengers ahead of him, he noticed what dishes they were given. Besides, he's heard some strange mumbling from some of the m - 1 passengers ahead of him, similar to phrase 'I'm disappointed'. That happened when a passenger asked for some dish but the stewardess gave him a polite smile and said that they had run out of that dish. In that case the passenger needed to choose some other dish that was available. If Polycarp heard no more sounds from a passenger, that meant that the passenger chose his dish at the first try.
Help Polycarp to find out for each dish: whether they could have run out of the dish by the moment Polyarp was served or that dish was definitely available.
|
Each test in this problem consists of one or more input sets. First goes a string that contains a single integer t (1 ≤ t ≤ 100 000) — the number of input data sets in the test. Then the sets follow, each set is preceded by an empty line.
The first line of each set of the input contains integers m, k (2 ≤ m ≤ 100 000, 1 ≤ k ≤ 100 000) — the number of Polycarp's seat and the number of dishes, respectively.
The second line contains a sequence of k integers a1, a2, ..., ak (1 ≤ ai ≤ 100 000), where ai is the initial number of portions of the i-th dish.
Then m - 1 lines follow, each line contains the description of Polycarp's observations about giving food to a passenger sitting in front of him: the j-th line contains a pair of integers tj, rj (0 ≤ tj ≤ k, 0 ≤ rj ≤ 1), where tj is the number of the dish that was given to the j-th passenger (or 0, if Polycarp didn't notice what dish was given to the passenger), and rj — a 1 or a 0, depending on whether the j-th passenger was or wasn't disappointed, respectively.
We know that sum ai equals at least m, that is,Polycarp will definitely get some dish, even if it is the last thing he wanted. It is guaranteed that the data is consistent.
Sum m for all input sets doesn't exceed 100 000. Sum k for all input sets doesn't exceed 100 000.
|
For each input set print the answer as a single line. Print a string of k letters "Y" or "N". Letter "Y" in position i should be printed if they could have run out of the i-th dish by the time the stewardess started serving Polycarp.
| null |
In the first input set depending on the choice of the second passenger the situation could develop in different ways:
- If he chose the first dish, then by the moment the stewardess reaches Polycarp, they will have run out of the first dish;
- If he chose the fourth dish, then by the moment the stewardess reaches Polycarp, they will have run out of the fourth dish;
- Otherwise, Polycarp will be able to choose from any of the four dishes.
Thus, the answer is "YNNY".
In the second input set there is, for example, the following possible scenario. First, the first passenger takes the only third dish, then the second passenger takes the second dish. Then, the third passenger asks for the third dish, but it is not available, so he makes disappointed muttering and ends up with the second dish. Then the fourth passenger takes the fourth dish, and Polycarp ends up with the choice between the first, fourth and fifth dish.
Likewise, another possible scenario is when by the time the stewardess comes to Polycarp, they will have run out of either the first or the fifth dish (this can happen if one of these dishes is taken by the second passenger). It is easy to see that there is more than enough of the fourth dish, so Polycarp can always count on it. Thus, the answer is "YYYNY".
|
[{"input": "2\n3 4\n2 3 2 1\n1 0\n0 0\n5 5\n1 2 1 3 1\n3 0\n0 0\n2 1\n4 0", "output": "YNNY\nYYYNY"}]
| 2,100 |
["greedy"]
| 48 |
[{"input": "2\r\n\r\n3 4\r\n2 3 2 1\r\n1 0\r\n0 0\r\n\r\n5 5\r\n1 2 1 3 1\r\n3 0\r\n0 0\r\n2 1\r\n4 0\r\n", "output": "YNNY\r\nYYYNY\r\n"}, {"input": "4\r\n\r\n2 1\r\n42\r\n0 0\r\n\r\n2 1\r\n2\r\n0 0\r\n\r\n2 1\r\n42\r\n1 0\r\n\r\n2 1\r\n2\r\n1 0\r\n", "output": "N\r\nN\r\nN\r\nN\r\n"}, {"input": "5\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n2 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 0\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 1\r\n", "output": "YYY\r\nYYN\r\nYYY\r\nYYY\r\nYYY\r\n"}, {"input": "1\r\n\r\n4 2\r\n2 2\r\n0 0\r\n0 0\r\n1 1\r\n", "output": "NY\r\n"}]
| false |
stdio
| null | true |
713/C
|
713
|
C
|
PyPy 3-64
|
TESTS
| 0 | 30 | 0 |
200936448
|
import heapq
n = int(input())
a = list(map(int, input().split()))
INF = 10 ** 18
dp = [INF] * n
dp[0] = 0
min_last = a[0]
heap_left = [-min_last]
heap_right = []
for i in range(1, n):
heapq.heappush(heap_right, a[i] - i)
if len(heap_right) > len(heap_left):
x = heapq.heappop(heap_right)
heapq.heappush(heap_left, -x)
if heap_left[0] < min_last:
min_last = heapq.heappop(heap_left)
heapq.heappush(heap_right, -min_last + i)
dp[i] = dp[i - 1] + a[i] - min_last - i
print(dp[-1])
| 57 | 202 | 4,096,000 |
73327542
|
from bisect import insort
class Graph:
def __init__(_):
_.change = [-10**27] # increment slope at ...
_.a = _.y = 0 # last line has slope a, starts from y
_.dx = 0 # the whole graph is shifted right by ...
def __repr__(_): return f"<{[x+_.dx for x in _.change]}; {_.a} {_.y}>"
def shiftx(_, v): _.dx+= v
def shifty(_, v): _.y+= v
def addleft(_, v):
if _.change[-1] < v-_.dx:
dx = v-_.dx - _.change[-1]
_.y+= _.a*dx
insort(_.change, v-_.dx)
def addright(_, v):
if _.change[-1] < v-_.dx:
dx = v-_.dx - _.change[-1]
_.y+= _.a*dx; _.a+= 1
insort(_.change, v-_.dx)
return
insort(_.change, v-_.dx)
_.a+= 1; _.y+= _.change[-1]-(v-_.dx)
#def remleft(_, v): change.remove(v-_.dx)
def cutright(_):
dx = _.change.pop()-_.change[-1]
_.a-= 1; _.y-= _.a*dx
n = int(input())
G = Graph()
for x in map(int,input().split()):
G.shiftx(1)
G.addleft(x)
G.addright(x)
while G.a > 0: G.cutright()
print(G.y)
|
Codeforces Round 371 (Div. 1)
|
CF
| 2,016 | 5 | 256 |
Sonya and Problem Wihtout a Legend
|
Sonya was unable to think of a story for this problem, so here comes the formal description.
You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.
|
The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.
Next line contains n integer ai (1 ≤ ai ≤ 109).
|
Print the minimum number of operation required to make the array strictly increasing.
| null |
In the first sample, the array is going to look as follows:
2 3 5 6 7 9 11
|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9
And for the second sample:
1 2 3 4 5
|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12
|
[{"input": "7\n2 1 5 11 5 9 11", "output": "9"}, {"input": "5\n5 4 3 2 1", "output": "12"}]
| 2,300 |
["dp", "sortings"]
| 57 |
[{"input": "7\r\n2 1 5 11 5 9 11\r\n", "output": "9\r\n"}, {"input": "5\r\n5 4 3 2 1\r\n", "output": "12\r\n"}, {"input": "2\r\n1 1000\r\n", "output": "0\r\n"}, {"input": "2\r\n1000 1\r\n", "output": "1000\r\n"}, {"input": "5\r\n100 80 60 70 90\r\n", "output": "54\r\n"}, {"input": "10\r\n10 16 17 11 1213 1216 1216 1209 3061 3062\r\n", "output": "16\r\n"}, {"input": "20\r\n103 103 110 105 107 119 113 121 116 132 128 124 128 125 138 137 140 136 154 158\r\n", "output": "43\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}, {"input": "5\r\n1 1 1 2 3\r\n", "output": "3\r\n"}, {"input": "1\r\n1000\r\n", "output": "0\r\n"}, {"input": "50\r\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319\r\n", "output": "12423\r\n"}, {"input": "75\r\n392 593 98 533 515 448 220 310 386 79 539 294 208 828 75 534 875 493 94 205 656 105 546 493 60 188 222 108 788 504 809 621 934 455 307 212 630 298 938 62 850 421 839 134 950 256 934 817 209 559 866 67 990 835 534 672 468 768 757 516 959 893 275 315 692 927 321 554 801 805 885 12 67 245 495\r\n", "output": "17691\r\n"}, {"input": "10\r\n26 723 970 13 422 968 875 329 234 983\r\n", "output": "2546\r\n"}, {"input": "20\r\n245 891 363 6 193 704 420 447 237 947 664 894 512 194 513 616 671 623 686 378\r\n", "output": "3208\r\n"}, {"input": "5\r\n850 840 521 42 169\r\n", "output": "1485\r\n"}]
| false |
stdio
| null | true |
522/C
|
522
|
C
|
Python 3
|
PRETESTS
| 3 | 61 | 0 |
10196356
|
def algorithm():
try:
sum_m = 0
sum_k = 0
set_count = int(input())
if not (1 <= set_count <= 100000):
return None
result = list()
for i in range(0, set_count):
input()
first = input().split(' ')
i_count = int(first[0])
k_count = int(first[1])
sum_m += i_count
sum_k += i_count
if sum_m > 100000 or sum_k > 100000:
return None
if not(2 <= i_count <= 100000) or not(1 <= k_count <= 100000):
return None
food = input().split(' ')
arr_food = [int(food[k]) for k in range(0, k_count)]
mask = ['N' for k in range(0, k_count)]
del food
for j in range(0, i_count - 1):
item = int(input().split(' ')[0])
if not(0 <= item <= k_count):
return None
if item > 0:
arr_food[item - 1] -= 1
if arr_food[item - 1] <= 0:
mask[item - 1] = 'Y'
else:
for k in range(0, k_count):
arr_food[k] -= 1
if arr_food[k] <= 0:
mask[k] = 'Y'
res = ''
for m in range(0, k_count):
res += mask[m]
result.append(res)
for i in range(0, set_count):
print(result[i])
except BaseException:
return None
algorithm()
| 48 | 998 | 13,926,400 |
10199791
|
t = int(input())
for i in range(t):
input()
m,k = map(int,input().split())
ak = list(map(int,input().split()))
ak2 = [0]*k
tjrj = [list(map(int,input().split())) for j in range(m-1)]
num = 0
num2 = 0
num3 = 100002
for j in range(m-1):
if num2 == 1 or tjrj[j][1] == 0:
if tjrj[j][0] != 0:
ak[tjrj[j][0]-1] -= 1
else:
num += 1
else:
for z in range(k):
if ak[z] - num < 1:
ak2[z] = 1
num2 = 1
if tjrj[j][0] != 0:
ak[tjrj[j][0]-1] -= 1
else:
num += 1
for f in range(j,m-1):
if tjrj[f][0] != 0:
ak2[tjrj[f][0]-1] = 0
for f in range(k):
if ak2[f] == 1:
if num3 > ak[f]:
num3 = ak[f]
num -= num3
for z in range(k):
if ak[z] - num < 1 or ak2[z] == 1:
print("Y",end="")
else:
print("N",end="")
print()
|
VK Cup 2015 - Qualification Round 1
|
CF
| 2,015 | 1 | 256 |
Chicken or Fish?
|
Polycarp is flying in the airplane. Finally, it is his favorite time — the lunchtime. The BerAvia company stewardess is giving food consecutively to all the passengers from the 1-th one to the last one. Polycarp is sitting on seat m, that means, he will be the m-th person to get food.
The flight menu has k dishes in total and when Polycarp boarded the flight, he had time to count the number of portions of each dish on board. Thus, he knows values a1, a2, ..., ak, where ai is the number of portions of the i-th dish.
The stewardess has already given food to m - 1 passengers, gave Polycarp a polite smile and asked him what he would prefer. That's when Polycarp realized that they might have run out of some dishes by that moment. For some of the m - 1 passengers ahead of him, he noticed what dishes they were given. Besides, he's heard some strange mumbling from some of the m - 1 passengers ahead of him, similar to phrase 'I'm disappointed'. That happened when a passenger asked for some dish but the stewardess gave him a polite smile and said that they had run out of that dish. In that case the passenger needed to choose some other dish that was available. If Polycarp heard no more sounds from a passenger, that meant that the passenger chose his dish at the first try.
Help Polycarp to find out for each dish: whether they could have run out of the dish by the moment Polyarp was served or that dish was definitely available.
|
Each test in this problem consists of one or more input sets. First goes a string that contains a single integer t (1 ≤ t ≤ 100 000) — the number of input data sets in the test. Then the sets follow, each set is preceded by an empty line.
The first line of each set of the input contains integers m, k (2 ≤ m ≤ 100 000, 1 ≤ k ≤ 100 000) — the number of Polycarp's seat and the number of dishes, respectively.
The second line contains a sequence of k integers a1, a2, ..., ak (1 ≤ ai ≤ 100 000), where ai is the initial number of portions of the i-th dish.
Then m - 1 lines follow, each line contains the description of Polycarp's observations about giving food to a passenger sitting in front of him: the j-th line contains a pair of integers tj, rj (0 ≤ tj ≤ k, 0 ≤ rj ≤ 1), where tj is the number of the dish that was given to the j-th passenger (or 0, if Polycarp didn't notice what dish was given to the passenger), and rj — a 1 or a 0, depending on whether the j-th passenger was or wasn't disappointed, respectively.
We know that sum ai equals at least m, that is,Polycarp will definitely get some dish, even if it is the last thing he wanted. It is guaranteed that the data is consistent.
Sum m for all input sets doesn't exceed 100 000. Sum k for all input sets doesn't exceed 100 000.
|
For each input set print the answer as a single line. Print a string of k letters "Y" or "N". Letter "Y" in position i should be printed if they could have run out of the i-th dish by the time the stewardess started serving Polycarp.
| null |
In the first input set depending on the choice of the second passenger the situation could develop in different ways:
- If he chose the first dish, then by the moment the stewardess reaches Polycarp, they will have run out of the first dish;
- If he chose the fourth dish, then by the moment the stewardess reaches Polycarp, they will have run out of the fourth dish;
- Otherwise, Polycarp will be able to choose from any of the four dishes.
Thus, the answer is "YNNY".
In the second input set there is, for example, the following possible scenario. First, the first passenger takes the only third dish, then the second passenger takes the second dish. Then, the third passenger asks for the third dish, but it is not available, so he makes disappointed muttering and ends up with the second dish. Then the fourth passenger takes the fourth dish, and Polycarp ends up with the choice between the first, fourth and fifth dish.
Likewise, another possible scenario is when by the time the stewardess comes to Polycarp, they will have run out of either the first or the fifth dish (this can happen if one of these dishes is taken by the second passenger). It is easy to see that there is more than enough of the fourth dish, so Polycarp can always count on it. Thus, the answer is "YYYNY".
|
[{"input": "2\n3 4\n2 3 2 1\n1 0\n0 0\n5 5\n1 2 1 3 1\n3 0\n0 0\n2 1\n4 0", "output": "YNNY\nYYYNY"}]
| 2,100 |
["greedy"]
| 48 |
[{"input": "2\r\n\r\n3 4\r\n2 3 2 1\r\n1 0\r\n0 0\r\n\r\n5 5\r\n1 2 1 3 1\r\n3 0\r\n0 0\r\n2 1\r\n4 0\r\n", "output": "YNNY\r\nYYYNY\r\n"}, {"input": "4\r\n\r\n2 1\r\n42\r\n0 0\r\n\r\n2 1\r\n2\r\n0 0\r\n\r\n2 1\r\n42\r\n1 0\r\n\r\n2 1\r\n2\r\n1 0\r\n", "output": "N\r\nN\r\nN\r\nN\r\n"}, {"input": "5\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n2 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 0\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 1\r\n", "output": "YYY\r\nYYN\r\nYYY\r\nYYY\r\nYYY\r\n"}, {"input": "1\r\n\r\n4 2\r\n2 2\r\n0 0\r\n0 0\r\n1 1\r\n", "output": "NY\r\n"}]
| false |
stdio
| null | true |
522/C
|
522
|
C
|
Python 3
|
PRETESTS
| 3 | 62 | 0 |
10207046
|
tests = int(input())
for i in range(tests):
input()
inp = input().split()
m = int(inp[0])
k = int(inp[1])
a = []
inp = input().split()
sum = 0
for j in range(k):
a.append(int(inp[j]))
sum += a[j]
has_zero = False
for j in range(m-1):
inp = input().split()
t = int(inp[0])
r = int(inp[1])
if r == 1:
if not has_zero:
has_zero = True
for z in range(k):
if a[z] != 0:
a[z] += 1
if t == 0:
for z in range(k):
if a[z] != 0:
a[z] -= 1
if t != 0:
if a[t-1] != 0:
a[t-1] -= 1
if a[t-1] == 0:
has_zero = True
sum -= 1
if sum < k:
has_zero = True
res = ""
for j in range(k):
if a[j] == 0:
res += 'Y'
else:
res += 'N'
print(res)
| 48 | 654 | 10,342,400 |
10207272
|
import sys
test_count = int(sys.stdin.readline())
for test in range(test_count):
sys.stdin.readline()
m, k = map(int, sys.stdin.readline().split())
counts = list(map(int, sys.stdin.readline().split()))
took = []
unhappy = []
for i in range(m - 1):
t, r = map(int, sys.stdin.readline().split())
t -= 1 # -1 means unknown
took.append(t)
unhappy.append(r)
took_in_total = [0 for dish_count in counts]
for i in range(m - 1):
if took[i] != -1:
took_in_total[took[i]] += 1
took_already = [0 for dish_count in counts]
left = [dish_count for dish_count in counts]
answer = [False for dish_count in counts]
unknown = 0
all_present = True
for i in range(m - 1):
if unhappy[i] == 1 and all_present:
could_exhaust = []
for j in range(k):
if took_already[j] < took_in_total[j]:
continue
if left[j] > unknown:
continue
could_exhaust.append(j)
if len(could_exhaust) == 0:
raise AssertionError
for j in could_exhaust:
answer[j] = True
unknown -= min(map(lambda j: left[j], could_exhaust))
all_present = False
if took[i] != -1:
left[took[i]] -= 1
took_already[took[i]] += 1
if left[took[i]] == 0:
all_present = False
else:
unknown += 1
for j in range(k):
if left[j] <= unknown:
answer[j] = True
sys.stdout.write(
''.join(map(lambda x: 'Y' if x else 'N', answer)) + '\n'
)
|
VK Cup 2015 - Qualification Round 1
|
CF
| 2,015 | 1 | 256 |
Chicken or Fish?
|
Polycarp is flying in the airplane. Finally, it is his favorite time — the lunchtime. The BerAvia company stewardess is giving food consecutively to all the passengers from the 1-th one to the last one. Polycarp is sitting on seat m, that means, he will be the m-th person to get food.
The flight menu has k dishes in total and when Polycarp boarded the flight, he had time to count the number of portions of each dish on board. Thus, he knows values a1, a2, ..., ak, where ai is the number of portions of the i-th dish.
The stewardess has already given food to m - 1 passengers, gave Polycarp a polite smile and asked him what he would prefer. That's when Polycarp realized that they might have run out of some dishes by that moment. For some of the m - 1 passengers ahead of him, he noticed what dishes they were given. Besides, he's heard some strange mumbling from some of the m - 1 passengers ahead of him, similar to phrase 'I'm disappointed'. That happened when a passenger asked for some dish but the stewardess gave him a polite smile and said that they had run out of that dish. In that case the passenger needed to choose some other dish that was available. If Polycarp heard no more sounds from a passenger, that meant that the passenger chose his dish at the first try.
Help Polycarp to find out for each dish: whether they could have run out of the dish by the moment Polyarp was served or that dish was definitely available.
|
Each test in this problem consists of one or more input sets. First goes a string that contains a single integer t (1 ≤ t ≤ 100 000) — the number of input data sets in the test. Then the sets follow, each set is preceded by an empty line.
The first line of each set of the input contains integers m, k (2 ≤ m ≤ 100 000, 1 ≤ k ≤ 100 000) — the number of Polycarp's seat and the number of dishes, respectively.
The second line contains a sequence of k integers a1, a2, ..., ak (1 ≤ ai ≤ 100 000), where ai is the initial number of portions of the i-th dish.
Then m - 1 lines follow, each line contains the description of Polycarp's observations about giving food to a passenger sitting in front of him: the j-th line contains a pair of integers tj, rj (0 ≤ tj ≤ k, 0 ≤ rj ≤ 1), where tj is the number of the dish that was given to the j-th passenger (or 0, if Polycarp didn't notice what dish was given to the passenger), and rj — a 1 or a 0, depending on whether the j-th passenger was or wasn't disappointed, respectively.
We know that sum ai equals at least m, that is,Polycarp will definitely get some dish, even if it is the last thing he wanted. It is guaranteed that the data is consistent.
Sum m for all input sets doesn't exceed 100 000. Sum k for all input sets doesn't exceed 100 000.
|
For each input set print the answer as a single line. Print a string of k letters "Y" or "N". Letter "Y" in position i should be printed if they could have run out of the i-th dish by the time the stewardess started serving Polycarp.
| null |
In the first input set depending on the choice of the second passenger the situation could develop in different ways:
- If he chose the first dish, then by the moment the stewardess reaches Polycarp, they will have run out of the first dish;
- If he chose the fourth dish, then by the moment the stewardess reaches Polycarp, they will have run out of the fourth dish;
- Otherwise, Polycarp will be able to choose from any of the four dishes.
Thus, the answer is "YNNY".
In the second input set there is, for example, the following possible scenario. First, the first passenger takes the only third dish, then the second passenger takes the second dish. Then, the third passenger asks for the third dish, but it is not available, so he makes disappointed muttering and ends up with the second dish. Then the fourth passenger takes the fourth dish, and Polycarp ends up with the choice between the first, fourth and fifth dish.
Likewise, another possible scenario is when by the time the stewardess comes to Polycarp, they will have run out of either the first or the fifth dish (this can happen if one of these dishes is taken by the second passenger). It is easy to see that there is more than enough of the fourth dish, so Polycarp can always count on it. Thus, the answer is "YYYNY".
|
[{"input": "2\n3 4\n2 3 2 1\n1 0\n0 0\n5 5\n1 2 1 3 1\n3 0\n0 0\n2 1\n4 0", "output": "YNNY\nYYYNY"}]
| 2,100 |
["greedy"]
| 48 |
[{"input": "2\r\n\r\n3 4\r\n2 3 2 1\r\n1 0\r\n0 0\r\n\r\n5 5\r\n1 2 1 3 1\r\n3 0\r\n0 0\r\n2 1\r\n4 0\r\n", "output": "YNNY\r\nYYYNY\r\n"}, {"input": "4\r\n\r\n2 1\r\n42\r\n0 0\r\n\r\n2 1\r\n2\r\n0 0\r\n\r\n2 1\r\n42\r\n1 0\r\n\r\n2 1\r\n2\r\n1 0\r\n", "output": "N\r\nN\r\nN\r\nN\r\n"}, {"input": "5\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n2 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 0\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 1\r\n", "output": "YYY\r\nYYN\r\nYYY\r\nYYY\r\nYYY\r\n"}, {"input": "1\r\n\r\n4 2\r\n2 2\r\n0 0\r\n0 0\r\n1 1\r\n", "output": "NY\r\n"}]
| false |
stdio
| null | true |
522/C
|
522
|
C
|
Python 3
|
PRETESTS
| 3 | 46 | 0 |
10195992
|
for i in range(int(input())):
_ = input()
m, k = [int(x) for x in input().split()]
foods = [[int(x), 0] for x in input().split()]
n = 0
for i in range(m-1):
eat, sad = [int(x) for x in input().split()]
if sad:
maybe = [(i, x) for i, x in enumerate(foods) if x[0] - x[1] - n <= 0]
if len(maybe) == 1:
i = maybe[0][0]
x = maybe[0][1]
dif = x[0] - x[1]
foods[i] = [x[0], x[0]]
n -= dif
if not eat: n += 1
else: foods[eat-1][1] += 1
print(''.join(['N' if x[0] - x[1] - n > 0 else 'Y' for x in foods]))
| 48 | 967 | 7,475,200 |
10211431
|
t = int(input())
for j in range(t):
e = input()
m, k = map(int, input().split())
arr = [int(i) for i in input().split()]
sum, bfail = [0] * k, [0] * k
ffail, undef = -1, 0
used = [False] * k
ubfail = 0
for i in range(m - 1):
c, ns = map(int, input().split())
if c == 0:
undef += 1
if ns == 0 and ffail == -1:
ubfail += 1
else:
sum[c - 1] += 1
if ns == 0 and ffail == -1:
bfail[c - 1] += 1
if ns and ffail == -1:
ffail = i
if ffail != -1 and c > 0:
used[c - 1] = True
if ffail == -1:
for i in range(k):
if sum[i] + undef >= arr[i]:
print('Y', end = '')
else:
print('N', end = '')
print()
continue
minu = 10 ** 6
for i in range(k):
if not used[i] and arr[i] - bfail[i] < minu:
minu = arr[i] - bfail[i]
best = i
for i in range(k):
if i == best or undef - minu + sum[i] >= arr[i]:
print('Y', end = '')
elif bfail[i] + ubfail >= arr[i] and not used[i]:
print('Y', end = '')
else:
print('N', end = '')
print()
|
VK Cup 2015 - Qualification Round 1
|
CF
| 2,015 | 1 | 256 |
Chicken or Fish?
|
Polycarp is flying in the airplane. Finally, it is his favorite time — the lunchtime. The BerAvia company stewardess is giving food consecutively to all the passengers from the 1-th one to the last one. Polycarp is sitting on seat m, that means, he will be the m-th person to get food.
The flight menu has k dishes in total and when Polycarp boarded the flight, he had time to count the number of portions of each dish on board. Thus, he knows values a1, a2, ..., ak, where ai is the number of portions of the i-th dish.
The stewardess has already given food to m - 1 passengers, gave Polycarp a polite smile and asked him what he would prefer. That's when Polycarp realized that they might have run out of some dishes by that moment. For some of the m - 1 passengers ahead of him, he noticed what dishes they were given. Besides, he's heard some strange mumbling from some of the m - 1 passengers ahead of him, similar to phrase 'I'm disappointed'. That happened when a passenger asked for some dish but the stewardess gave him a polite smile and said that they had run out of that dish. In that case the passenger needed to choose some other dish that was available. If Polycarp heard no more sounds from a passenger, that meant that the passenger chose his dish at the first try.
Help Polycarp to find out for each dish: whether they could have run out of the dish by the moment Polyarp was served or that dish was definitely available.
|
Each test in this problem consists of one or more input sets. First goes a string that contains a single integer t (1 ≤ t ≤ 100 000) — the number of input data sets in the test. Then the sets follow, each set is preceded by an empty line.
The first line of each set of the input contains integers m, k (2 ≤ m ≤ 100 000, 1 ≤ k ≤ 100 000) — the number of Polycarp's seat and the number of dishes, respectively.
The second line contains a sequence of k integers a1, a2, ..., ak (1 ≤ ai ≤ 100 000), where ai is the initial number of portions of the i-th dish.
Then m - 1 lines follow, each line contains the description of Polycarp's observations about giving food to a passenger sitting in front of him: the j-th line contains a pair of integers tj, rj (0 ≤ tj ≤ k, 0 ≤ rj ≤ 1), where tj is the number of the dish that was given to the j-th passenger (or 0, if Polycarp didn't notice what dish was given to the passenger), and rj — a 1 or a 0, depending on whether the j-th passenger was or wasn't disappointed, respectively.
We know that sum ai equals at least m, that is,Polycarp will definitely get some dish, even if it is the last thing he wanted. It is guaranteed that the data is consistent.
Sum m for all input sets doesn't exceed 100 000. Sum k for all input sets doesn't exceed 100 000.
|
For each input set print the answer as a single line. Print a string of k letters "Y" or "N". Letter "Y" in position i should be printed if they could have run out of the i-th dish by the time the stewardess started serving Polycarp.
| null |
In the first input set depending on the choice of the second passenger the situation could develop in different ways:
- If he chose the first dish, then by the moment the stewardess reaches Polycarp, they will have run out of the first dish;
- If he chose the fourth dish, then by the moment the stewardess reaches Polycarp, they will have run out of the fourth dish;
- Otherwise, Polycarp will be able to choose from any of the four dishes.
Thus, the answer is "YNNY".
In the second input set there is, for example, the following possible scenario. First, the first passenger takes the only third dish, then the second passenger takes the second dish. Then, the third passenger asks for the third dish, but it is not available, so he makes disappointed muttering and ends up with the second dish. Then the fourth passenger takes the fourth dish, and Polycarp ends up with the choice between the first, fourth and fifth dish.
Likewise, another possible scenario is when by the time the stewardess comes to Polycarp, they will have run out of either the first or the fifth dish (this can happen if one of these dishes is taken by the second passenger). It is easy to see that there is more than enough of the fourth dish, so Polycarp can always count on it. Thus, the answer is "YYYNY".
|
[{"input": "2\n3 4\n2 3 2 1\n1 0\n0 0\n5 5\n1 2 1 3 1\n3 0\n0 0\n2 1\n4 0", "output": "YNNY\nYYYNY"}]
| 2,100 |
["greedy"]
| 48 |
[{"input": "2\r\n\r\n3 4\r\n2 3 2 1\r\n1 0\r\n0 0\r\n\r\n5 5\r\n1 2 1 3 1\r\n3 0\r\n0 0\r\n2 1\r\n4 0\r\n", "output": "YNNY\r\nYYYNY\r\n"}, {"input": "4\r\n\r\n2 1\r\n42\r\n0 0\r\n\r\n2 1\r\n2\r\n0 0\r\n\r\n2 1\r\n42\r\n1 0\r\n\r\n2 1\r\n2\r\n1 0\r\n", "output": "N\r\nN\r\nN\r\nN\r\n"}, {"input": "5\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n2 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 0\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 1\r\n", "output": "YYY\r\nYYN\r\nYYY\r\nYYY\r\nYYY\r\n"}, {"input": "1\r\n\r\n4 2\r\n2 2\r\n0 0\r\n0 0\r\n1 1\r\n", "output": "NY\r\n"}]
| false |
stdio
| null | true |
522/C
|
522
|
C
|
Python 3
|
PRETESTS
| 3 | 61 | 0 |
10208721
|
t = int(input())
for v in range(t):
q = 0
empty_line = input()
m,k = map(int,input().split())
res = [0]*k
t,r = [0]*(m-1),[0]*(m-1)
help = [0]*k
amount_of_food = list(map(int,input().split()))
for i in range(m-1):
t[i],r[i] = map(int,input().split())
for i in range(m-1):
if t[i]!=0:
if r[i]==0:
amount_of_food[t[i]-1] -=1
else:
res = ['Y']*k
amount_of_food[t[i]-1] -=1
if amount_of_food[t[i]-1]>0:
res[t[i]-1] = 'N'
unknown = t.count(0)
#print(amount_of_food)
#print(res)
for i in range(k):
if amount_of_food[i]>unknown:
res[i] = 'N'
else:
res[i] = 'Y'
print(''.join(map(str,res)))
| 48 | 998 | 13,926,400 |
10199791
|
t = int(input())
for i in range(t):
input()
m,k = map(int,input().split())
ak = list(map(int,input().split()))
ak2 = [0]*k
tjrj = [list(map(int,input().split())) for j in range(m-1)]
num = 0
num2 = 0
num3 = 100002
for j in range(m-1):
if num2 == 1 or tjrj[j][1] == 0:
if tjrj[j][0] != 0:
ak[tjrj[j][0]-1] -= 1
else:
num += 1
else:
for z in range(k):
if ak[z] - num < 1:
ak2[z] = 1
num2 = 1
if tjrj[j][0] != 0:
ak[tjrj[j][0]-1] -= 1
else:
num += 1
for f in range(j,m-1):
if tjrj[f][0] != 0:
ak2[tjrj[f][0]-1] = 0
for f in range(k):
if ak2[f] == 1:
if num3 > ak[f]:
num3 = ak[f]
num -= num3
for z in range(k):
if ak[z] - num < 1 or ak2[z] == 1:
print("Y",end="")
else:
print("N",end="")
print()
|
VK Cup 2015 - Qualification Round 1
|
CF
| 2,015 | 1 | 256 |
Chicken or Fish?
|
Polycarp is flying in the airplane. Finally, it is his favorite time — the lunchtime. The BerAvia company stewardess is giving food consecutively to all the passengers from the 1-th one to the last one. Polycarp is sitting on seat m, that means, he will be the m-th person to get food.
The flight menu has k dishes in total and when Polycarp boarded the flight, he had time to count the number of portions of each dish on board. Thus, he knows values a1, a2, ..., ak, where ai is the number of portions of the i-th dish.
The stewardess has already given food to m - 1 passengers, gave Polycarp a polite smile and asked him what he would prefer. That's when Polycarp realized that they might have run out of some dishes by that moment. For some of the m - 1 passengers ahead of him, he noticed what dishes they were given. Besides, he's heard some strange mumbling from some of the m - 1 passengers ahead of him, similar to phrase 'I'm disappointed'. That happened when a passenger asked for some dish but the stewardess gave him a polite smile and said that they had run out of that dish. In that case the passenger needed to choose some other dish that was available. If Polycarp heard no more sounds from a passenger, that meant that the passenger chose his dish at the first try.
Help Polycarp to find out for each dish: whether they could have run out of the dish by the moment Polyarp was served or that dish was definitely available.
|
Each test in this problem consists of one or more input sets. First goes a string that contains a single integer t (1 ≤ t ≤ 100 000) — the number of input data sets in the test. Then the sets follow, each set is preceded by an empty line.
The first line of each set of the input contains integers m, k (2 ≤ m ≤ 100 000, 1 ≤ k ≤ 100 000) — the number of Polycarp's seat and the number of dishes, respectively.
The second line contains a sequence of k integers a1, a2, ..., ak (1 ≤ ai ≤ 100 000), where ai is the initial number of portions of the i-th dish.
Then m - 1 lines follow, each line contains the description of Polycarp's observations about giving food to a passenger sitting in front of him: the j-th line contains a pair of integers tj, rj (0 ≤ tj ≤ k, 0 ≤ rj ≤ 1), where tj is the number of the dish that was given to the j-th passenger (or 0, if Polycarp didn't notice what dish was given to the passenger), and rj — a 1 or a 0, depending on whether the j-th passenger was or wasn't disappointed, respectively.
We know that sum ai equals at least m, that is,Polycarp will definitely get some dish, even if it is the last thing he wanted. It is guaranteed that the data is consistent.
Sum m for all input sets doesn't exceed 100 000. Sum k for all input sets doesn't exceed 100 000.
|
For each input set print the answer as a single line. Print a string of k letters "Y" or "N". Letter "Y" in position i should be printed if they could have run out of the i-th dish by the time the stewardess started serving Polycarp.
| null |
In the first input set depending on the choice of the second passenger the situation could develop in different ways:
- If he chose the first dish, then by the moment the stewardess reaches Polycarp, they will have run out of the first dish;
- If he chose the fourth dish, then by the moment the stewardess reaches Polycarp, they will have run out of the fourth dish;
- Otherwise, Polycarp will be able to choose from any of the four dishes.
Thus, the answer is "YNNY".
In the second input set there is, for example, the following possible scenario. First, the first passenger takes the only third dish, then the second passenger takes the second dish. Then, the third passenger asks for the third dish, but it is not available, so he makes disappointed muttering and ends up with the second dish. Then the fourth passenger takes the fourth dish, and Polycarp ends up with the choice between the first, fourth and fifth dish.
Likewise, another possible scenario is when by the time the stewardess comes to Polycarp, they will have run out of either the first or the fifth dish (this can happen if one of these dishes is taken by the second passenger). It is easy to see that there is more than enough of the fourth dish, so Polycarp can always count on it. Thus, the answer is "YYYNY".
|
[{"input": "2\n3 4\n2 3 2 1\n1 0\n0 0\n5 5\n1 2 1 3 1\n3 0\n0 0\n2 1\n4 0", "output": "YNNY\nYYYNY"}]
| 2,100 |
["greedy"]
| 48 |
[{"input": "2\r\n\r\n3 4\r\n2 3 2 1\r\n1 0\r\n0 0\r\n\r\n5 5\r\n1 2 1 3 1\r\n3 0\r\n0 0\r\n2 1\r\n4 0\r\n", "output": "YNNY\r\nYYYNY\r\n"}, {"input": "4\r\n\r\n2 1\r\n42\r\n0 0\r\n\r\n2 1\r\n2\r\n0 0\r\n\r\n2 1\r\n42\r\n1 0\r\n\r\n2 1\r\n2\r\n1 0\r\n", "output": "N\r\nN\r\nN\r\nN\r\n"}, {"input": "5\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n2 1\r\n\r\n3 3\r\n1 1 1\r\n1 0\r\n0 1\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 0\r\n\r\n3 3\r\n1 1 1\r\n0 0\r\n1 1\r\n", "output": "YYY\r\nYYN\r\nYYY\r\nYYY\r\nYYY\r\n"}, {"input": "1\r\n\r\n4 2\r\n2 2\r\n0 0\r\n0 0\r\n1 1\r\n", "output": "NY\r\n"}]
| false |
stdio
| null | true |
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