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class |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
227538365
|
n=int(input())
for i in range(n):
angle=int(input())
if angle in [60,90,120,135,144,150,160,162,165,168,170,171,174,175,176,177,178,179]:
print('YES')
else:
print('NO')
| 3 | 30 | 0 |
144631819
|
for i in range(int(input())):
n=int(input())
if n<60:
print("NO")
else:
if (360%(180-n))==0:
print("YES")
else:
print("NO")
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
193129242
|
n = int(input())
l = [-180, 0, 60, 90, 108, 120, 135, 140, 144, 150, 156, 160,
162, 165, 168, 170, 171, 172, 174, 175, 176, 178, 179]
for i in range(n):
x = int(input())
if x in l:
print('YES')
else: print('NO')
| 3 | 30 | 0 |
146321238
|
def solve():
t = int(input())
q = [int(input()) for i in range(t)]
angles = {}
for i in q:
n = -360/(i-180)
if n.is_integer():
print('YES')
else:
print('NO')
solve()
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
786/A
|
786
|
A
|
Python 3
|
TESTS
| 0 | 30 | 0 |
227030291
|
n = int(input())
rick_set = list(map(int, input().split()))[1:]
morty_set = list(map(int, input().split()))[1:]
win_rick = [False] * (n - 1)
win_morty = [False] * (n - 1)
for i in range(n - 1, 0, -1):
if i + rick_set[0] >= n:
win_rick[i - 1] = True
for i in range(n - 1, 0, -1):
for x in morty_set:
if i + x >= n or not win_rick[i + x - 1]:
win_morty[i - 1] = True
break
for i in range(n - 1):
if win_rick[i]:
print("Win", end=" ")
elif win_morty[i]:
print("Lose", end=" ")
else:
print("Loop", end=" ")
print()
for i in range(n - 1):
if win_morty[i]:
print("Win", end=" ")
elif win_rick[i]:
print("Lose", end=" ")
else:
print("Loop", end=" ")
| 49 | 1,309 | 28,979,200 |
33321494
|
f = lambda: list(map(int, input().split()))[1:]
n = int(input())
s, p, q = [], [], []
for x in [0, 1]:
r = f()
s.append(r)
t = [len(r)] * n
t[0] = 0
p.append(t)
q.append((x, 0))
while q:
x, i = q.pop()
y = 1 - x
for d in s[y]:
j = (i - d) % n
if p[y][j] < 1: continue
p[y][j] = -1
for d in s[x]:
k = (j - d) % n
if p[x][k] < 1: continue
p[x][k] -= 1
if p[x][k] == 0: q.append((x, k))
for x in [0, 1]:
print(*[['Lose', 'Loop', 'Win'][min(q, 1)] for q in p[x][1:]])
|
Codeforces Round 406 (Div. 1)
|
CF
| 2,017 | 4 | 256 |
Berzerk
|
Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.
In this game there are n objects numbered from 1 to n arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There's a monster in one of the planet. Rick and Morty don't know on which one yet, only that he's not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.
Each one of them has a set of numbers between 1 and n - 1 (inclusive). Rick's set is s1 with k1 elements and Morty's is s2 with k2 elements. One of them goes first and the player changes alternatively. In each player's turn, he should choose an arbitrary number like x from his set and the monster will move to his x-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.
Your task is that for each of monster's initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.
|
The first line of input contains a single integer n (2 ≤ n ≤ 7000) — number of objects in game.
The second line contains integer k1 followed by k1 distinct integers s1, 1, s1, 2, ..., s1, k1 — Rick's set.
The third line contains integer k2 followed by k2 distinct integers s2, 1, s2, 2, ..., s2, k2 — Morty's set
1 ≤ ki ≤ n - 1 and 1 ≤ si, 1, si, 2, ..., si, ki ≤ n - 1 for 1 ≤ i ≤ 2.
|
In the first line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Rick plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Similarly, in the second line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Morty plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.
| null | null |
[{"input": "5\n2 3 2\n3 1 2 3", "output": "Lose Win Win Loop\nLoop Win Win Win"}, {"input": "8\n4 6 2 3 4\n2 3 6", "output": "Win Win Win Win Win Win Win\nLose Win Lose Lose Win Lose Lose"}]
| 2,000 |
["dfs and similar", "dp", "games"]
| 49 |
[{"input": "5\r\n2 3 2\r\n3 1 2 3\r\n", "output": "Lose Win Win Loop\r\nLoop Win Win Win\r\n"}, {"input": "8\r\n4 6 2 3 4\r\n2 3 6\r\n", "output": "Win Win Win Win Win Win Win\r\nLose Win Lose Lose Win Lose Lose\r\n"}, {"input": "10\r\n3 4 7 5\r\n2 8 5\r\n", "output": "Win Win Win Win Win Win Win Loop Win\r\nLose Win Loop Lose Win Lose Lose Lose Lose\r\n"}, {"input": "17\r\n1 10\r\n1 12\r\n", "output": "Win Win Win Win Win Win Win Win Win Win Win Lose Win Win Win Win\r\nLose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose\r\n"}, {"input": "23\r\n1 20\r\n3 9 2 12\r\n", "output": "Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose\r\nWin Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win\r\n"}, {"input": "85\r\n12 76 7 75 51 43 41 66 13 59 48 81 73\r\n3 65 60 25\r\n", "output": "Loop Loop Loop Win Loop Loop Loop Loop Win Win Loop Win Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Win Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop \nLoop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop \n"}, {"input": "100\r\n84 80 73 28 76 21 44 97 63 59 6 77 41 2 8 71 57 19 33 46 92 5 61 88 53 68 94 56 14 35 4 47 17 79 84 10 67 58 45 38 13 12 87 3 91 30 15 11 24 55 62 39 83 43 89 1 81 75 50 86 72 18 52 78 7 29 64 42 70 49 37 25 66 74 95 36 85 48 99 60 51 98 27 40 93\r\n47 52 76 9 4 25 8 63 29 74 97 61 93 35 49 62 5 10 57 73 42 3 19 23 71 70 43 67 48 2 34 31 41 90 18 6 40 83 98 72 14 51 38 46 21 99 65 37\r\n", "output": "Win Win Win Loop Win Win Win Win Win Loop Win Win Win Win Win Win Win Loop Win Win Win Win Win Win Win Win Win Win Win Win Loop Win Win Win Loop Win Win Win Win Win Win Win Win Win Win Loop Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Loop Win Loop Loop Win Win Win Win Loop Win Win Loop Loop Win Loop Win Win Win Loop Win Win Win Win Win Win Loop Win Win Win Win Win Win Win Win \nWin Win Win Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Win Loop Win Win Win Win Win Loop Loop Win Loop Win Loop Win Win Win Loop Loop Loop Win Loop Loop Loop Loop Win Win Loop Win Win Loop Win Loop Loop Win Win Win Win Loop Win Win Loop Win Win Loop Loop Win Loop Win Loop Loop Loop Win Loop Win Loop Win Loop Win Win Loop Loop Loop Win Loop Loop Loop Win Win Win Loop Win Win Win Win Win Loop \n"}, {"input": "100\r\n66 70 54 10 72 81 84 56 15 27 19 43 55 49 44 52 33 63 40 95 17 58 2 51 39 22 18 82 1 16 99 32 29 24 94 9 98 5 37 47 14 42 73 41 31 79 64 12 6 53 26 68 67 89 13 90 4 21 93 46 74 75 88 66 57 23 7\r\n18 8 47 76 39 34 52 62 5 36 19 22 80 32 71 55 7 37 57\r\n", "output": "Win Win Loop Loop Win Win Win Loop Loop Win Win Win Loop Loop Loop Win Loop Win Win Loop Win Loop Loop Loop Win Win Win Win Loop Win Loop Win Win Win Loop Win Win Loop Loop Loop Loop Win Win Win Win Win Win Win Win Loop Win Loop Win Win Loop Win Win Win Win Win Win Loop Win Loop Loop Loop Win Win Win Loop Win Loop Win Win Loop Win Win Win Win Loop Win Win Win Win Win Win Win Win Loop Win Win Loop Win Win Win Win Loop Win Win \nLoop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Win Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Win Loop Win Loop Loop Win Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Win Loop Win Win Loop Win Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Win Loop Win Loop Loop Loop Loop \n"}, {"input": "300\r\n1 179\r\n2 293 180\r\n", "output": "Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose \nWin Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win \n"}, {"input": "1000\r\n14 77 649 670 988 469 453 445 885 101 58 728 474 488 230\r\n8 83 453 371 86 834 277 847 958\r\n", "output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop \nLoop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop \n"}, {"input": "2\r\n1 1\r\n1 1\r\n", "output": "Win\r\nWin\r\n"}, {"input": "2\r\n1 1\r\n1 1\r\n", "output": "Win\r\nWin\r\n"}, {"input": "3\r\n1 1\r\n1 2\r\n", "output": "Loop Win\r\nWin Loop\r\n"}, {"input": "20\r\n1 1\r\n1 11\r\n", "output": "Loop Loop Win Lose Loop Loop Win Lose Loop Loop Win Lose Loop Loop Win Lose Loop Loop Win\r\nWin Loop Loop Lose Win Loop Loop Lose Win Loop Loop Lose Win Loop Loop Lose Win Loop Loop\r\n"}, {"input": "309\r\n30 197 38 142 159 163 169 263 70 151 288 264 41 285 225 216 306 128 242 221 94 39 43 292 54 157 78 272 257 97 57\r\n3 97 172 165\r\n", "output": "Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Win Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Win Win Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop 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Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose Loop Loop Loop Win Loop Loop Loop Lose 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Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop Lose Loop Win Loop Loop Loop Loop \n"}, {"input": "7000\r\n1 3041\r\n1 6128\r\n", "output": "Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win 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Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win 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Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Lose Win Win Win Win Win Win Win Win Win Win Win Win Win Win Win \n"}, {"input": "7000\r\n5 5080 4890 1201 4903 1360\r\n5 2415 6678 5200 2282 4648\r\n", "output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop 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Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop \n"}, {"input": "7000\r\n3 6965 1271 5818\r\n3 6331 5681 6636\r\n", "output": "Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Win Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop 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Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop Loop \n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
200430380
|
a=int(input())
b=[]
for i in range(a):
c=int(input())
if c>=60 and c<=179:
b.append("YES")
else:
b.append("NO")
for i in b:
print(i)
| 3 | 30 | 0 |
175516098
|
# -*- coding: utf-8 -*-
"""
Created on Tue Oct 11 15:21:27 2022
@author: lpf
"""
t=int(input())
for i in range(t):
a=int(input())
print('YES'if 360%(180-a)==0 else 'NO')
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
990/F
|
990
|
F
|
Python 3
|
TESTS
| 0 | 31 | 0 |
158501113
|
print("Impossible")
# 凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数凑字数
| 37 | 1,388 | 80,486,400 |
39209819
|
import sys
from time import time
def i_ints():
return list(map(int, sys.stdin.readline().split()))
def main():
limit =10**10
n, = i_ints()
s = [0] + i_ints()
if sum(s):
print("Impossible")
return
print("Possible")
m, = i_ints()
es = [i_ints() for _ in range(m)]
nb = [[] for i in range(n+1)]
for i, (x, y) in enumerate(es):
nb[x].append((y, i, 1))
nb[y].append((x, i, -1))
path = []
def make_path():
stack = []
seen = [False] * (n+1)
stack.append(1)
seen[1] = True
while stack:
x = stack.pop()
for y, i, factor in nb[x]:
if not seen[y]:
seen[y] = True
stack.append(y)
path.append((x, y, i, factor))
make_path()
f = [0] * m
for x, y, i,factor in reversed(path):
f[i] = factor * s[y]
s[x] += s[y]
s[y] = 0
print("\n".join(map(str, f)))
return
main()
|
Educational Codeforces Round 45 (Rated for Div. 2)
|
ICPC
| 2,018 | 2 | 256 |
Flow Control
|
You have to handle a very complex water distribution system. The system consists of $$$n$$$ junctions and $$$m$$$ pipes, $$$i$$$-th pipe connects junctions $$$x_i$$$ and $$$y_i$$$.
The only thing you can do is adjusting the pipes. You have to choose $$$m$$$ integer numbers $$$f_1$$$, $$$f_2$$$, ..., $$$f_m$$$ and use them as pipe settings. $$$i$$$-th pipe will distribute $$$f_i$$$ units of water per second from junction $$$x_i$$$ to junction $$$y_i$$$ (if $$$f_i$$$ is negative, then the pipe will distribute $$$|f_i|$$$ units of water per second from junction $$$y_i$$$ to junction $$$x_i$$$). It is allowed to set $$$f_i$$$ to any integer from $$$-2 \cdot 10^9$$$ to $$$2 \cdot 10^9$$$.
In order for the system to work properly, there are some constraints: for every $$$i \in [1, n]$$$, $$$i$$$-th junction has a number $$$s_i$$$ associated with it meaning that the difference between incoming and outcoming flow for $$$i$$$-th junction must be exactly $$$s_i$$$ (if $$$s_i$$$ is not negative, then $$$i$$$-th junction must receive $$$s_i$$$ units of water per second; if it is negative, then $$$i$$$-th junction must transfer $$$|s_i|$$$ units of water per second to other junctions).
Can you choose the integers $$$f_1$$$, $$$f_2$$$, ..., $$$f_m$$$ in such a way that all requirements on incoming and outcoming flows are satisfied?
|
The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of junctions.
The second line contains $$$n$$$ integers $$$s_1, s_2, \dots, s_n$$$ ($$$-10^4 \le s_i \le 10^4$$$) — constraints for the junctions.
The third line contains an integer $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) — the number of pipes.
$$$i$$$-th of the next $$$m$$$ lines contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$1 \le x_i, y_i \le n$$$, $$$x_i \ne y_i$$$) — the description of $$$i$$$-th pipe. It is guaranteed that each unordered pair $$$(x, y)$$$ will appear no more than once in the input (it means that there won't be any pairs $$$(x, y)$$$ or $$$(y, x)$$$ after the first occurrence of $$$(x, y)$$$). It is guaranteed that for each pair of junctions there exists a path along the pipes connecting them.
|
If you can choose such integer numbers $$$f_1, f_2, \dots, f_m$$$ in such a way that all requirements on incoming and outcoming flows are satisfied, then output "Possible" in the first line. Then output $$$m$$$ lines, $$$i$$$-th line should contain $$$f_i$$$ — the chosen setting numbers for the pipes. Pipes are numbered in order they appear in the input.
Otherwise output "Impossible" in the only line.
| null | null |
[{"input": "4\n3 -10 6 1\n5\n1 2\n3 2\n2 4\n3 4\n3 1", "output": "Possible\n4\n-6\n8\n-7\n7"}, {"input": "4\n3 -10 6 4\n5\n1 2\n3 2\n2 4\n3 4\n3 1", "output": "Impossible"}]
| 2,400 |
["dfs and similar", "dp", "greedy", "trees"]
| 37 |
[{"input": "4\r\n3 -10 6 1\r\n5\r\n1 2\r\n3 2\r\n2 4\r\n3 4\r\n3 1\r\n", "output": "Possible\r\n-3\r\n-6\r\n1\r\n0\r\n0\r\n"}, {"input": "4\r\n3 -10 6 4\r\n5\r\n1 2\r\n3 2\r\n2 4\r\n3 4\r\n3 1\r\n", "output": "Impossible\r\n"}, {"input": "1\r\n0\r\n0\r\n", "output": "Possible\r\n"}, {"input": "1\r\n123\r\n0\r\n", "output": "Impossible\r\n"}, {"input": "2\r\n-1 1\r\n1\r\n1 2\r\n", "output": "Possible\r\n1\r\n"}, {"input": "2\r\n-1 1\r\n1\r\n2 1\r\n", "output": "Possible\r\n-1\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline())
s = list(map(int, f.readline().split()))
m = int(f.readline())
pipes = [tuple(map(int, f.readline().split())) for _ in range(m)]
sum_s = sum(s)
possible = (sum_s == 0)
with open(submission_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip() != '']
if not lines:
print(0)
return
first_line = lines[0]
if possible:
if first_line != "Possible":
print(0)
return
if len(lines) != m + 1:
print(0)
return
try:
f_values = [int(line) for line in lines[1:]]
except:
print(0)
return
flows_in = [0] * (n + 1)
flows_out = [0] * (n + 1)
for i in range(m):
x, y = pipes[i]
f = f_values[i]
flows_in[y] += f
flows_out[x] += f
for u in range(1, n + 1):
if (flows_in[u] - flows_out[u]) != s[u - 1]:
print(0)
return
print(1)
else:
if first_line == "Impossible" and len(lines) == 1:
print(1)
else:
print(0)
if __name__ == '__main__':
main()
| true |
785/D
|
785
|
D
|
PyPy 3-64
|
TESTS
| 0 | 46 | 3,072,000 |
222605411
|
R=10**5
M=10**9+7
F=[1]*(R+1)
for i in range(1,R+1):
F[i]=(F[i-1]*i)%M
Fhyp=[1]*(R+1)
Fhyp[R]=pow(F[R],M-2,M)
for i in range(R-1,-1,-1):
Fhyp[i]=(Fhyp[i+1]*(i+1))%M
def C(n,k):
if n<0 or k<0 or n<k:
return 0
return (F[n]*Fhyp[n-k]*Fhyp[k])%M
s=[e==')' for e in input()]
n=len(s)
L=[0]*n
L[0]=int(s[0]==0)
for i in range(1,n):
L[i]=L[i-1]+(s[i]==0)
R=[0]*n
R[-1]=int(s[-1]==1)
for i in range(n-2,-1,-1):
R[i]=R[i+1]+(s[i]==1)
s=0
for i in range(n-1):
s+=C(L[i]+R[i+1]-1,L[i])
print(L[i],R[i+1],C(L[i]+R[i+1]-1,L[i])*(L[i]>0 and R[i+1]>0))
print(s%M)
| 60 | 561 | 25,702,400 |
197613643
|
t = input()
n, m = len(t) + 1, 1000000007
a, b = 0, t.count(')') - 1
f = [1] * n
for i in range(2, n): f[i] = i * f[i - 1] % m
g = [pow(q, m - 2, m) for q in f]
s = 0
for q in t:
if b < 0: break
if q == '(':
a += 1
s += f[a + b] * g[a] * g[b]
else: b -= 1
print(s % m)
|
Codeforces Round 404 (Div. 2)
|
CF
| 2,017 | 2 | 256 |
Anton and School - 2
|
As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn different sequences that consist of round brackets (characters "(" and ")" (without quotes)).
On the last lesson Anton learned about the regular simple bracket sequences (RSBS). A bracket sequence s of length n is an RSBS if the following conditions are met:
- It is not empty (that is n ≠ 0).
- The length of the sequence is even.
- First $$\frac{\pi}{2}$$ charactes of the sequence are equal to "(".
- Last $$\frac{\pi}{2}$$ charactes of the sequence are equal to ")".
For example, the sequence "((()))" is an RSBS but the sequences "((())" and "(()())" are not RSBS.
Elena Ivanovna, Anton's teacher, gave him the following task as a homework. Given a bracket sequence s. Find the number of its distinct subsequences such that they are RSBS. Note that a subsequence of s is a string that can be obtained from s by deleting some of its elements. Two subsequences are considered distinct if distinct sets of positions are deleted.
Because the answer can be very big and Anton's teacher doesn't like big numbers, she asks Anton to find the answer modulo 109 + 7.
Anton thought of this task for a very long time, but he still doesn't know how to solve it. Help Anton to solve this task and write a program that finds the answer for it!
|
The only line of the input contains a string s — the bracket sequence given in Anton's homework. The string consists only of characters "(" and ")" (without quotes). It's guaranteed that the string is not empty and its length doesn't exceed 200 000.
|
Output one number — the answer for the task modulo 109 + 7.
| null |
In the first sample the following subsequences are possible:
- If we delete characters at the positions 1 and 5 (numbering starts with one), we will get the subsequence "(())".
- If we delete characters at the positions 1, 2, 3 and 4, we will get the subsequence "()".
- If we delete characters at the positions 1, 2, 4 and 5, we will get the subsequence "()".
- If we delete characters at the positions 1, 2, 5 and 6, we will get the subsequence "()".
- If we delete characters at the positions 1, 3, 4 and 5, we will get the subsequence "()".
- If we delete characters at the positions 1, 3, 5 and 6, we will get the subsequence "()".
The rest of the subsequnces are not RSBS. So we got 6 distinct subsequences that are RSBS, so the answer is 6.
|
[{"input": ")(()()", "output": "6"}, {"input": "()()()", "output": "7"}, {"input": ")))", "output": "0"}]
| 2,300 |
["combinatorics", "dp", "math", "number theory"]
| 60 |
[{"input": ")(()()\r\n", "output": "6\r\n"}, {"input": "()()()\r\n", "output": "7\r\n"}, {"input": ")))\r\n", "output": "0\r\n"}, {"input": ")())((((((()))()((()))()((()()()((()()()))\r\n", "output": "4344426\r\n"}, {"input": "(\r\n", "output": "0\r\n"}, {"input": ")\r\n", "output": "0\r\n"}, {"input": "((\r\n", "output": "0\r\n"}, {"input": "))\r\n", "output": "0\r\n"}, {"input": "()\r\n", "output": "1\r\n"}, {"input": ")(\r\n", "output": "0\r\n"}, {"input": "(()\r\n", "output": "2\r\n"}, {"input": "()(\r\n", "output": "1\r\n"}, {"input": ")()\r\n", "output": "1\r\n"}, {"input": ")((\r\n", "output": "0\r\n"}, {"input": "))(\r\n", "output": "0\r\n"}, {"input": "((((()())))())((((((()(()())(()())()((()()))(((())()()()(())))))()())())))(()()(()))))()(((()(((())()))())())))))(()()()((((((())(()())())()())(()))))()(()))(((()()()))())()(((()(()(((()))(()))()(()))((((((((())))))()))())(())))))()(()(()(((()((()()()())(((())())((()())(((((()()(())()()(()())()()))((()()())(())))))(()()(((()()()))(()())((()))()())))))()))))()(()()()))(()(()))))))(())))))()(()(())()()(\r\n", "output": "702597744\r\n"}, {"input": "))())(()()(((()))(()((())(()()(()()(((((()))((()())()())()()()))(((())()(()((())))(())())(((())))(()))())(((()(((())(()()))())))()((((())()()())(((((((())((())()(()(())(())()()(()())()))()()(()))()())()()())())))(()())(())())))(())))))(())())(()((((()()(()\r\n", "output": "546547243\r\n"}, {"input": "))())(()((())()((((()((())(()()()))((()())()(()))))()))()))()))(())))((()))())()(())()))()()())(())(()())())((()())())()())(((((()())((((())()()((()))))())()))(())())())))()())))())()((((()(()(())((()((())))(((())))((()()())()))\r\n", "output": "587599278\r\n"}, {"input": "())(()(())(()()))()()(()((()(())()))(()()(()()(())()()())))()()(()(((((()())(((())(((())((()(()()())()(()())()((((())()()(((()(((()(())()()((()(()(((()(((()()())(()((())()))())((()()())(()())(((((((()((((()())(()(\r\n", "output": "933002737\r\n"}, {"input": ")(()()))((\r\n", "output": "24\r\n"}, {"input": ")())))(((())))())))))))()())(())))))))())))(((())))())()()))(())())((()())(((((()()()())()()()))((()\r\n", "output": "99037306\r\n"}, {"input": ")()))()(()))\r\n", "output": "42\r\n"}, {"input": ")))()(())(((((()))\r\n", "output": "256\r\n"}, {"input": ")(())()())(())))\r\n", "output": "404\r\n"}, {"input": "((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((()\r\n", "output": "199\r\n"}, {"input": ")))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))())))))()))))))))))))))))))))))))))))))))))))))))))))))))))))())))))))())))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "286866\r\n"}, {"input": "(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((())\r\n", "output": "19899\r\n"}, {"input": "(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((\r\n", "output": "0\r\n"}, {"input": "()))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "199\r\n"}, {"input": "(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((())))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "407336794\r\n"}, {"input": "(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((())))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "211518470\r\n"}, {"input": "))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
241/A
|
241
|
A
|
Python 3
|
TESTS
| 2 | 186 | 0 |
58515097
|
import sys
def solution(k, d_arr, s_arr):
hours = 0
liters = 0
for i in range(len(d_arr)):
liters += s_arr[i]
d_curr = d_arr[i]
while liters <= d_curr:
liters += s_arr[i]
hours += k
hours += d_curr
return hours
if __name__ == '__main__':
line = sys.stdin.readline()
while line:
m, k = line.split()
d_arr = [int(d) for d in sys.stdin.readline().split()]
s_arr = [int(s) for s in sys.stdin.readline().split()]
print(solution(int(k), d_arr, s_arr))
line = sys.stdin.readline()
| 52 | 124 | 5,120,000 |
17870511
|
# -*- coding: utf-8 -*-
def main():
m, k = [int(x) for x in input().split()]
d = [int(x) for x in input().split()]
s = [int(x) for x in input().split()]
ans = 0 # total time spent
fuel = 0 # amount of fuel in the tank
mx = 0 # largest depo seen so far
for i in range(m):
fuel += s[i]
mx = max(mx, s[i])
if d[i] > fuel:
diff = d[i] - fuel
refill = diff // mx
if refill*mx < diff: refill += 1
fuel += (refill * mx)
ans += (k * refill)
fuel -= d[i]
ans += d[i]
#print(ans)
print(ans)
if __name__ == '__main__':
main()
|
Bayan 2012-2013 Elimination Round (ACM ICPC Rules, English statements)
|
ICPC
| 2,012 | 2 | 256 |
Old Peykan
|
There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city ci to city ci + 1 and is di kilometers long.
The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.
Each city ci (except for the last city cn) has a supply of si liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.
Initially (at time zero) the Old Peykan is at city c1 and s1 liters of fuel is transferred to it's empty tank from c1's supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.
Find the minimum time the Old Peykan needs to reach city cn.
|
The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.
The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).
|
In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from city c1.
| null |
In the second sample above, the Old Peykan stays in c1 for 3 hours.
|
[{"input": "4 6\n1 2 5 2\n2 3 3 4", "output": "10"}, {"input": "2 3\n5 6\n5 5", "output": "14"}]
| 1,300 |
["greedy"]
| 52 |
[{"input": "4 6\r\n1 2 5 2\r\n2 3 3 4\r\n", "output": "10\r\n"}, {"input": "2 3\r\n5 6\r\n5 5\r\n", "output": "14\r\n"}, {"input": "24 3\r\n11 8 8 12 17 4 4 25 39 37 31 32 38 34 29 29 34 39 39 39 17 9 24 6\r\n3 5 4 3 3 3 4 3 4 3 3 3 3 4 3 3 4 3 4 3 3 3 3 3\r\n", "output": "862\r\n"}, {"input": "43 5\r\n6 7 15 12 15 7 22 33 38 15 7 23 31 21 26 41 25 14 26 33 5 28 22 6 35 17 19 32 41 27 20 25 5 32 37 19 40 9 25 22 10 24 9\r\n3 5 3 6 5 4 5 3 3 3 3 6 6 3 3 3 3 3 3 3 3 6 3 3 4 3 4 3 6 4 3 6 3 4 6 3 4 5 4 4 3 3 5\r\n", "output": "1566\r\n"}]
| false |
stdio
| null | true |
241/A
|
241
|
A
|
PyPy 3
|
TESTS
| 2 | 278 | 1,228,800 |
93616866
|
a,b=map(int,input().split())
d=list(map(int,input().split()))
p=list(map(int,input().split()));k=int(1e9)
for i in range(a):
if sum(d[:i+1])>sum(p[:i+1]):k=min(k,b*(sum(p[:i+1])+max(d[:i+1])-1-sum(p[:i+1]))//max(p[:i+1]))
print(sum(d)+[k,0][k==int(1e9)])
| 52 | 154 | 1,536,000 |
209795812
|
import math
'''
4 6
1 2 5 2
2 3 3 4
'''
m, k = map(int, input().split())
n = m+1
distance = list(map(int, input().split()))
supply = list(map(int, input().split()))
currentTank = supply[0]
currentCity = 1
biggestSupply = supply[0]
time = sum(distance)
#the next road is distance[currentCity-1]
#the supply of currentCity is supply[currentCity-1]
while currentCity < n:
if currentTank < distance[currentCity-1]:
fuelNeeded = distance[currentCity-1] - currentTank
refills = math.ceil(fuelNeeded / biggestSupply)
currentTank+= refills * biggestSupply
time += refills * k
currentTank -= distance[currentCity-1]
currentCity += 1
if currentCity < n:
biggestSupply = max(biggestSupply, supply[currentCity-1])
currentTank += supply[currentCity-1]
print(time)
|
Bayan 2012-2013 Elimination Round (ACM ICPC Rules, English statements)
|
ICPC
| 2,012 | 2 | 256 |
Old Peykan
|
There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city ci to city ci + 1 and is di kilometers long.
The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.
Each city ci (except for the last city cn) has a supply of si liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.
Initially (at time zero) the Old Peykan is at city c1 and s1 liters of fuel is transferred to it's empty tank from c1's supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.
Find the minimum time the Old Peykan needs to reach city cn.
|
The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.
The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).
|
In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from city c1.
| null |
In the second sample above, the Old Peykan stays in c1 for 3 hours.
|
[{"input": "4 6\n1 2 5 2\n2 3 3 4", "output": "10"}, {"input": "2 3\n5 6\n5 5", "output": "14"}]
| 1,300 |
["greedy"]
| 52 |
[{"input": "4 6\r\n1 2 5 2\r\n2 3 3 4\r\n", "output": "10\r\n"}, {"input": "2 3\r\n5 6\r\n5 5\r\n", "output": "14\r\n"}, {"input": "24 3\r\n11 8 8 12 17 4 4 25 39 37 31 32 38 34 29 29 34 39 39 39 17 9 24 6\r\n3 5 4 3 3 3 4 3 4 3 3 3 3 4 3 3 4 3 4 3 3 3 3 3\r\n", "output": "862\r\n"}, {"input": "43 5\r\n6 7 15 12 15 7 22 33 38 15 7 23 31 21 26 41 25 14 26 33 5 28 22 6 35 17 19 32 41 27 20 25 5 32 37 19 40 9 25 22 10 24 9\r\n3 5 3 6 5 4 5 3 3 3 3 6 6 3 3 3 3 3 3 3 3 6 3 3 4 3 4 3 6 4 3 6 3 4 6 3 4 5 4 4 3 3 5\r\n", "output": "1566\r\n"}]
| false |
stdio
| null | true |
241/A
|
241
|
A
|
PyPy 3-64
|
TESTS
| 2 | 92 | 0 |
164830634
|
import sys
input = sys.stdin.readline
from math import ceil
m, k = map(int, input().split())
w = list(map(int, input().split()))
s = list(map(int, input().split()))
c = 0
t = 0
for i in range(m):
t += s[i]
if t < w[i]:
x = ceil((w[i] - t)/s[i])
t += x*s[i]
c += x*k
t -= w[i]
c += w[i]
print(c)
| 52 | 218 | 6,963,200 |
80074357
|
from math import ceil
n, k = map(int,input().split())
d = list(map(int,input().split()))
s = list(map(int,input().split()))
fuel = 0
stor = 0
time = 0
for i in range(n):
fuel+= s[i]
stor = max(stor, s[i])
if fuel >= d[i]:
fuel-= d[i]
time+= d[i]
else:
yy = ceil((d[i] - fuel) / stor)
time+= k * yy
time+= d[i]
fuel+= stor * yy
fuel-= d[i]
print(time)
|
Bayan 2012-2013 Elimination Round (ACM ICPC Rules, English statements)
|
ICPC
| 2,012 | 2 | 256 |
Old Peykan
|
There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city ci to city ci + 1 and is di kilometers long.
The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.
Each city ci (except for the last city cn) has a supply of si liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.
Initially (at time zero) the Old Peykan is at city c1 and s1 liters of fuel is transferred to it's empty tank from c1's supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.
Find the minimum time the Old Peykan needs to reach city cn.
|
The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.
The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).
|
In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from city c1.
| null |
In the second sample above, the Old Peykan stays in c1 for 3 hours.
|
[{"input": "4 6\n1 2 5 2\n2 3 3 4", "output": "10"}, {"input": "2 3\n5 6\n5 5", "output": "14"}]
| 1,300 |
["greedy"]
| 52 |
[{"input": "4 6\r\n1 2 5 2\r\n2 3 3 4\r\n", "output": "10\r\n"}, {"input": "2 3\r\n5 6\r\n5 5\r\n", "output": "14\r\n"}, {"input": "24 3\r\n11 8 8 12 17 4 4 25 39 37 31 32 38 34 29 29 34 39 39 39 17 9 24 6\r\n3 5 4 3 3 3 4 3 4 3 3 3 3 4 3 3 4 3 4 3 3 3 3 3\r\n", "output": "862\r\n"}, {"input": "43 5\r\n6 7 15 12 15 7 22 33 38 15 7 23 31 21 26 41 25 14 26 33 5 28 22 6 35 17 19 32 41 27 20 25 5 32 37 19 40 9 25 22 10 24 9\r\n3 5 3 6 5 4 5 3 3 3 3 6 6 3 3 3 3 3 3 3 3 6 3 3 4 3 4 3 6 4 3 6 3 4 6 3 4 5 4 4 3 3 5\r\n", "output": "1566\r\n"}]
| false |
stdio
| null | true |
489/A
|
489
|
A
|
PyPy 3-64
|
TESTS
| 0 | 46 | 0 |
201606017
|
n=int(input())
x = list(map(lambda q:int(q), input().split(" ")))
y=x.copy()
y.sort()
a=[]
t=0
for i in range(n):
for j in range(n):
if y[i]==x[j]:
t+=1
f=(j,i)
a.append(f)
break
if x==y:
break
print(t)
for i in range(t):
print(' '.join(map(str,a[i])))
| 22 | 218 | 409,600 |
8719997
|
import operator
n = int(input())
xs = list(map(int, str.split(input())))
if xs == sorted(xs):
print(0)
else:
swaps = []
counter = 0
while xs:
i = xs.index(min(xs))
if i:
swaps.append(str.format("{} {}", counter, i + counter))
xs[0], xs[i] = xs[i], xs[0]
xs.pop(0)
counter += 1
print(len(swaps))
print(str.join("\n", swaps))
|
Codeforces Round 277.5 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
SwapSort
|
In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.
Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n.
|
The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a0, a1, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai is the i-th element of the array. The elements are numerated from 0 to n - 1 from left to right. Some integers may appear in the array more than once.
|
In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the k swaps, one per line. Each swap should be printed as a pair of integers i, j (0 ≤ i, j ≤ n - 1), representing the swap of elements ai and aj. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = j and swap the same pair of elements multiple times.
If there are multiple answers, print any of them. It is guaranteed that at least one answer exists.
| null | null |
[{"input": "5\n5 2 5 1 4", "output": "2\n0 3\n4 2"}, {"input": "6\n10 20 20 40 60 60", "output": "0"}, {"input": "2\n101 100", "output": "1\n0 1"}]
| 1,200 |
["greedy", "implementation", "sortings"]
| 22 |
[{"input": "5\r\n5 2 5 1 4\r\n", "output": "2\r\n0 3\r\n4 2\r\n"}, {"input": "6\r\n10 20 20 40 60 60\r\n", "output": "0\r\n"}, {"input": "2\r\n101 100\r\n", "output": "1\r\n0 1\r\n"}, {"input": "1\r\n1000\r\n", "output": "0\r\n"}, {"input": "2\r\n1000000000 -1000000000\r\n", "output": "1\r\n0 1\r\n"}, {"input": "8\r\n5 2 6 8 3 1 6 8\r\n", "output": "4\r\n0 5\r\n4 2\r\n5 3\r\n6 5\r\n"}, {"input": "2\r\n200000000 199999999\r\n", "output": "1\r\n0 1\r\n"}, {"input": "3\r\n100000000 100000002 100000001\r\n", "output": "1\r\n1 2\r\n"}, {"input": "5\r\n1000000000 -10000000 0 8888888 7777777\r\n", "output": "3\r\n0 1\r\n2 1\r\n4 2\r\n"}, {"input": "5\r\n10 30 20 50 40\r\n", "output": "2\r\n1 2\r\n4 3\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
sub_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline().strip())
arr = list(map(int, f.readline().split()))
with open(sub_path) as f:
lines = f.readlines()
if not lines:
print(0)
return
try:
k = int(lines[0].strip())
swaps = []
for line in lines[1:1 + k]:
i, j = map(int, line.strip().split())
swaps.append((i, j))
except (ValueError, IndexError):
print(0)
return
if not (0 <= k <= n):
print(0)
return
for i, j in swaps:
if not (0 <= i < n and 0 <= j < n):
print(0)
return
current = arr.copy()
for i, j in swaps:
current[i], current[j] = current[j], current[i]
sorted_arr = sorted(arr)
if current == sorted_arr:
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
379/C
|
379
|
C
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
187473498
|
n=int(input())
a=list(map(int,input().split()))
b=[]
for i in range(n):
if a[i] not in b:
b.append(a[i])
else:
while a[i] in b:
a[i]+=1
b.append(a[i])
a[i]=i
print(*a)
| 41 | 670 | 53,862,400 |
167040474
|
import sys
input = sys.stdin.readline
n = int(input())
w = sorted(enumerate(map(int, input().split())), key=lambda x:x[1])
x = [0]*n
c = 0
for i in range(n):
if w[i][1] > c:
x[w[i][0]] = w[i][1]
c = w[i][1] + 1
else:
x[w[i][0]] = c
c += 1
print(' '.join(map(str, x)))
|
Good Bye 2013
|
CF
| 2,013 | 1 | 256 |
New Year Ratings Change
|
One very well-known internet resource site (let's call it X) has come up with a New Year adventure. Specifically, they decided to give ratings to all visitors.
There are n users on the site, for each user we know the rating value he wants to get as a New Year Present. We know that user i wants to get at least ai rating units as a present.
The X site is administered by very creative and thrifty people. On the one hand, they want to give distinct ratings and on the other hand, the total sum of the ratings in the present must be as small as possible.
Help site X cope with the challenging task of rating distribution. Find the optimal distribution.
|
The first line contains integer n (1 ≤ n ≤ 3·105) — the number of users on the site. The next line contains integer sequence a1, a2, ..., an (1 ≤ ai ≤ 109).
|
Print a sequence of integers b1, b2, ..., bn. Number bi means that user i gets bi of rating as a present. The printed sequence must meet the problem conditions.
If there are multiple optimal solutions, print any of them.
| null | null |
[{"input": "3\n5 1 1", "output": "5 1 2"}, {"input": "1\n1000000000", "output": "1000000000"}]
| 1,400 |
["greedy", "sortings"]
| 41 |
[{"input": "3\r\n5 1 1\r\n", "output": "5 1 2\r\n"}, {"input": "1\r\n1000000000\r\n", "output": "1000000000\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 1 1\r\n", "output": "1 2 3 4 5 6 7 8 9 10\r\n"}, {"input": "10\r\n1 10 1 10 1 1 7 8 6 7\r\n", "output": "1 10 2 11 3 4 7 9 6 8\r\n"}, {"input": "10\r\n20 19 12 1 12 15 2 12 6 10\r\n", "output": "20 19 12 1 13 15 2 14 6 10\r\n"}, {"input": "10\r\n4 5 10 5 2 14 15 6 10 6\r\n", "output": "4 5 10 6 2 14 15 7 11 8\r\n"}]
| false |
stdio
| null | true |
394/C
|
394
|
C
|
Python 3
|
TESTS
| 0 | 93 | 0 |
45285029
|
a,b=map(int,input().split())
c=list()
cn=list()
x00=0
x01=0
x11=0
for i in range(a):
c.append(list(input().split()))
x11=c[-1].count('11')
x01=c[-1].count('01')+c[-1].count('10')
d=x11//b
for i in range(d):
print((b-1)*'11 '+'11')
d=x11%b
e=x01//2
f=(d+e)//b
for i in range(f):
print((d//f)*'11 '+(e//f-1)*'01 '+'01')
print((d//f)*'00 '+(e//f-1)*'10 '+'10')
for i in range(a-d-2*f):
print(b*'00')
| 27 | 951 | 45,977,600 |
48909369
|
a,b=map(int,input().split())
c=list()
x00=0
x01=0
x11=0
for i in range(a):
c.append(list(input().split()))
x11+=c[-1].count('11')
x01+=c[-1].count('01')+c[-1].count('10')
x00=a*b-x11-x01
new=[[]for i in range(b)]
i=0
while x11>0:
x11-=1
new[i].append('11')
i+=1
i%=b
step=1
reverse=False
while x01>0:
x01-=1
if reverse:
new[i].append('01')
else:
new[i].append('10')
i+=step
if i==-1 or i==b:
step=-step
i+=step
reverse=1-reverse
for r in new:
r+=['00']*(a-len(r))
for i in range(a):
print(' '.join([new[k][i]for k in range(b)]))
|
Codeforces Round 231 (Div. 2)
|
CF
| 2,014 | 2 | 256 |
Dominoes
|
During the break, we decided to relax and play dominoes. Our box with Domino was empty, so we decided to borrow the teacher's dominoes.
The teacher responded instantly at our request. He put nm dominoes on the table as an n × 2m rectangle so that each of the n rows contained m dominoes arranged horizontally. Each half of each domino contained number (0 or 1).
We were taken aback, and the teacher smiled and said: "Consider some arrangement of dominoes in an n × 2m matrix. Let's count for each column of the matrix the sum of numbers in this column. Then among all such sums find the maximum one. Can you rearrange the dominoes in the matrix in such a way that the maximum sum will be minimum possible? Note that it is prohibited to change the orientation of the dominoes, they all need to stay horizontal, nevertheless dominoes are allowed to rotate by 180 degrees. As a reward I will give you all my dominoes".
We got even more taken aback. And while we are wondering what was going on, help us make an optimal matrix of dominoes.
|
The first line contains integers n, m (1 ≤ n, m ≤ 103).
In the next lines there is a description of the teachers' matrix. Each of next n lines contains m dominoes. The description of one domino is two integers (0 or 1), written without a space — the digits on the left and right half of the domino.
|
Print the resulting matrix of dominoes in the format: n lines, each of them contains m space-separated dominoes.
If there are multiple optimal solutions, print any of them.
| null |
Consider the answer for the first sample. There, the maximum sum among all columns equals 1 (the number of columns is 6, and not 3). Obviously, this maximum can't be less than 1, then such matrix is optimal.
Note that the dominoes can be rotated by 180 degrees.
|
[{"input": "2 3\n01 11 00\n00 01 11", "output": "11 11 10\n00 00 01"}, {"input": "4 1\n11\n10\n01\n00", "output": "11\n10\n01\n00"}]
| null |
["constructive algorithms", "greedy"]
| 27 |
[{"input": "2 3\r\n01 11 00\r\n00 01 11\r\n", "output": "11 11 10\r\n00 00 01\r\n"}, {"input": "4 1\r\n11\r\n10\r\n01\r\n00\r\n", "output": "11\r\n10\r\n01\r\n00\r\n"}, {"input": "1 1\r\n00\r\n", "output": "00\r\n"}, {"input": "1 1\r\n01\r\n", "output": "10\r\n"}, {"input": "1 1\r\n11\r\n", "output": "11\r\n"}, {"input": "9 9\r\n01 00 00 01 00 01 11 11 11\r\n10 10 10 01 10 01 11 01 10\r\n10 00 10 00 11 01 00 10 00\r\n01 00 01 01 11 00 00 11 11\r\n11 00 10 11 01 01 11 00 01\r\n01 10 00 00 11 10 01 01 10\r\n11 10 11 00 11 11 01 10 10\r\n10 00 01 00 00 00 11 01 01\r\n00 11 01 00 10 01 10 00 01\r\n", "output": "11 11 11 11 11 11 11 11 11\r\n11 11 11 11 11 11 11 11 11\r\n10 10 10 10 10 10 10 10 10\r\n10 10 10 10 10 10 10 10 10\r\n10 10 10 01 01 01 01 01 01\r\n01 01 01 01 01 01 01 01 01\r\n01 01 01 00 00 00 00 01 01\r\n00 00 00 00 00 00 00 00 00\r\n00 00 00 00 00 00 00 00 00\r\n"}, {"input": "9 9\r\n10 10 10 01 10 11 11 01 10\r\n11 00 10 10 11 10 01 00 00\r\n10 00 11 01 00 01 01 11 10\r\n10 11 10 00 01 11 11 10 11\r\n01 11 11 01 11 00 10 00 01\r\n01 00 00 10 01 01 10 00 01\r\n11 10 11 10 01 00 00 11 00\r\n10 11 10 10 01 10 10 10 01\r\n10 10 10 10 11 11 01 00 11\r\n", "output": "11 11 11 11 11 11 11 11 11\r\n11 11 11 11 11 11 11 11 11\r\n11 11 10 10 10 10 10 10 10\r\n10 10 10 10 10 10 10 10 10\r\n10 10 10 10 10 10 10 01 01\r\n01 01 01 01 01 01 01 01 01\r\n01 01 01 01 01 01 01 01 01\r\n00 00 00 00 01 01 01 00 00\r\n00 00 00 00 00 00 00 00 00\r\n"}, {"input": "9 1\r\n01\r\n00\r\n01\r\n01\r\n00\r\n00\r\n00\r\n01\r\n11\r\n", "output": "11\r\n10\r\n10\r\n01\r\n01\r\n00\r\n00\r\n00\r\n00\r\n"}, {"input": "2 9\r\n11 10 11 10 10 11 00 10 00\r\n10 00 00 10 10 00 11 01 01\r\n", "output": "11 11 11 11 10 10 10 10 10\r\n00 00 00 00 00 01 01 01 01\r\n"}, {"input": "2 8\r\n10 01 01 11 10 10 01 10\r\n01 11 01 01 11 10 01 01\r\n", "output": "11 11 11 10 10 10 10 10\r\n10 10 01 01 01 01 01 01\r\n"}, {"input": "3 5\r\n00 10 10 11 01\r\n11 01 11 11 10\r\n10 11 00 00 00\r\n", "output": "11 11 11 11 11\r\n10 10 10 01 01\r\n00 00 01 00 00\r\n"}, {"input": "2 3\r\n00 10 01\r\n01 01 00\r\n", "output": "10 10 01\r\n00 01 00\r\n"}, {"input": "2 5\r\n01 00 01 01 00\r\n11 01 11 11 10\r\n", "output": "11 11 11 10 10\r\n10 00 00 01 01\r\n"}]
| false |
stdio
|
import sys
from collections import Counter
def read_dominoes(file, n, m):
dominoes = []
grid = []
for _ in range(n):
line = file.readline().strip()
if not line:
return [], []
row_dominoes = line.split()
grid_row = []
for d in row_dominoes:
a, b = d[0], d[1]
sorted_d = tuple(sorted((a, b)))
dominoes.append(sorted_d)
grid_row.append(a)
grid_row.append(b)
grid.append(grid_row)
return dominoes, grid
def compute_max_col_sum(grid, m):
columns = 2 * m
col_sums = [0] * columns
for row in grid:
for i in range(columns):
col_sums[i] += int(row[i])
return max(col_sums)
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as input_file, \
open(output_path, 'r') as output_file, \
open(submission_path, 'r') as submission_file:
# Read input
n_line = input_file.readline().strip()
if not n_line:
print(0)
return
n, m = map(int, n_line.split())
input_dominoes, _ = read_dominoes(input_file, n, m)
input_counter = Counter(input_dominoes)
# Read submission
submission_dominoes, submission_grid = read_dominoes(submission_file, n, m)
if not submission_dominoes or len(submission_grid) != n:
print(0)
return
submission_counter = Counter(submission_dominoes)
# Check dominoes
if submission_counter != input_counter:
print(0)
return
# Compute submission max column sum
submission_max = compute_max_col_sum(submission_grid, m)
# Read reference output to compute max
_, reference_grid = read_dominoes(output_file, n, m)
if not reference_grid:
print(0)
return
reference_max = compute_max_col_sum(reference_grid, m)
# Compare maxima
if submission_max == reference_max:
print(1)
else:
print(0)
if __name__ == '__main__':
main()
| true |
367/B
|
367
|
B
|
Python 3
|
TESTS
| 2 | 61 | 0 |
116267494
|
st=input().split()
n=int(st[0])
m=int(st[1])
p=int(st[2])
a=list(map(int,input().split()))
b=list(map(int,input().split()))
q=n-(m-1)*p
print(2)
print(1,q)
| 59 | 452 | 27,750,400 |
5995052
|
from collections import defaultdict
n, m, p = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
u = defaultdict(int)
for i in b: u[i] += 1
ans = []
for q in range(p):
c = a[q: n: p]
if len(c) < m: break
v = defaultdict(int)
for i in c[: m]: v[i] += 1
d = q + 1
if u == v: ans.append(d)
for j, k in zip(c[: len(c) - m], c[m: ]):
v[j] -= 1
if v[j] == 0: v.pop(j)
v[k] += 1
d += p
if u == v: ans.append(d)
ans.sort()
print(len(ans))
print(' '.join(map(str, ans)))
|
Codeforces Round 215 (Div. 1)
|
CF
| 2,013 | 1 | 256 |
Sereja ans Anagrams
|
Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, ..., an. Similarly, sequence b consists of m integers b1, b2, ..., bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q (q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, ..., aq + (m - 1)p by rearranging elements.
Sereja needs to rush to the gym, so he asked to find all the described positions of q.
|
The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109).
|
In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.
| null | null |
[{"input": "5 3 1\n1 2 3 2 1\n1 2 3", "output": "2\n1 3"}, {"input": "6 3 2\n1 3 2 2 3 1\n1 2 3", "output": "2\n1 2"}]
| 1,900 |
["binary search", "data structures"]
| 45 |
[{"input": "5 3 1\r\n1 2 3 2 1\r\n1 2 3\r\n", "output": "2\r\n1 3\r\n"}, {"input": "6 3 2\r\n1 3 2 2 3 1\r\n1 2 3\r\n", "output": "2\r\n1 2\r\n"}, {"input": "68 16 3\r\n5 3 4 3 3 3 2 2 2 3 2 4 2 2 2 2 4 3 5 1 1 2 2 2 3 1 5 1 2 2 1 5 1 5 3 2 3 5 2 1 1 4 2 3 4 3 4 3 3 1 3 4 1 5 2 5 3 4 4 1 4 5 5 1 1 2 2 2\r\n5 4 4 3 5 1 1 2 3 2 2 1 3 3 2 2\r\n", "output": "3\r\n2 13 15\r\n"}, {"input": "44 11 4\r\n4 3 3 3 4 3 4 5 1 3 4 2 4 4 2 2 1 5 3 1 5 2 3 2 4 4 5 3 2 2 2 4 2 2 2 5 4 2 3 5 4 3 1 1\r\n4 4 1 4 4 1 2 4 2 5 4\r\n", "output": "1\r\n1\r\n"}, {"input": "54 6 4\r\n5 4 1 2 2 2 1 3 3 1 5 5 2 2 2 5 4 4 1 3 4 3 4 2 1 4 2 2 4 3 3 2 5 5 3 5 2 2 1 4 2 3 5 3 5 5 5 5 1 2 5 2 4 5\r\n2 3 5 5 3 2\r\n", "output": "3\r\n4 8 30\r\n"}, {"input": "75 54 1\r\n1 1 1 5 5 4 2 1 1 1 5 1 5 1 2 1 5 3 1 2 1 3 2 3 4 1 3 5 1 1 3 5 4 1 3 4 3 3 1 2 3 3 1 4 1 4 1 4 3 2 4 3 3 1 2 4 4 4 1 3 4 1 3 1 5 4 4 1 2 3 5 1 4 4 4\r\n1 3 3 4 3 4 1 4 4 3 1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1\r\n", "output": "3\r\n10 11 12\r\n"}, {"input": "31 28 1\r\n1 4 1 2 5 1 1 4 2 2 5 2 4 5 5 2 4 1 5 3 5 4 1 2 4 3 1 2 5 2 1\r\n2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3\r\n", "output": "1\r\n2\r\n"}, {"input": "59 2 3\r\n2 4 5 3 2 4 4 5 5 5 4 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 4 4 1 5 5 3 1 1 1 1 5 3 4 3 5 1 1\r\n5 4\r\n", "output": "12\r\n3 6 7 8 10 15 16 21 24 41 43 44\r\n"}, {"input": "74 33 1\r\n4 5 5 2 1 2 2 2 2 2 3 2 3 4 2 2 1 4 4 4 5 4 1 2 4 5 4 2 4 2 5 1 2 1 5 3 5 4 1 4 1 1 2 4 4 5 1 4 2 4 2 3 3 2 5 5 4 3 5 1 3 5 5 4 4 4 2 5 4 2 2 3 4 4\r\n2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3\r\n", "output": "1\r\n20\r\n"}, {"input": "70 7 3\r\n4 2 5 5 2 3 1 1 2 5 3 1 5 1 1 4 2 4 3 4 4 5 2 3 2 3 5 3 5 5 1 2 2 4 5 1 3 1 2 2 1 5 4 2 4 1 5 1 1 4 5 2 3 4 4 2 4 2 1 3 5 5 1 3 3 2 1 5 5 5\r\n4 1 1 3 2 3 4\r\n", "output": "2\r\n6 28\r\n"}, {"input": "82 50 1\r\n4 2 2 4 4 3 2 2 3 3 4 3 4 1 4 1 4 3 5 3 4 2 4 4 5 2 5 4 2 4 2 5 3 2 3 4 2 5 4 3 3 4 3 4 1 1 5 2 4 1 3 3 2 4 1 2 2 5 4 4 5 3 3 1 2 3 4 5 3 1 2 3 3 5 5 2 4 1 4 2 3 1\r\n4 3 3 3 4 1 4 4 5 5 5 2 5 4 1 2 2 1 3 3 3 3 2 5 3 2 4 3 2 2 4 4 4 3 4 2 1 2 3 2 1 1 5 2 3 5 4 5 3 4\r\n", "output": "2\r\n24 28\r\n"}, {"input": "41 6 1\r\n1 3 1 1 3 4 2 3 1 5 2 3 5 2 3 1 4 4 3 4 2 5 1 3 3 4 2 3 1 3 4 3 1 5 2 3 4 2 1 5 3\r\n3 2 3 1 4 5\r\n", "output": "10\r\n5 12 19 20 21 22 30 31 32 36\r\n"}, {"input": "3 5 1\r\n1 1 1\r\n1 1 1 1 1\r\n", "output": "0\r\n"}, {"input": "3 5 1\r\n1 1 1\r\n1 2 3 4 5\r\n", "output": "0\r\n"}, {"input": "1 1 1\r\n1\r\n1\r\n", "output": "1\r\n1\r\n"}, {"input": "1 1 2\r\n1\r\n2\r\n", "output": "0\r\n"}, {"input": "2 2 1\r\n1 2\r\n2 1\r\n", "output": "1\r\n1\r\n"}, {"input": "2 2 1\r\n1 2\r\n1 2\r\n", "output": "1\r\n1\r\n"}, {"input": "5 10 1\r\n1 1 1 1 1\r\n1 1 1 1 1 1 1 1 1 1\r\n", "output": "0\r\n"}, {"input": "5 10 1\r\n1 1 1 1 2\r\n1 1 1 1 1 1 1 1 1 2\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
149414735
|
t = int(input())
works = [60,90,108,120,135,140,144,150,156,160,165,168,170,171,172,174,175,176,177,178,179]
for i in range(t):
angle = int(input())
if angle in works:
print("YES")
else: print("NO")
| 3 | 30 | 0 |
176961603
|
iter = int(input())
for i in range(iter):
a = int(input())
if (360 % (180 - a) == 0):
print("YES\n")
else:
print("NO\n")
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 62 | 0 |
228567216
|
x = int(input())
for _ in range(x):
a = int(input())
if 59 < a < 180:
print("YES")
else:
print("NO")
| 3 | 30 | 0 |
206103584
|
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- ############
def Fancy_Fence():
t = inp()
outputStr = ''
for i in range(t):
a = inp()
n = 360/(180-a)
if n == int(n):
outputStr += 'YES' + '\n'
else:
outputStr += 'NO' + '\n'
outputStr = outputStr.strip()
print(outputStr)
return
Fancy_Fence()
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 62 | 0 |
227569325
|
t=int(input())
for _ in range(t):
s=int(input())
i=3
flag=0
while 1:
tmp=(i-2)*180//i
i+=1
if s==tmp:
print("YES");flag=1;break
if tmp>s:
break
if not flag:
print("NO")
| 3 | 30 | 0 |
223965776
|
t=int(input())
output=[]
for _ in range(t):
a=int(input())
line=3
while True:
if (180-a)*line==360:
output.append("YES")
break
else:
line+=1
if (180-a)*line>360:
output.append("NO")
break
[print(i) for i in output]
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
144931394
|
for i in range(int(input())):
n = int(input())
t= 0
i=3
while t<n:
t = int(((i-2)*180)/i)
i+=1
if t>n:
print('NO')
else:
print('YES')
| 3 | 30 | 0 |
227500778
|
t=int(input())
for i in range(t):
a=int(input())
b=360/(180-a)
if b%1==0 and b>=3:
print('YES')
else:
print('NO')
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
126/D
|
126
|
D
|
PyPy 3
|
TESTS
| 2 | 155 | 0 |
92453952
|
import sys,bisect as bi
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
#Fibonacci Sums
n,=I()
fi=[1,2]
an=[1,1]
for i in range(82):
fi.append(fi[-1]+fi[-2])
cr=2
for i in range(82):
an.append(cr)
if i%2:
cr+=1
#print(fi[:20])
for _ in range(n):
x,=I()
te=x
cr=82;ar=[]
ans=1;pr=0
while cr>=3:
cr=bi.bisect(fi,te)
if pr:
ans*=(pr-cr+1)//2
#print(pr,cr)
if cr>0:
ar.append(fi[cr-1])
te-=fi[cr-1]
pr=cr
#print(_,x,cr,ar)
print(ans)
| 55 | 218 | 12,288,000 |
185884877
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = int(input())
ans = []
f = [1, 2]
inf = pow(10, 18) + 1
while f[-1] + f[-2] < inf:
f.append(f[-1] + f[-2])
l = len(f)
for _ in range(t):
n = int(input())
x = []
for i in range(l - 1, -1, -1):
if n >= f[i]:
n -= f[i]
x.append(i)
x.append(-1)
x.reverse()
m = len(x)
dp1, dp2 = [0] * m, [0] * m
dp1[0] = 1
for i in range(1, m):
d = x[i] - x[i - 1] - 1
d1, d2 = d // 2, (d + 1) // 2
dp1[i] = dp1[i - 1] + dp2[i - 1]
dp2[i] = dp1[i - 1] * d1 + dp2[i - 1] * d2
ans0 = dp1[-1] + dp2[-1]
ans.append(ans0)
sys.stdout.write("\n".join(map(str, ans)))
|
Codeforces Beta Round 93 (Div. 1 Only)
|
CF
| 2,011 | 1 | 256 |
Fibonacci Sums
|
Fibonacci numbers have the following form:
F1 = 1,
F2 = 2,
Fi = Fi - 1 + Fi - 2, i > 2.
Let's consider some non-empty set S = {s1, s2, ..., sk}, consisting of different Fibonacci numbers. Let's find the sum of values of this set's elements:
$$\sum_{i=1}^{k}s_{i}=n$$
Let's call the set S a number n's decomposition into Fibonacci sum.
It's easy to see that several numbers have several decompositions into Fibonacci sum. For example, for 13 we have 13, 5 + 8, 2 + 3 + 8 — three decompositions, and for 16: 3 + 13, 1 + 2 + 13, 3 + 5 + 8, 1 + 2 + 5 + 8 — four decompositions.
By the given number n determine the number of its possible different decompositions into Fibonacci sum.
|
The first line contains an integer t — the number of tests (1 ≤ t ≤ 105). Each of the following t lines contains one test.
Each test is an integer n (1 ≤ n ≤ 1018).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
|
For each input data test print a single number on a single line — the answer to the problem.
| null |
Two decompositions are different if there exists a number that is contained in the first decomposition, but is not contained in the second one. Decompositions that differ only in the order of summands are considered equal.
|
[{"input": "2\n13\n16", "output": "3\n4"}]
| 2,300 |
["dp", "math"]
| 56 |
[{"input": "2\r\n13\r\n16\r\n", "output": "3\r\n4\r\n"}, {"input": "10\r\n1\r\n2\r\n3\r\n4\r\n5\r\n6\r\n7\r\n8\r\n9\r\n10\r\n", "output": "1\r\n1\r\n2\r\n1\r\n2\r\n2\r\n1\r\n3\r\n2\r\n2\r\n"}, {"input": "10\r\n24\r\n37\r\n42\r\n58\r\n60\r\n73\r\n79\r\n84\r\n92\r\n99\r\n", "output": "5\r\n6\r\n6\r\n7\r\n6\r\n6\r\n8\r\n7\r\n8\r\n6\r\n"}, {"input": "5\r\n484775665757\r\n968685776575\r\n687675666422\r\n348587665784\r\n373875686864\r\n", "output": "117120\r\n155520\r\n320160\r\n115920\r\n84600\r\n"}, {"input": "1\r\n123456789101112\r\n", "output": "2868360\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
130251087
|
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 29 17:44:39 2021
@author: Boush
"""
output = []
t = int(input())
for i in range(t):
a = int(input())
n = 3
while 1:
q = int(180*(n-2)/n)
if a == q : output.append("YES");break
elif a > q : n += 1
else : output.append("NO");break
for i in output:
print(i)
| 3 | 30 | 0 |
227569211
|
'''程文奇 2100015898'''
def fence(a):
if 360%(180-a)==0:
return 'YES'
else:
return 'NO'
n=int(input())
answerlist=[]
for i in range(n):
answerlist.append(fence(int(input())))
for j in answerlist:
print(j)
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
652/F
|
652
|
F
|
Python 3
|
TESTS
| 1 | 46 | 0 |
114146139
|
n, m, t = map(int, input().split())
saida = []
for i in range(n):
entrada = input().split()
p, d = entrada
p = int(p)
if d == 'R':
saida.append((p + t) % m)
else:
saida.append((p - t) % m)
print(" ".join([str(x) for x in saida]))
| 30 | 1,372 | 46,284,800 |
144096677
|
I = lambda: [int(i) for i in input().split()]
import sys
input = sys.stdin.readline
n, L, T = I()
a, b, ans, cnt = [0]*n, [0]*n, [0]*n, 0
for i in range(n):
x, ch = input().split()
a[i] = int(x) - 1
b[i] = (a[i], i)
dr = 1 if ch == "R" else -1
w, u = divmod(a[i] + T * dr, L)
a[i], cnt = u, cnt + w
a, b = sorted(a), sorted(b)
for i in range(n):
ans[b[i][1]] = a[(i + cnt) % n]+1
print(" ".join(str(x) for x in ans))
|
Educational Codeforces Round 10
|
ICPC
| 2,016 | 2 | 256 |
Ants on a Circle
|
n ants are on a circle of length m. An ant travels one unit of distance per one unit of time. Initially, the ant number i is located at the position si and is facing in the direction di (which is either L or R). Positions are numbered in counterclockwise order starting from some point. Positions of the all ants are distinct.
All the ants move simultaneously, and whenever two ants touch, they will both switch their directions. Note that it is possible for an ant to move in some direction for a half of a unit of time and in opposite direction for another half of a unit of time.
Print the positions of the ants after t time units.
|
The first line contains three integers n, m and t (2 ≤ n ≤ 3·105, 2 ≤ m ≤ 109, 0 ≤ t ≤ 1018) — the number of ants, the length of the circle and the number of time units.
Each of the next n lines contains integer si and symbol di (1 ≤ si ≤ m and di is either L or R) — the position and the direction of the i-th ant at the start. The directions L and R corresponds to the clockwise and counterclockwise directions, respectively.
It is guaranteed that all positions si are distinct.
|
Print n integers xj — the position of the j-th ant after t units of time. The ants are numbered from 1 to n in order of their appearing in input.
| null | null |
[{"input": "2 4 8\n1 R\n3 L", "output": "1 3"}, {"input": "4 8 6\n6 R\n5 L\n1 R\n8 L", "output": "7 4 2 7"}, {"input": "4 8 2\n1 R\n5 L\n6 L\n8 R", "output": "3 3 4 2"}]
| 2,800 |
["constructive algorithms", "math"]
| 30 |
[{"input": "2 4 8\r\n1 R\r\n3 L\r\n", "output": "1 3\r\n"}, {"input": "4 8 6\r\n6 R\r\n5 L\r\n1 R\r\n8 L\r\n", "output": "7 4 2 7\r\n"}, {"input": "4 8 2\r\n1 R\r\n5 L\r\n6 L\r\n8 R\r\n", "output": "3 3 4 2\r\n"}, {"input": "10 10 90\r\n2 R\r\n1 R\r\n3 L\r\n4 R\r\n7 L\r\n8 L\r\n6 R\r\n9 R\r\n5 R\r\n10 L\r\n", "output": "10 9 1 2 5 6 4 7 3 8\r\n"}, {"input": "10 20 85\r\n6 L\r\n12 R\r\n2 L\r\n20 R\r\n18 L\r\n8 R\r\n16 R\r\n14 L\r\n10 L\r\n4 R\r\n", "output": "5 13 1 1 17 9 17 13 9 5\r\n"}, {"input": "10 20 59\r\n1 R\r\n15 L\r\n7 L\r\n13 R\r\n5 R\r\n17 R\r\n3 L\r\n9 R\r\n11 L\r\n19 L\r\n", "output": "20 16 8 12 4 16 4 8 12 20\r\n"}, {"input": "2 2 0\r\n2 L\r\n1 R\r\n", "output": "2 1\r\n"}, {"input": "2 2 0\r\n2 L\r\n1 R\r\n", "output": "2 1\r\n"}, {"input": "4 8 6\r\n6 R\r\n5 L\r\n1 R\r\n8 R\r\n", "output": "7 7 6 4\r\n"}]
| false |
stdio
| null | true |
270/A
|
270
|
A
|
Python 3
|
TESTS
| 2 | 60 | 0 |
221950002
|
A=[60,90,108,120,135,140,144,150,156,160,165]
for x in range(1,12):
if 360%x==0:
A.append(180-x)
for _ in range(int(input())):
if int(input()) in A:
print("YES")
else:
print("NO")
| 3 | 60 | 0 |
20656108
|
n = int(input())
for i in range(n):
t = int(input())
if 360%(180-t)==0: print('YES')
else: print('NO')
|
Codeforces Round 165 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Fancy Fence
|
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle a.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to a?
|
The first line of input contains an integer t (0 < t < 180) — the number of tests. Each of the following t lines contains a single integer a (0 < a < 180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
| null |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle $$30^\circ$$.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
[{"input": "3\n30\n60\n90", "output": "NO\nYES\nYES"}]
| 1,100 |
["geometry", "implementation", "math"]
| 3 |
[{"input": "3\r\n30\r\n60\r\n90\r\n", "output": "NO\r\nYES\r\nYES\r\n"}, {"input": "6\r\n1\r\n2\r\n3\r\n170\r\n179\r\n25\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\n"}]
| false |
stdio
| null | true |
1212/D
|
977
|
D
|
PyPy 3-64
|
TESTS
| 0 | 31 | 0 |
227965961
|
n = int(input())
a = list(map(int, input().split()))
start = max(a, key=lambda x: (-x, sum(1 for i in range(60) if x % (3**i) == 0)))
ans = [start]
a.remove(start)
while len(ans) < n:
next_value = ans[-1] // 3 if ans[-1] % 3 == 0 else ans[-1] * 2
if next_value in a:
ans.append(next_value)
a.remove(next_value)
else:
break # If the next value is not in 'a', break the loop.
print(*ans)
| 26 | 62 | 0 |
225011252
|
import sys
# sys.setrecursionlimit(10**8)设置最大递归次数
class FastIO:
def __init__(self):
return
@staticmethod
def read_int():
return int(sys.stdin.readline().strip())
@staticmethod
def read_list_ints():
return list(map(int, sys.stdin.readline().strip().split()))
@staticmethod
def lst(x):
return sys.stdout.write(" ".join(str(w) for w in x) + '\n')
class Solution:
def __init__(self):
return
@staticmethod
def main(ac=FastIO()):
ac.read_int()
def check(x):
res = 0
while x % 3 == 0:
x //= 3
res += 1
return [-res, x]
nums = ac.read_list_ints()
nums.sort(key=check)
ac.lst(nums)
return
Solution().main()
|
Kotlin Heroes: Practice 2
|
ICPC
| 2,019 | 1 | 256 |
Divide by three, multiply by two
|
Polycarp likes to play with numbers. He takes some integer number $$$x$$$, writes it down on the board, and then performs with it $$$n - 1$$$ operations of the two kinds:
- divide the number $$$x$$$ by $$$3$$$ ($$$x$$$ must be divisible by $$$3$$$);
- multiply the number $$$x$$$ by $$$2$$$.
After each operation, Polycarp writes down the result on the board and replaces $$$x$$$ by the result. So there will be $$$n$$$ numbers on the board after all.
You are given a sequence of length $$$n$$$ — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists.
|
The first line of the input contatins an integer number $$$n$$$ ($$$2 \le n \le 100$$$) — the number of the elements in the sequence. The second line of the input contains $$$n$$$ integer numbers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 3 \cdot 10^{18}$$$) — rearranged (reordered) sequence that Polycarp can wrote down on the board.
|
Print $$$n$$$ integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists.
| null |
In the first example the given sequence can be rearranged in the following way: $$$[9, 3, 6, 12, 4, 8]$$$. It can match possible Polycarp's game which started with $$$x = 9$$$.
|
[{"input": "6\n4 8 6 3 12 9", "output": "9 3 6 12 4 8"}, {"input": "4\n42 28 84 126", "output": "126 42 84 28"}, {"input": "2\n1000000000000000000 3000000000000000000", "output": "3000000000000000000 1000000000000000000"}]
| 1,400 |
["*special", "math"]
| 26 |
[{"input": "6\r\n4 8 6 3 12 9\r\n", "output": "9 3 6 12 4 8 \r\n"}, {"input": "4\r\n42 28 84 126\r\n", "output": "126 42 84 28 \r\n"}, {"input": "2\r\n1000000000000000000 3000000000000000000\r\n", "output": "3000000000000000000 1000000000000000000 \r\n"}, {"input": "19\r\n46875000000000000 732421875000000 5859375000000000 11444091796875 2929687500000000 187500000000000000 91552734375000 11718750000000000 3000000000000000000 22888183593750 1464843750000000 375000000000000000 45776367187500 183105468750000 93750000000000000 366210937500000 23437500000000000 750000000000000000 1500000000000000000\r\n", "output": "11444091796875 22888183593750 45776367187500 91552734375000 183105468750000 366210937500000 732421875000000 1464843750000000 2929687500000000 5859375000000000 11718750000000000 23437500000000000 46875000000000000 93750000000000000 187500000000000000 375000000000000000 750000000000000000 1500000000000000000 3000000000000000000 \r\n"}, {"input": "6\r\n558 744 1488 279 2232 1116\r\n", "output": "279 558 1116 2232 744 1488 \r\n"}, {"input": "17\r\n2985984 2239488 7077888 5971968 10616832 746496 28311552 3538944 7962624 3145728 15925248 1492992 14155776 5308416 3981312 11943936 9437184\r\n", "output": "2239488 746496 1492992 2985984 5971968 11943936 3981312 7962624 15925248 5308416 10616832 3538944 7077888 14155776 28311552 9437184 3145728 \r\n"}, {"input": "18\r\n47775744 7077888 5971968 3538944 4478976 3145728 2985984 4718592 1572864 5308416 1048576 1492992 23887872 10616832 2239488 11943936 15925248 14155776\r\n", "output": "2239488 4478976 1492992 2985984 5971968 11943936 23887872 47775744 15925248 5308416 10616832 3538944 7077888 14155776 4718592 1572864 3145728 1048576 \r\n"}, {"input": "3\r\n9000 1000 3000\r\n", "output": "9000 3000 1000 \r\n"}, {"input": "2\r\n3000 9000\r\n", "output": "9000 3000 \r\n"}, {"input": "2\r\n3000000000000000000 1000000000000000000\r\n", "output": "3000000000000000000 1000000000000000000 \r\n"}, {"input": "2\r\n1 3\r\n", "output": "3 1 \r\n"}, {"input": "2\r\n1500000000000000000 3000000000000000000\r\n", "output": "1500000000000000000 3000000000000000000 \r\n"}, {"input": "3\r\n4 1 2\r\n", "output": "1 2 4 \r\n"}, {"input": "2\r\n2000000000000000004 1000000000000000002\r\n", "output": "1000000000000000002 2000000000000000004 \r\n"}, {"input": "2\r\n2999999999999999997 999999999999999999\r\n", "output": "2999999999999999997 999999999999999999 \r\n"}, {"input": "2\r\n999999999999999999 1999999999999999998\r\n", "output": "999999999999999999 1999999999999999998 \r\n"}, {"input": "2\r\n1999999999999999998 999999999999999999\r\n", "output": "999999999999999999 1999999999999999998 \r\n"}, {"input": "2\r\n10 5\r\n", "output": "5 10 \r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
Python 3
|
TESTS
| 2 | 124 | 0 |
77721340
|
t = int(input())
first = []
sec = []
for i in range(0, t-1):
u, v = map(int, input().split())
first.append(u)
sec.append(v)
first = list(dict.fromkeys(first))
sec = list(dict.fromkeys(sec))
count = 0
for i in sec:
if i not in first:
count += 1
else:
continue
print(count)
| 9 | 31 | 0 |
194344376
|
N = int(input())
planets = []
ans = 0
for i in range(N - 1):
tunnel = [int(x) for x in input().split()]
planets.append(tunnel[0])
planets.append(tunnel[1])
planets.sort()
for i in range(len(planets)):
if i == 0:
if planets[i] != planets[i + 1]:
ans += 1
elif i == len(planets) - 1:
if planets[i] != planets[i - 1]:
ans += 1
else:
if planets[i] != planets[i + 1] and planets[i] != planets[i - 1]:
ans += 1
print(ans)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/A2
|
958
|
A2
|
Python 3
|
TESTS
| 1 | 124 | 8,192,000 |
37316715
|
N, M = input().split(' ')
N = int(N)
M = int(M)
matrix_1 = []
matrix_2 = []
for i in range(N):
matrix_1.append(input())
for i in range(M):
matrix_2.append(input())
def find(word):
if word in matrix_2[0]:
return matrix_2[0].index(word)
return False
def go():
for i in range(N):
x = find(matrix_1[i])
if x is not False:
found = True
for j in range(1, M):
if matrix_2[j][x:x+5] != matrix_1[i + j]:
found = False
break
if found:
return '{} {}'.format(i + 1, x + 1)
print(go())
| 28 | 1,278 | 43,929,600 |
38869030
|
n,m=list(map(int,input().split()))
s=[]
h=[]
mod=10**9
mod+=7
for i in range(n):
s=(input())
s=s.lower()
hw=0
for i in range(m):
hw+=ord(s[i])*(10**(m-i))
h.append(hw%mod)
h1=[[] for i in range(n-m+1)]
for i in range(m):
hw=0
s=input()
s=s.lower()
aux=[]
for i in range(m):
hw+=ord(s[i])*(10**(m-i))
hw%=mod
h1[0].append(hw)
yy=1
for i in range(m,n):
hw-=(ord(s[i-m])*(10**(m)))
hw*=10
hw+=(ord(s[i])*(10))
hw%=mod
h1[yy].append(hw)
yy+=1
t=False
y=0
for i in range(len(h1)):
x=0
while x<n-m+1:
if h1[i][0]==h[x]:
y=0
ans=[x+1,i+1]
while y<m and h1[i][y]==h[x]:
x+=1
y+=1
else:
x+=1
if y==m:
break
if y==m:
break
print(ans[0],ans[1])
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2.5 | 256 |
Death Stars (medium)
|
The stardate is 1983, and Princess Heidi is getting better at detecting the Death Stars. This time, two Rebel spies have yet again given Heidi two maps with the possible locations of the Death Star. Since she got rid of all double agents last time, she knows that both maps are correct, and indeed show the map of the solar system that contains the Death Star. However, this time the Empire has hidden the Death Star very well, and Heidi needs to find a place that appears on both maps in order to detect the Death Star.
The first map is an N × M grid, each cell of which shows some type of cosmic object that is present in the corresponding quadrant of space. The second map is an M × N grid. Heidi needs to align those two maps in such a way that they overlap over some M × M section in which all cosmic objects are identical. Help Heidi by identifying where such an M × M section lies within both maps.
|
The first line of the input contains two space-separated integers N and M (1 ≤ N ≤ 2000, 1 ≤ M ≤ 200, M ≤ N). The next N lines each contain M lower-case Latin characters (a-z), denoting the first map. Different characters correspond to different cosmic object types. The next M lines each contain N characters, describing the second map in the same format.
|
The only line of the output should contain two space-separated integers i and j, denoting that the section of size M × M in the first map that starts at the i-th row is equal to the section of the second map that starts at the j-th column. Rows and columns are numbered starting from 1.
If there are several possible ways to align the maps, Heidi will be satisfied with any of those. It is guaranteed that a solution exists.
| null |
The 5-by-5 grid for the first test case looks like this:
|
[{"input": "10 5\nsomer\nandom\nnoise\nmayth\neforc\nebewi\nthyou\nhctwo\nagain\nnoise\nsomermayth\nandomeforc\nnoiseebewi\nagainthyou\nnoisehctwo", "output": "4 6"}]
| 2,000 |
["hashing", "strings"]
| 28 |
[{"input": "10 5\r\nsomer\r\nandom\r\nnoise\r\nmayth\r\neforc\r\nebewi\r\nthyou\r\nhctwo\r\nagain\r\nnoise\r\nsomermayth\r\nandomeforc\r\nnoiseebewi\r\nagainthyou\r\nnoisehctwo\r\n", "output": "4 6\r\n"}, {"input": "1 1\r\ng\r\ng\r\n", "output": "1 1\r\n"}]
| false |
stdio
| null | true |
690/D1
|
690
|
D1
|
PyPy 3
|
TESTS
| 1 | 93 | 23,040,000 |
32975171
|
from sys import *
print((stdin.readlines()[-1] + '.').count('B.'))
| 119 | 62 | 0 |
19000257
|
import sys, math
n,m = map(int,input().split())
for i in range(n-1):
input()
w=input()
w+='.'
ans = 0
now=1
for i in w:
if i == 'B':
if now:
now = 0
ans+=1
else:
continue
else:
now = 1
print(ans)
|
Helvetic Coding Contest 2016 online mirror (teams, unrated)
|
ICPC
| 2,016 | 0.5 | 256 |
The Wall (easy)
|
"The zombies are lurking outside. Waiting. Moaning. And when they come..."
"When they come?"
"I hope the Wall is high enough."
Zombie attacks have hit the Wall, our line of defense in the North. Its protection is failing, and cracks are showing. In places, gaps have appeared, splitting the wall into multiple segments. We call on you for help. Go forth and explore the wall! Report how many disconnected segments there are.
The wall is a two-dimensional structure made of bricks. Each brick is one unit wide and one unit high. Bricks are stacked on top of each other to form columns that are up to R bricks high. Each brick is placed either on the ground or directly on top of another brick. Consecutive non-empty columns form a wall segment. The entire wall, all the segments and empty columns in-between, is C columns wide.
|
The first line of the input consists of two space-separated integers R and C, 1 ≤ R, C ≤ 100. The next R lines provide a description of the columns as follows:
- each of the R lines contains a string of length C,
- the c-th character of line r is B if there is a brick in column c and row R - r + 1, and . otherwise.
|
The number of wall segments in the input configuration.
| null |
In the first sample case, the 2nd and 3rd columns define the first wall segment, and the 5th column defines the second.
|
[{"input": "3 7\n.......\n.......\n.BB.B..", "output": "2"}, {"input": "4 5\n..B..\n..B..\nB.B.B\nBBB.B", "output": "2"}, {"input": "4 6\n..B...\nB.B.BB\nBBB.BB\nBBBBBB", "output": "1"}, {"input": "1 1\nB", "output": "1"}, {"input": "10 7\n.......\n.......\n.......\n.......\n.......\n.......\n.......\n.......\n...B...\nB.BB.B.", "output": "3"}, {"input": "8 8\n........\n........\n........\n........\n.B......\n.B.....B\n.B.....B\n.BB...BB", "output": "2"}]
| 1,200 |
[]
| 119 |
[{"input": "3 7\r\n.......\r\n.......\r\n.BB.B..\r\n", "output": "2\r\n"}, {"input": "4 5\r\n..B..\r\n..B..\r\nB.B.B\r\nBBB.B\r\n", "output": "2\r\n"}, {"input": "4 6\r\n..B...\r\nB.B.BB\r\nBBB.BB\r\nBBBBBB\r\n", "output": "1\r\n"}, {"input": "1 1\r\nB\r\n", "output": "1\r\n"}, {"input": "10 7\r\n.......\r\n.......\r\n.......\r\n.......\r\n.......\r\n.......\r\n.......\r\n.......\r\n...B...\r\nB.BB.B.\r\n", "output": "3\r\n"}, {"input": "8 8\r\n........\r\n........\r\n........\r\n........\r\n.B......\r\n.B.....B\r\n.B.....B\r\n.BB...BB\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
958/A2
|
958
|
A2
|
Python 3
|
TESTS
| 5 | 624 | 27,545,600 |
37322117
|
n,m=map(int,input().strip().split(' '))
Dicti={}
M=[]
DP=[]
for i in range(m):
DP.append([[0]]*n)
for i in range(n):
s=input()
if s in Dicti:
Dicti[s].append(i+1)
else:
Dicti[s]=[i+1]
for i in range(m):
M.append(input())
for i in range(m):
j=0
while j+m<=n :
a=M[i][j:j+m]
if a in Dicti:
DP[i][j]=Dicti[a]
else :
DP[i][j]=[-5]
j=j+1
for i in range(n):
j=1
d=DP[0][i][0]
e=True
while j<m and e==True:
e=False
for k in range(len(DP[j][i])):
if DP[j][i][k]==d+1:
d=DP[j][i][k]
e=True
break
j=j+1
if j==m:
print(d-m+1,end=' ')
print(i+1)
| 28 | 1,481 | 10,956,800 |
37327090
|
n, m = list(map(int, input().strip().split(' ')))
mat1, mat2 = [], []
for i in range(0, n): mat1.append(tuple(input().strip()))
for i in range(0, m): mat2.append(tuple(input().strip()))
ix, jx, flg = -1, -1, 0
matr, matc = [], []
for i in range(0, n-m+1):
si, se = i, i+m
matr.append(hash(tuple(mat1[si:se])))
matcur2 = []
for c2i in range(0, m): matcur2.append(tuple(mat2[c2i][si:se]))
matc.append(hash(tuple(matcur2)))
nx = len(matr)
ix, jx = -1, -1
for ix in range(0, nx):
flg=0
for jx in range(0, nx):
if matr[ix]==matc[jx]:
flg=1
break
if flg==1: break
print(str(ix+1)+" "+str(jx+1))
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2.5 | 256 |
Death Stars (medium)
|
The stardate is 1983, and Princess Heidi is getting better at detecting the Death Stars. This time, two Rebel spies have yet again given Heidi two maps with the possible locations of the Death Star. Since she got rid of all double agents last time, she knows that both maps are correct, and indeed show the map of the solar system that contains the Death Star. However, this time the Empire has hidden the Death Star very well, and Heidi needs to find a place that appears on both maps in order to detect the Death Star.
The first map is an N × M grid, each cell of which shows some type of cosmic object that is present in the corresponding quadrant of space. The second map is an M × N grid. Heidi needs to align those two maps in such a way that they overlap over some M × M section in which all cosmic objects are identical. Help Heidi by identifying where such an M × M section lies within both maps.
|
The first line of the input contains two space-separated integers N and M (1 ≤ N ≤ 2000, 1 ≤ M ≤ 200, M ≤ N). The next N lines each contain M lower-case Latin characters (a-z), denoting the first map. Different characters correspond to different cosmic object types. The next M lines each contain N characters, describing the second map in the same format.
|
The only line of the output should contain two space-separated integers i and j, denoting that the section of size M × M in the first map that starts at the i-th row is equal to the section of the second map that starts at the j-th column. Rows and columns are numbered starting from 1.
If there are several possible ways to align the maps, Heidi will be satisfied with any of those. It is guaranteed that a solution exists.
| null |
The 5-by-5 grid for the first test case looks like this:
|
[{"input": "10 5\nsomer\nandom\nnoise\nmayth\neforc\nebewi\nthyou\nhctwo\nagain\nnoise\nsomermayth\nandomeforc\nnoiseebewi\nagainthyou\nnoisehctwo", "output": "4 6"}]
| 2,000 |
["hashing", "strings"]
| 28 |
[{"input": "10 5\r\nsomer\r\nandom\r\nnoise\r\nmayth\r\neforc\r\nebewi\r\nthyou\r\nhctwo\r\nagain\r\nnoise\r\nsomermayth\r\nandomeforc\r\nnoiseebewi\r\nagainthyou\r\nnoisehctwo\r\n", "output": "4 6\r\n"}, {"input": "1 1\r\ng\r\ng\r\n", "output": "1 1\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
Python 3
|
TESTS
| 2 | 77 | 7,168,000 |
37323384
|
import sys
from collections import Counter
def main(args):
n = int(input())
counter = Counter()
for _ in range(n - 1):
u, v = map(int, input().split())
counter[u] += 1
counter[v] += 1
print(counter[1])
if __name__ == '__main__':
sys.exit(main(sys.argv))
| 9 | 31 | 0 |
215291281
|
N = int(input())
a = [0] * N
for _ in range(N-1):
u, v = list(map(int, input().split()))
a[u-1] += 1
a[v-1] += 1
print(a.count(1))
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
PyPy 3
|
TESTS
| 2 | 109 | 20,172,800 |
90813432
|
#author : SanskarxRawat
n=int(input())
s=set()
for i in range(n-1):
u,v=map(int,input().strip().split())
s.add(u)
print(n-len(s))
| 9 | 46 | 0 |
135895593
|
def solve(l1,l2):
ans = 0
M = {}
for i in range(len(l1)):
if l1[i] in M:
M[l1[i]] += 1
else:
M[l1[i]] = 1
for i in range(len(l2)):
if l2[i] in M:
M[l2[i]] += 1
else:
M[l2[i]] = 1
for i in M:
if M[i] < 2:
ans += 1
print(ans)
n = int(input())
l1 = []
l2 = []
while (n-1):
input_string = input()
l = input_string.split()
l1.append(l[0])
l2.append(l[1])
n -= 1
solve(l1,l2)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/A2
|
958
|
A2
|
Python 3
|
TESTS
| 5 | 545 | 9,932,800 |
37318146
|
n,m=map(int,input().strip().split(' '))
Dicti={}
M=[]
DP=[]
for i in range(m):
DP.append([0]*n)
for i in range(n):
Dicti[input()]=i+1
for i in range(m):
M.append(input())
for i in range(m):
j=0
k=0
while j!=n:
a=M[i][j:j+m]
if a in Dicti:
DP[i][j]=Dicti[a]
else :
DP[i][j]=0
j=j+1
for i in range(n):
j=1
d=DP[0][i]
while j<m and DP[j][i]==d+1:
d=DP[j][i]
j=j+1
if j==m:
print(DP[0][i],end=' ')
print(i+1)
| 28 | 1,715 | 83,558,400 |
37339782
|
n, m = list(map(int, input().strip().split(' ')))
L, M = [], []
for i in range(n):
L.append(tuple(input().strip()))
for i in range(0, m):
M.append(tuple(input().strip()))
k=0
row, col = [], []
for i in range(n-m+1):
init, end = i, i+m
row.append(hash(tuple(L[init:end])))
D = []
for j in range(0, m):
D.append(tuple(M[j][init:end]))
col.append(hash(tuple(D)))
for ix in range(len(row)):
k=0
for jx in range(len(row)):
if row[ix]==col[jx]:
k=1
break
if k==1:
break
print(ix+1,end=' ')
print(jx+1)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2.5 | 256 |
Death Stars (medium)
|
The stardate is 1983, and Princess Heidi is getting better at detecting the Death Stars. This time, two Rebel spies have yet again given Heidi two maps with the possible locations of the Death Star. Since she got rid of all double agents last time, she knows that both maps are correct, and indeed show the map of the solar system that contains the Death Star. However, this time the Empire has hidden the Death Star very well, and Heidi needs to find a place that appears on both maps in order to detect the Death Star.
The first map is an N × M grid, each cell of which shows some type of cosmic object that is present in the corresponding quadrant of space. The second map is an M × N grid. Heidi needs to align those two maps in such a way that they overlap over some M × M section in which all cosmic objects are identical. Help Heidi by identifying where such an M × M section lies within both maps.
|
The first line of the input contains two space-separated integers N and M (1 ≤ N ≤ 2000, 1 ≤ M ≤ 200, M ≤ N). The next N lines each contain M lower-case Latin characters (a-z), denoting the first map. Different characters correspond to different cosmic object types. The next M lines each contain N characters, describing the second map in the same format.
|
The only line of the output should contain two space-separated integers i and j, denoting that the section of size M × M in the first map that starts at the i-th row is equal to the section of the second map that starts at the j-th column. Rows and columns are numbered starting from 1.
If there are several possible ways to align the maps, Heidi will be satisfied with any of those. It is guaranteed that a solution exists.
| null |
The 5-by-5 grid for the first test case looks like this:
|
[{"input": "10 5\nsomer\nandom\nnoise\nmayth\neforc\nebewi\nthyou\nhctwo\nagain\nnoise\nsomermayth\nandomeforc\nnoiseebewi\nagainthyou\nnoisehctwo", "output": "4 6"}]
| 2,000 |
["hashing", "strings"]
| 28 |
[{"input": "10 5\r\nsomer\r\nandom\r\nnoise\r\nmayth\r\neforc\r\nebewi\r\nthyou\r\nhctwo\r\nagain\r\nnoise\r\nsomermayth\r\nandomeforc\r\nnoiseebewi\r\nagainthyou\r\nnoisehctwo\r\n", "output": "4 6\r\n"}, {"input": "1 1\r\ng\r\ng\r\n", "output": "1 1\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
PyPy 3
|
TESTS
| 2 | 77 | 0 |
108081635
|
def dfs(edges,source,path = [],count = 0):
if source not in path:
path.append(source)
if(source not in edges):
return [path,count+1]
for n in edges[source]:
path,count = dfs(edges,n,path,count)
return [path,count]
n = int(input())
edges = {}
root =[]
for i in range(n-1):
source,point = [int(x) for x in input().split()]
root.append(point)
if(source not in edges):
edges[source] = [point]
continue
edges[source] = edges[source] + [point]
#find root
for i in range(1,n+1):
if(i not in root):
root = i
break
print(dfs(edges,root)[1])
| 9 | 46 | 0 |
136744943
|
n=int(input())
a=[0]*(n+1)
for i in range(n-1):
u,v=map(int,input().split())
a[u]+=1
a[v]+=1
cnt=0
for i in range(1,n+1):
if a[i]==1:
cnt+=1
print(cnt)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
Python 3
|
TESTS
| 2 | 46 | 0 |
105234619
|
resole = []
result = 0
for i in range(int(input()) - 1):
resole.extend(list(map(lambda x: int(x), input().split(' '))))
resole.sort()
_temp = 0
for i in range(0, len(resole)):
if not resole[i] == _temp:
result += 1
_temp = resole[i]
print(len(resole) - result)
| 9 | 46 | 0 |
156078340
|
import sys
input = sys.stdin.readline
n = int(input())
d = [0]*n
for _ in range(n-1):
a, b = map(int, input().split())
d[a-1] += 1
d[b-1] += 1
c = 0
for i in d:
if i < 2:
c += 1
print(c)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
PyPy 3
|
TESTS
| 2 | 77 | 0 |
109409974
|
n = int(input())
a = 0
p = {x: 0 for x in range(1, n+1)}
for _ in range(n-1):
u, v = map(int, input().split())
p[u] += 1
p[v] += 1
if p[u] > 1:
a += 1
if p[v] > 1:
a += 1
print(a)
| 9 | 46 | 0 |
166115802
|
n = int(input())
li = list([[0]*n])
li = li[0]
count = n
for i in range(n-1):
a, b = map(int, input().split())
if li[a-1] == 0:
li[a-1] = b
elif li[a-1] == -1:
li[a-1] = li[a-1]
else:
count -= 1
li[a-1] = -1
if li[b-1] == 0:
li[b-1] = a
elif li[b-1] == -1:
li[b-1] = li[b-1]
else:
count -= 1
li[b-1] = -1
print(count)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
Python 3
|
TESTS
| 2 | 31 | 0 |
208281795
|
res = []
t = int(input())
for _ in range(t-1):
a,b = list(map(int,input().split()))
res.append(a)
res.append(b)
dc = {}
fd = []
for i in res:
dc.update({i:res.count(i)})
fd.append(res.count(i))
mx = max(fd)
ans = 0
for k,v in dc.items():
if v==mx:
ans=v
print(ans)
# print(dc)
| 9 | 46 | 0 |
172245267
|
n=int(input())
l=[0]*(n)
for _ in range(n-1):
a,b=list(map(int,input().split()))
l[a-1]+=1
l[b-1]+=1
print(l.count(1))
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
958/B1
|
958
|
B1
|
Python 3
|
TESTS
| 2 | 46 | 0 |
120642713
|
n=int(input())
f=[0 for i in range(n+1)]
for i in range(n-1):
x,y=map(int,input().split())
f[x]+=1
f[y]+=1
print(max(f))
| 9 | 46 | 0 |
179980889
|
"""
https://codeforces.com/problemset/problem/958/B1
"""
planets = int(input())
links = dict()
for _ in range(planets - 1):
a, b = [x for x in input().split()]
links.setdefault(a, []).append(b)
links.setdefault(b, []).append(a)
compte=0
for v in links.values():
if len(v)==1:
compte+=1
print(compte)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 92 | 0 |
113590925
|
import math
t,l=0,[]
n,p=map(int,input().split(' '))
for i in range(n):
x,y=map(int,input().split(' '))
l.append(math.sqrt(x**2 + y**2))
for i in range(1,n):
t+=abs(l[i]-l[i-1])
print((t/50)*p)
| 42 | 62 | 0 |
210675325
|
import math
n, k = map(int, input().split()) # Read the values of n and k
points = []
for _ in range(n):
x, y = map(int, input().split()) # Read the coordinates of each point
points.append((x, y))
total_distance = 0
for i in range(n - 1):
x1, y1 = points[i]
x2, y2 = points[i + 1]
distance = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2) # Calculate the distance between two points
total_distance += distance
total_time = total_distance / 50 * k # Calculate the total time wasted
print("{:.9f}".format(total_time)) # Print the result with 9 decimal places
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
990/F
|
990
|
F
|
Python 3
|
TESTS
| 0 | 77 | 0 |
39208122
|
def i_ints():
return list(map(int, input().split()))
#############
n, = i_ints()
s = [0] + i_ints()
if sum(s):
print("Impossible")
else:
print("Possible")
m, = i_ints()
es = [i_ints() for _ in range(m)]
nb = [[] for i in range(n+1)]
for i, (x, y) in enumerate(es):
nb[x].append((y, i, 1))
nb[y].append((x, i, -1))
f = [0] * m
def walk_tree(x, seen):
print("A:", x, seen)
seen.add(x)
for y, i, factor in nb[x]:
if y not in seen:
walk_tree(y, seen)
f[i] = factor * s[y]
s[x] += s[y]
s[y] = 0
walk_tree(1, set())
for x in f:
print(x)
| 37 | 1,216 | 96,768,000 |
155788846
|
from collections import deque
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def get_root(s):
v = []
while not s == root[s]:
v.append(s)
s = root[s]
for i in v:
root[i] = s
return s
def unite(s, t):
rs, rt = get_root(s), get_root(t)
if not rs ^ rt:
return
if rank[s] == rank[t]:
rank[rs] += 1
if rank[s] >= rank[t]:
root[rt] = rs
size[rs] += size[rt]
else:
root[rs] = rt
size[rt] += size[rs]
return
def same(s, t):
return True if get_root(s) == get_root(t) else False
def bfs(s):
q = deque()
q.append(s)
visit = [0] * (n + 1)
visit[s] = 1
parent = [-1] * (n + 1)
p = []
while q:
i = q.popleft()
for j, k, l in G[i]:
if not visit[j]:
visit[j] = 1
q.append(j)
parent[j] = i
p.append((j, k, l))
return parent, p
n = int(input())
s = [0] + list(map(int, input().split()))
m = int(input())
e = [tuple(map(int, input().split())) for _ in range(m)]
if sum(s):
ans = "Impossible"
print(ans)
exit()
ans = "Possible"
print(ans)
root = [i for i in range(n + 1)]
rank = [1 for _ in range(n + 1)]
size = [1 for _ in range(n + 1)]
G = [[] for _ in range(n + 1)]
for i in range(m):
x, y = e[i]
if not same(x, y):
unite(x, y)
G[x].append((y, 1, i))
G[y].append((x, -1, i))
parent, p = bfs(1)
ans = [0] * m
while p:
i, k, l = p.pop()
j = parent[i]
ans[l] = k * s[i]
s[j] += s[i]
sys.stdout.write("\n".join(map(str, ans)))
|
Educational Codeforces Round 45 (Rated for Div. 2)
|
ICPC
| 2,018 | 2 | 256 |
Flow Control
|
You have to handle a very complex water distribution system. The system consists of $$$n$$$ junctions and $$$m$$$ pipes, $$$i$$$-th pipe connects junctions $$$x_i$$$ and $$$y_i$$$.
The only thing you can do is adjusting the pipes. You have to choose $$$m$$$ integer numbers $$$f_1$$$, $$$f_2$$$, ..., $$$f_m$$$ and use them as pipe settings. $$$i$$$-th pipe will distribute $$$f_i$$$ units of water per second from junction $$$x_i$$$ to junction $$$y_i$$$ (if $$$f_i$$$ is negative, then the pipe will distribute $$$|f_i|$$$ units of water per second from junction $$$y_i$$$ to junction $$$x_i$$$). It is allowed to set $$$f_i$$$ to any integer from $$$-2 \cdot 10^9$$$ to $$$2 \cdot 10^9$$$.
In order for the system to work properly, there are some constraints: for every $$$i \in [1, n]$$$, $$$i$$$-th junction has a number $$$s_i$$$ associated with it meaning that the difference between incoming and outcoming flow for $$$i$$$-th junction must be exactly $$$s_i$$$ (if $$$s_i$$$ is not negative, then $$$i$$$-th junction must receive $$$s_i$$$ units of water per second; if it is negative, then $$$i$$$-th junction must transfer $$$|s_i|$$$ units of water per second to other junctions).
Can you choose the integers $$$f_1$$$, $$$f_2$$$, ..., $$$f_m$$$ in such a way that all requirements on incoming and outcoming flows are satisfied?
|
The first line contains an integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of junctions.
The second line contains $$$n$$$ integers $$$s_1, s_2, \dots, s_n$$$ ($$$-10^4 \le s_i \le 10^4$$$) — constraints for the junctions.
The third line contains an integer $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) — the number of pipes.
$$$i$$$-th of the next $$$m$$$ lines contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$1 \le x_i, y_i \le n$$$, $$$x_i \ne y_i$$$) — the description of $$$i$$$-th pipe. It is guaranteed that each unordered pair $$$(x, y)$$$ will appear no more than once in the input (it means that there won't be any pairs $$$(x, y)$$$ or $$$(y, x)$$$ after the first occurrence of $$$(x, y)$$$). It is guaranteed that for each pair of junctions there exists a path along the pipes connecting them.
|
If you can choose such integer numbers $$$f_1, f_2, \dots, f_m$$$ in such a way that all requirements on incoming and outcoming flows are satisfied, then output "Possible" in the first line. Then output $$$m$$$ lines, $$$i$$$-th line should contain $$$f_i$$$ — the chosen setting numbers for the pipes. Pipes are numbered in order they appear in the input.
Otherwise output "Impossible" in the only line.
| null | null |
[{"input": "4\n3 -10 6 1\n5\n1 2\n3 2\n2 4\n3 4\n3 1", "output": "Possible\n4\n-6\n8\n-7\n7"}, {"input": "4\n3 -10 6 4\n5\n1 2\n3 2\n2 4\n3 4\n3 1", "output": "Impossible"}]
| 2,400 |
["dfs and similar", "dp", "greedy", "trees"]
| 37 |
[{"input": "4\r\n3 -10 6 1\r\n5\r\n1 2\r\n3 2\r\n2 4\r\n3 4\r\n3 1\r\n", "output": "Possible\r\n-3\r\n-6\r\n1\r\n0\r\n0\r\n"}, {"input": "4\r\n3 -10 6 4\r\n5\r\n1 2\r\n3 2\r\n2 4\r\n3 4\r\n3 1\r\n", "output": "Impossible\r\n"}, {"input": "1\r\n0\r\n0\r\n", "output": "Possible\r\n"}, {"input": "1\r\n123\r\n0\r\n", "output": "Impossible\r\n"}, {"input": "2\r\n-1 1\r\n1\r\n1 2\r\n", "output": "Possible\r\n1\r\n"}, {"input": "2\r\n-1 1\r\n1\r\n2 1\r\n", "output": "Possible\r\n-1\r\n"}]
| false |
stdio
|
import sys
def main():
input_path = sys.argv[1]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline())
s = list(map(int, f.readline().split()))
m = int(f.readline())
pipes = [tuple(map(int, f.readline().split())) for _ in range(m)]
sum_s = sum(s)
possible = (sum_s == 0)
with open(submission_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip() != '']
if not lines:
print(0)
return
first_line = lines[0]
if possible:
if first_line != "Possible":
print(0)
return
if len(lines) != m + 1:
print(0)
return
try:
f_values = [int(line) for line in lines[1:]]
except:
print(0)
return
flows_in = [0] * (n + 1)
flows_out = [0] * (n + 1)
for i in range(m):
x, y = pipes[i]
f = f_values[i]
flows_in[y] += f
flows_out[x] += f
for u in range(1, n + 1):
if (flows_in[u] - flows_out[u]) != s[u - 1]:
print(0)
return
print(1)
else:
if first_line == "Impossible" and len(lines) == 1:
print(1)
else:
print(0)
if __name__ == '__main__':
main()
| true |
958/B1
|
958
|
B1
|
Python 3
|
TESTS
| 2 | 30 | 0 |
208283359
|
res = []
t = int(input())
for _ in range(t-1):
a,b = list(map(int,input().split()))
res.append(a)
res.append(b)
print(res.count(a))
| 9 | 46 | 0 |
196616253
|
# from dust i have come dust i will be
n=int(input())
a=[0]*(n+1)
for i in range(n-1):
u,v=map(int,input().split())
a[u]+=1
a[v]+=1
cnt=0
for i in range(1,n+1):
if a[i]==1:
cnt+=1
print(cnt)
|
Helvetic Coding Contest 2018 online mirror (teams allowed, unrated)
|
ICPC
| 2,018 | 2 | 256 |
Maximum Control (easy)
|
The Resistance is trying to take control over all planets in a particular solar system. This solar system is shaped like a tree. More precisely, some planets are connected by bidirectional hyperspace tunnels in such a way that there is a path between every pair of the planets, but removing any tunnel would disconnect some of them.
The Resistance already has measures in place that will, when the time is right, enable them to control every planet that is not remote. A planet is considered to be remote if it is connected to the rest of the planets only via a single hyperspace tunnel.
How much work is there left to be done: that is, how many remote planets are there?
|
The first line of the input contains an integer N (2 ≤ N ≤ 1000) – the number of planets in the galaxy.
The next N - 1 lines describe the hyperspace tunnels between the planets. Each of the N - 1 lines contains two space-separated integers u and v (1 ≤ u, v ≤ N) indicating that there is a bidirectional hyperspace tunnel between the planets u and v. It is guaranteed that every two planets are connected by a path of tunnels, and that each tunnel connects a different pair of planets.
|
A single integer denoting the number of remote planets.
| null |
In the first example, only planets 2, 3 and 5 are connected by a single tunnel.
In the second example, the remote planets are 2 and 3.
Note that this problem has only two versions – easy and medium.
|
[{"input": "5\n4 1\n4 2\n1 3\n1 5", "output": "3"}, {"input": "4\n1 2\n4 3\n1 4", "output": "2"}]
| 1,000 |
["implementation"]
| 9 |
[{"input": "5\r\n4 1\r\n4 2\r\n1 3\r\n1 5\r\n", "output": "3\r\n"}, {"input": "4\r\n1 2\r\n4 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "10\r\n4 3\r\n2 6\r\n10 1\r\n5 7\r\n5 8\r\n10 6\r\n5 9\r\n9 3\r\n2 9\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
831/B
|
831
|
B
|
Python 3
|
TESTS
| 0 | 15 | 0 |
205403071
|
# Read input
layout1 = input().strip()
layout2 = input().strip()
text = input().strip()
# Create a dictionary to map each letter from layout1 to its corresponding letter in layout2
mapping = {}
for i in range(len(layout1)):
mapping[layout1[i]] = layout2[i]
# Apply the mapping to each letter in the text, and keep all other characters as is
new_text = ''
for c in text:
if c in mapping:
new_text += mapping[c]
else:
new_text += c
# Print the new text
print(new_text)
| 19 | 31 | 0 |
147198336
|
s=input()
s1=input()
d={}
for i in range(26):
d[s[i]]=s1[i]
s2=input()
r=""
for i in s2:
if(i.isupper()):
r=r+d[i.lower()].upper()
elif(i.isalpha()):
r=r+d[i]
else:
r=r+i
print(r)
|
Codeforces Round 424 (Div. 2, rated, based on VK Cup Finals)
|
CF
| 2,017 | 1 | 256 |
Keyboard Layouts
|
There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.
You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.
You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.
Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.
|
The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.
The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.
The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.
|
Print the text if the same keys were pressed in the second layout.
| null | null |
[{"input": "qwertyuiopasdfghjklzxcvbnm\nveamhjsgqocnrbfxdtwkylupzi\nTwccpQZAvb2017", "output": "HelloVKCup2017"}, {"input": "mnbvcxzlkjhgfdsapoiuytrewq\nasdfghjklqwertyuiopzxcvbnm\n7abaCABAABAcaba7", "output": "7uduGUDUUDUgudu7"}]
| 800 |
["implementation", "strings"]
| 19 |
[{"input": "qwertyuiopasdfghjklzxcvbnm\r\nveamhjsgqocnrbfxdtwkylupzi\r\nTwccpQZAvb2017\r\n", "output": "HelloVKCup2017\r\n"}, {"input": "mnbvcxzlkjhgfdsapoiuytrewq\r\nasdfghjklqwertyuiopzxcvbnm\r\n7abaCABAABAcaba7\r\n", "output": "7uduGUDUUDUgudu7\r\n"}, {"input": "ayvguplhjsoiencbkxdrfwmqtz\r\nkhzvtbspcndierqumlojyagfwx\r\n3\r\n", "output": "3\r\n"}, {"input": "oaihbljgekzsxucwnqyrvfdtmp\r\nwznqcfvrthjibokeglmudpayxs\r\ntZ8WI33UZZytE8A99EvJjck228LxUQtL5A8q7O217KrmdhpmdhN7JEdVXc8CRm07TFidlIou9AKW9cCl1c4289rfU87oXoSCwHpZO7ggC2GmmDl0KGuA2IimDco2iKaBKl46H089r2tw16mhzI44d2X6g3cnoD0OU5GvA8l89nhNpzTbY9FtZ2wE3Y2a5EC7zXryudTZhXFr9EEcX8P71fp6694aa02B4T0w1pDaVml8FM3N2qB78DBrS723Vpku105sbTJEdBpZu77b1C47DujdoR7rjm5k2nsaPBqX93EfhW95Mm0sBnFtgo12gS87jegSR5u88tM5l420dkt1l1b18UjatzU7P2i9KNJA528caiEpE3JtRw4m4TJ7M1zchxO53skt3Fqvxk2C51gD8XEY7YJC2xmTUqyEUFmPX581Gow2HWq4jaP8FK87\r\n", "output": "yJ8EN33OJJmyT8Z99TdVvkh228FbOLyF5Z8l7W217HuxaqsxaqG7VTaDBk8KUx07YPnafNwo9ZHE9kKf1k4289upO87wBwIKeQsJW7rrK2RxxAf0HRoZ2NnxAkw2nHzCHf46Q089u2ye16xqjN44a2B6r3kgwA0WO5RdZ8f89gqGsjYcM9PyJ2eT3M2z5TK7jBumoaYJqBPu9TTkB8S71ps6694zz02C4Y0e1sAzDxf8PX3G2lC78ACuI723Dsho105icYVTaCsJo77c1K47AovawU7uvx5h2gizSClB93TpqE95Xx0iCgPyrw12rI87vtrIU5o88yX5f420ahy1f1c18OvzyjO7S2n9HGVZ528kznTsT3VyUe4x4YV7X1jkqbW53ihy3Pldbh2K51rA8BTM7MVK2bxYOlmTOPxSB581Rwe2QEl4vzS8PH87\r\n"}, {"input": "aymrnptzhklcbuxfdvjsgqweio\r\nwzsavqryltmjnfgcedxpiokbuh\r\nB5\r\n", "output": "N5\r\n"}, {"input": "unbclszprgiqjodxeawkymvfth\r\ncxfwbdvuqlotkgparmhsyinjze\r\nk081O\r\n", "output": "s081G\r\n"}, {"input": "evfsnczuiodgbhqmlypkjatxrw\r\nhvsockwjxtgreqmyanlzidpbuf\r\n306QMPpaqZ\r\n", "output": "306MYLldmW\r\n"}, {"input": "pbfjtvryklwmuhxnqsoceiadgz\r\ntaipfdvlzemhjsnkwyocqgrxbu\r\nTm9H66Ux59PuGe3lEG94q18u11Dda6w59q1hAAIvHR1qquKI2Xf5ZFdKAPhcEnqKT6BF6Oh16P48YvrIKWGDlRcx9BZwwEF64o0As\r\n", "output": "Fh9S66Jn59TjBq3eQB94w18j11Xxr6m59w1sRRGdSV1wwjZG2Ni5UIxZRTscQkwZF6AI6Os16T48LdvGZMBXeVcn9AUmmQI64o0Ry\r\n"}, {"input": "rtqgahmkeoldsiynjbuwpvcxfz\r\noxqiuwflvebnapyrmcghtkdjzs\r\nJqNskelr3FNjbDhfKPfPXxlqOw72p9BVBwf0tN8Ucs48Vlfjxqo9V3ruU5205UgTYi3JKFbW91NLQ1683315VJ4RSLFW7s26s6uZKs5cO2wAT4JS8rCytZVlPWXdNXaCTq06F1v1Fj2zq7DeJbBSfM5Eko6vBndR75d46mf5Pq7Ark9NARTtQ176ukljBdaqXRsYxrBYl7hda1V7sy38hfbjz59HYM9U55P9eh1CX7tUE44NFlQu7zSjSBHyS3Tte2XaXD3O470Q8U20p8W5rViIh8lsn2TvmcdFdxrF3Ye26J2ZK0BR3KShN597WSJmHJTl4ZZ88IMhzHi6vFyr7MuGYNFGebTB573e6Crwj8P18h344yd8sR2NPge36Y3QC8Y2uW577CO2w4fz\r\n", "output": "MqRalvbo3ZRmcNwzLTzTJjbqEh72t9CKChz0xR8Gda48Kbzmjqe9K3ogG5205GiXYp3MLZcH91RBQ1683315KM4OABZH7a26a6gSLa5dE2hUX4MA8oDyxSKbTHJnRJuDXq06Z1k1Zm2sq7NvMcCAzF5Vle6kCrnO75n46fz5Tq7Uol9RUOXxQ176glbmCnuqJOaYjoCYb7wnu1K7ay38wzcms59WYF9G55T9vw1DJ7xGV44RZbQg7sAmACWyA3Xxv2JuJN3E470Q8G20t8H5oKpPw8bar2XkfdnZnjoZ3Yv26M2SL0CO3LAwR597HAMfWMXb4SS88PFwsWp6kZyo7FgIYRZIvcXC573v6Dohm8T18w344yn8aO2RTiv36Y3QD8Y2gH577DE2h4zs\r\n"}, {"input": "buneohqdgxjsafrmwtzickvlpy\r\nzblwamjxifyuqtnrgdkchpoves\r\n4RZf8YivG6414X1GdDfcCbc10GA0Wz8514LI9D647XzPb66UNh7lX1rDQv0hQvJ7aqhyh1Z39yABGKn24g185Y85ER5q9UqPFaQ2JeK97wHZ78CMSuU8Zf091mePl2OX61BLe5KdmUWodt4BXPiseOZkZ4SZ27qtBM4hT499mCirjy6nB0ZqjQie4Wr3uhW2mGqBlHyEZbW7A6QnsNX9d3j5aHQN0H6GF8J0365KWuAmcroutnJD6l6HI3kSSq17Sdo2htt9y967y8sc98ZAHbutH1m9MOVT1E9Mb5UIK3qNatk9A0m2i1fQl9A65204Q4z4O4rQf374YEq0s2sfmQNW9K7E1zSbj51sGINJVr5736Gw8aW6u9Cjr0sjffXctLopJ0YQ47xD1yEP6bB3odG7slgiM8hJ9BuwfGUwN8tbAgJU8wMI2L0P446MO\r\n", "output": "4NKt8ScoI6414F1IxXthHzh10IQ0Gk8514VC9X647FkEz66BLm7vF1nXJo0mJoY7qjmsm1K39sQZIPl24i185S85WN5j9BjETqJ2YwP97gMK78HRUbB8Kt091rwEv2AF61ZVw5PxrBGaxd4ZFEcuwAKpK4UK27jdZR4mD499rHcnys6lZ0KjyJcw4Gn3bmG2rIjZvMsWKzG7Q6JluLF9x3y5qMJL0M6IT8Y0365PGbQrhnabdlYX6v6MC3pUUj17Uxa2mdd9s967s8uh98KQMzbdM1r9RAOD1W9Rz5BCP3jLqdp9Q0r2c1tJv9Q65204J4k4A4nJt374SWj0u2utrJLG9P7W1kUzy51uICLYOn5736Ig8qG6b9Hyn0uyttFhdVaeY0SJ47fX1sWE6zZ3axI7uvicR8mY9ZbgtIBgL8dzQiYB8gRC2V0E446RA\r\n"}, {"input": "qwertyuiopasdfghjklzxcvbnm\r\nqwertyuiopasdfghjklzxcvbnm\r\nqwertyuiopasdfghjklzxcvbnmPOIUYTREWQLKJHGFDSAMNBVCXZ12345678900987654321ASDFGHJKLqwertyuiopZXCVBNM\r\n", "output": "qwertyuiopasdfghjklzxcvbnmPOIUYTREWQLKJHGFDSAMNBVCXZ12345678900987654321ASDFGHJKLqwertyuiopZXCVBNM\r\n"}, {"input": "qwertyuiopasdfghjklzxcvbnm\r\nmnbvcxzlkjhgfdsapoiuytrewq\r\nasdfghjklzxcvbnmqwertyuiopASDFGHJKLQWERTYUIOPZXCVBNM12345678900987654321QWSDFGVBNxcvghjkoWQEDFGHNJMzxcfghjkl\r\n", "output": "hgfdsapoiuytrewqmnbvcxzlkjHGFDSAPOIMNBVCXZLKJUYTREWQ12345678900987654321MNGFDSREWytrsapokNMBFDSAWPQuytdsapoi\r\n"}]
| false |
stdio
| null | true |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 186 | 6,963,200 |
89463256
|
from math import sqrt
n,k=input().split(" ")
n,k=int(n),int(k)
x,y=-0.5,-1
a,b=-1,-1
time=0
for i in range(n):
a,b=input().split(" ")
a,b=int(a),int(b)
if x!=-0.5:
time+=sqrt((x-a)**2+(y-b)**2)/50
x,y=a,b
print(time)
| 42 | 62 | 0 |
213562730
|
take = input().split()
n = int(take[0])
k = int(take[1])
X = []
Y = []
for i in range(n) :
point = input().split()
X.append(int(point[0]))
Y.append(int(point[1]))
import math
length = 0
for i in range(n-1) :
dist = math.sqrt((X[i+1]-X[i])**2+(Y[i+1]-Y[i])**2)
length += dist
print(length*k/50)
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 186 | 307,200 |
94580553
|
import math
n, k = [int(x) for x in input().split(' ')]
points = []
t = 0
for i in range(n):
points.append([int(x) for x in input().split(' ')])
for i in range(n - 1):
t += math.sqrt((points[i][0] - points[i + 1][0]) ** 2 + (points[i][1] - points[i + 1][1]) ** 2)
print(t / 50)
| 42 | 62 | 0 |
215806788
|
n, k = map(int, input().split())
x0, y0 = map(int, input().split())
a = 0
for _ in range(n-1):
x1, y1 = map(int, input().split())
a += ((x1-x0)**2+(y1-y0)**2)**0.5
x0, y0 = x1, y1
print(k*a/50)
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 186 | 0 |
48373556
|
n, k = map(int, input().split())
coords = []
for _ in range(n):
x, y = map(int, input().split())
coords.append((x, y))
px, py = coords[0]
space = 0
for i in range(1, n):
cx, cy = coords[i]
dist = ((cx - px) * (cx - px) + (cy - py) * (cy - py)) ** 0.5
space += dist
px, py = cx, cy
ans = space / 50
print(ans)
| 42 | 62 | 0 |
231468713
|
from math import sqrt
n, k = input().split()
n, k = int(n), int(k)
coordinates = []
for i in range(n):
coordinates += [tuple(map(int, input().split()))]
total = 0
for j in range(n - 1):
total += sqrt((coordinates[j][0] - coordinates[j + 1][0]) ** 2 + (coordinates[j][1] - coordinates[j + 1][1]) ** 2)
print(f'{(total / 50) * k:.9f}')
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 186 | 0 |
55098827
|
n, k = map(int, input().split())
x0 = 0
y0 = 0
r = 0
for i in range(n):
x,y = map(int, input().split())
if i > 0:
r += ((x-x0)**2 + (y-y0)**2)**0.5
x0 = x
y0 = y
t = r / 50
print(t)
| 42 | 92 | 0 |
4607148
|
n, k = map(int, input().split())
x, y = map(int, input().split())
l = 0
for i in range(n - 1):
u, v = map(int, input().split())
l += ((x - u) ** 2 + (y - v) ** 2) ** 0.5
x, y = u, v
print(k * l / 50)
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
127/A
|
127
|
A
|
PyPy 3
|
TESTS
| 1 | 186 | 20,172,800 |
125533270
|
"""
reading 9:16 -> 9:20
thinking 9:20 -> 9:24
coding 9:24 -> 9:40
debugging 9:40 ->
"""
"""
Se toma n veces las posiciones de los trazos,
como son líneas rectas, se calcula la distancia
entre A_k y A_k+1, luego se multiplica por k
y se divide entre su velocidad.
"""
def distance(a : tuple, b : tuple):
return ((a[0]-b[0])**2 + (a[1]-b[1]**2))**(1/2)
if __name__ == "__main__":
n, k = map(int, input().split())
speed = 50.0
old = tuple(map(int, input().split()))
dist = 0.0
for i in range(n-1):
new = tuple(map(int, input().split()))
dist += distance(old, new)
old = new
print(f'{k*dist/speed:.9f}')
| 42 | 92 | 0 |
136191539
|
nk = list(map(int, input().split(' ')))
n = nk[0]
k = nk[1]
dist = 0.0
xy = list(map(int, input().split(' ')))
x = xy[0]
y = xy[1]
for idx in range(n-1):
xy1 = list(map(int, input().split(' ')))
x1 = xy1[0]
y1 = xy1[1]
dist += ((((x1 - x) ** 2) + ((y1 - y) ** 2)) ** 0.5)
x = x1
y = y1
time = dist/50
print(time*k)
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
31/D
|
31
|
D
|
Python 3
|
TESTS
| 2 | 62 | 102,400 |
207016922
|
# LUOGU_RID: 111131923
from sys import stdin
input = stdin.readline
row = col = [[0] * 110 for i in range(110)]
vis = [[False] * 110 for i in range(110)]
ans = []
def get() :
return map(int,input().split())
W,H,N = get()
for _ in range(N) :
x1, y1, x2, y2 = get()
if x1 == x2 :
for i in range(y1 + 1, y2 + 1) :
row[x1][i] = 1
elif y1 == y2 :
for i in range(x1 + 1, x2 + 1) :
col[i][y1] = 1
for i in range(1, W + 1) :
for j in range(1, H + 1) :
if not vis[i][j] :
t1, t2, t3, t4 = i, j, 1, 1
while col[t1][t2] == 0 and t2 < H :
t2 += 1
t4 += 1
while row[t1][t2] == 0 and t1 < W :
t1 += 1
t3 += 1
for u in range(i, t1 + 1) :
for v in range(j, t2 + 1) :
vis[u][v] = True
ans.append(t3 * t4)
ans.sort()
for i in ans :
print(i, end = ' ')
| 33 | 278 | 4,505,600 |
146016755
|
from collections import deque
import sys
input = sys.stdin.readline
def bfs(x, y):
q = deque()
q.append((x, y))
visit[x][y] = 1
c = 0
while q:
i, j = q.popleft()
c += 1
for di, dj in G[i][j]:
ni, nj = i + di, j + dj
if 0 <= ni < w and 0 <= nj < h:
if not visit[ni][nj]:
q.append((ni, nj))
visit[ni][nj] = 1
return c
w, h, n = map(int, input().split())
v = [(1, 0), (-1, 0), (0, 1), (0, -1)]
G = [[set(v) for _ in range(h)] for _ in range(w)]
for _ in range(n):
x1, y1, x2, y2 = map(int, input().split())
if x1 == x2:
for y in range(y1, y2):
G[x1 - 1][y].remove((1, 0))
for y in range(y1, y2):
G[x1][y].remove((-1, 0))
else:
for x in range(x1, x2):
G[x][y1 - 1].remove((0, 1))
for x in range(x1, x2):
G[x][y1].remove((0, -1))
visit = [[0] * h for _ in range(w)]
ans = []
for x in range(w):
for y in range(h):
if not visit[x][y]:
c = bfs(x, y)
ans.append(c)
ans.sort()
sys.stdout.write(" ".join(map(str, ans)))
|
Codeforces Beta Round 31 (Div. 2, Codeforces format)
|
CF
| 2,010 | 2 | 256 |
Chocolate
|
Bob has a rectangular chocolate bar of the size W × H. He introduced a cartesian coordinate system so that the point (0, 0) corresponds to the lower-left corner of the bar, and the point (W, H) corresponds to the upper-right corner. Bob decided to split the bar into pieces by breaking it. Each break is a segment parallel to one of the coordinate axes, which connects the edges of the bar. More formally, each break goes along the line x = xc or y = yc, where xc and yc are integers. It should divide one part of the bar into two non-empty parts. After Bob breaks some part into two parts, he breaks the resulting parts separately and independently from each other. Also he doesn't move the parts of the bar. Bob made n breaks and wrote them down in his notebook in arbitrary order. At the end he got n + 1 parts. Now he wants to calculate their areas. Bob is lazy, so he asks you to do this task.
|
The first line contains 3 integers W, H and n (1 ≤ W, H, n ≤ 100) — width of the bar, height of the bar and amount of breaks. Each of the following n lines contains four integers xi, 1, yi, 1, xi, 2, yi, 2 — coordinates of the endpoints of the i-th break (0 ≤ xi, 1 ≤ xi, 2 ≤ W, 0 ≤ yi, 1 ≤ yi, 2 ≤ H, or xi, 1 = xi, 2, or yi, 1 = yi, 2). Breaks are given in arbitrary order.
It is guaranteed that the set of breaks is correct, i.e. there is some order of the given breaks that each next break divides exactly one part of the bar into two non-empty parts.
|
Output n + 1 numbers — areas of the resulting parts in the increasing order.
| null | null |
[{"input": "2 2 2\n1 0 1 2\n0 1 1 1", "output": "1 1 2"}, {"input": "2 2 3\n1 0 1 2\n0 1 1 1\n1 1 2 1", "output": "1 1 1 1"}, {"input": "2 4 2\n0 1 2 1\n0 3 2 3", "output": "2 2 4"}]
| 2,000 |
["dfs and similar", "implementation"]
| 33 |
[{"input": "2 2 2\r\n1 0 1 2\r\n0 1 1 1\r\n", "output": "1 1 2 "}, {"input": "2 2 3\r\n1 0 1 2\r\n0 1 1 1\r\n1 1 2 1\r\n", "output": "1 1 1 1 "}, {"input": "2 4 2\r\n0 1 2 1\r\n0 3 2 3\r\n", "output": "2 2 4 "}, {"input": "5 5 3\r\n2 1 2 5\r\n0 1 5 1\r\n4 0 4 1\r\n", "output": "1 4 8 12 "}, {"input": "10 10 4\r\n9 0 9 10\r\n4 0 4 10\r\n1 0 1 10\r\n1 4 4 4\r\n", "output": "10 10 12 18 50 "}, {"input": "9 8 5\r\n4 3 4 4\r\n0 4 9 4\r\n5 4 5 8\r\n0 3 9 3\r\n1 4 1 8\r\n", "output": "4 4 5 16 16 27 "}, {"input": "100 100 1\r\n0 14 100 14\r\n", "output": "1400 8600 "}]
| false |
stdio
| null | true |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 216 | 0 |
49985139
|
cnt=lambda s,i:s.count(i)
iip=lambda:int(input())
sip=lambda:input()
f=lambda:map(int,input().split())
lf=lambda:list(map(int,input().split()))
n,k=f()
l=lf()
d=0
for i in range(n-1):
l2=lf()
d+=((l2[0]-l[0])**2+(l2[1]-l[1])**2)**0.5
l=l2
print('%0.9f'%(d/50))
| 42 | 92 | 0 |
136781502
|
(N, K) = [int(i) for i in input().split()]
total_distance = 0
last_point = [int(i) for i in input().split()]
for i in range(N-1):
point = [int(i) for i in input().split()]
total_distance += (abs(point[0]-last_point[0])**2 + abs(point[1]-last_point[1])**2)**0.5
last_point = point
wasted_time = (total_distance / 50) * K
print(wasted_time)
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
127/A
|
127
|
A
|
Python 3
|
TESTS
| 1 | 60 | 0 |
174158343
|
a=input().split()
n=int(a[0])
k=int(a[1])
lista = []
for l in range(n):
coord = []
b=input().split()
for i in b: coord.append(int(i))
lista.append(coord)
total=0
for i in range(len(lista)-1):
x1=lista[i][0]
x2=lista[i+1][0]
y1=lista[i][1]
y2=lista[i+1][1]
dist = ( (x2 - x1)**2 + (y2 - y1)**2 )**(1/2)
total+=dist
print(total/50)
| 42 | 92 | 0 |
139640508
|
n, k = map(int, input().split());
x, y = map(int, input().split());
distance = 0
for i in range(n - 1):
a, b = map(int, input().split())
t = (a - x) ** 2 + (b - y) ** 2;
distance += t ** 0.5
x, y = a, b
print(distance * k / 50)
|
Codeforces Beta Round 93 (Div. 2 Only)
|
CF
| 2,011 | 2 | 256 |
Wasted Time
|
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
|
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
|
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
| null | null |
[{"input": "2 1\n0 0\n10 0", "output": "0.200000000"}, {"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0", "output": "6.032163204"}, {"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0", "output": "3.000000000"}]
| 900 |
["geometry"]
| 42 |
[{"input": "2 1\r\n0 0\r\n10 0\r\n", "output": "0.200000000"}, {"input": "5 10\r\n3 1\r\n-5 6\r\n-2 -1\r\n3 2\r\n10 0\r\n", "output": "6.032163204"}, {"input": "6 10\r\n5 0\r\n4 0\r\n6 0\r\n3 0\r\n7 0\r\n2 0\r\n", "output": "3.000000000"}, {"input": "10 95\r\n-20 -5\r\n2 -8\r\n14 13\r\n10 3\r\n17 11\r\n13 -12\r\n-6 11\r\n14 -15\r\n-13 14\r\n19 8\r\n", "output": "429.309294877"}, {"input": "30 1000\r\n4 -13\r\n14 13\r\n-14 -16\r\n-9 18\r\n17 11\r\n2 -8\r\n2 15\r\n8 -1\r\n-9 13\r\n8 -12\r\n-2 20\r\n11 -12\r\n19 8\r\n9 -15\r\n-20 -5\r\n-18 20\r\n-13 14\r\n-12 -17\r\n-4 3\r\n13 -12\r\n11 -10\r\n18 7\r\n-6 11\r\n10 13\r\n10 3\r\n6 -14\r\n-1 10\r\n14 -15\r\n2 11\r\n-8 10\r\n", "output": "13629.282573522"}, {"input": "2 1\r\n-20 -10\r\n-10 -6\r\n", "output": "0.215406592"}, {"input": "2 13\r\n13 -10\r\n-3 -2\r\n", "output": "4.651021393"}, {"input": "2 21\r\n13 8\r\n14 10\r\n", "output": "0.939148551"}, {"input": "2 75\r\n-3 12\r\n1 12\r\n", "output": "6.000000000"}, {"input": "2 466\r\n10 16\r\n-6 -3\r\n", "output": "231.503997374"}, {"input": "2 999\r\n6 16\r\n-17 -14\r\n", "output": "755.286284531"}, {"input": "2 1000\r\n-17 -14\r\n-14 -8\r\n", "output": "134.164078650"}, {"input": "3 384\r\n-4 -19\r\n-17 -2\r\n3 4\r\n", "output": "324.722285390"}, {"input": "5 566\r\n-11 8\r\n2 -7\r\n7 0\r\n-7 -9\r\n-7 5\r\n", "output": "668.956254495"}, {"input": "7 495\r\n-10 -13\r\n-9 -5\r\n4 9\r\n8 13\r\n-4 2\r\n2 10\r\n-18 15\r\n", "output": "789.212495576"}, {"input": "10 958\r\n7 13\r\n20 19\r\n12 -7\r\n10 -10\r\n-13 -15\r\n-10 -7\r\n20 -5\r\n-11 19\r\n-7 3\r\n-4 18\r\n", "output": "3415.618464093"}, {"input": "13 445\r\n-15 16\r\n-8 -14\r\n8 7\r\n4 15\r\n8 -13\r\n15 -11\r\n-12 -4\r\n2 -13\r\n-5 0\r\n-20 -14\r\n-8 -7\r\n-10 -18\r\n18 -5\r\n", "output": "2113.552527680"}, {"input": "18 388\r\n11 -8\r\n13 10\r\n18 -17\r\n-15 3\r\n-13 -15\r\n20 -7\r\n1 -10\r\n-13 -12\r\n-12 -15\r\n-17 -8\r\n1 -2\r\n3 -20\r\n-8 -9\r\n15 -13\r\n-19 -6\r\n17 3\r\n-17 2\r\n6 6\r\n", "output": "2999.497312668"}, {"input": "25 258\r\n-5 -3\r\n-18 -14\r\n12 3\r\n6 11\r\n4 2\r\n-19 -3\r\n19 -7\r\n-15 19\r\n-19 -12\r\n-11 -10\r\n-5 17\r\n10 15\r\n-4 1\r\n-3 -20\r\n6 16\r\n18 -19\r\n11 -19\r\n-17 10\r\n-17 17\r\n-2 -17\r\n-3 -9\r\n18 13\r\n14 8\r\n-2 -5\r\n-11 4\r\n", "output": "2797.756635934"}, {"input": "29 848\r\n11 -10\r\n-19 1\r\n18 18\r\n19 -19\r\n0 -5\r\n16 10\r\n-20 -14\r\n7 15\r\n6 8\r\n-15 -16\r\n9 3\r\n16 -20\r\n-12 12\r\n18 -1\r\n-11 14\r\n18 10\r\n11 -20\r\n-20 -16\r\n-1 11\r\n13 10\r\n-6 13\r\n-7 -10\r\n-11 -10\r\n-10 3\r\n15 -13\r\n-4 11\r\n-13 -11\r\n-11 -17\r\n11 -5\r\n", "output": "12766.080247922"}]
| false |
stdio
|
import sys
import math
def read_points(n, input_file):
points = []
for _ in range(n):
x, y = map(int, input_file.readline().split())
points.append((x, y))
return points
def compute_total_time(n, k, points):
total_length = 0.0
for i in range(1, n):
x0, y0 = points[i-1]
x1, y1 = points[i]
dx = x1 - x0
dy = y1 - y0
total_length += math.hypot(dx, dy)
return (total_length * k) / 50.0
def main(input_path, output_path, submission_path):
with open(input_path) as f_input:
n, k = map(int, f_input.readline().split())
points = read_points(n, f_input)
correct = compute_total_time(n, k, points)
with open(submission_path) as f_sub:
submission_line = f_sub.readline().strip()
try:
submission = float(submission_line)
except:
print(0)
return
absolute_error = abs(submission - correct)
if correct == 0:
is_correct = absolute_error <= 1e-6
else:
relative_error = absolute_error / correct
is_correct = absolute_error <= 1e-6 or relative_error <= 1e-6
print(100 if is_correct else 0)
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 |
526/D
|
526
|
D
|
Python 3
|
TESTS
| 0 | 15 | 0 |
193252335
|
# LUOGU_RID: 101971216
print("0")
| 57 | 140 | 22,528,000 |
223321082
|
def prefix_function(s: str):
n = len(s)
pi = [0] * n
k = 0
for i in range(1, n):
while k > 0 and s[i] != s[k]:
k = pi[k - 1]
if s[i] == s[k]:
k += 1
pi[i] = k
return pi
def z_function(s: str):
n = len(s)
z = [0] * n
l, r = 0, 0
for i in range(1, n):
z[i] = 0 if i >= r else min(r - i, z[i - l])
while i + z[i] < n and s[z[i]] == s[i + z[i]]:
z[i] += 1
if i + z[i] > r:
l, r = i, i + z[i]
return z
n, k = map(int, input().split())
s = input()
pi = prefix_function(s)
ans = [0] * n
for i in range(n):
L = i + 1
T = L - pi[i]
have, remain = divmod(L, T)
# print(L, T, have, remain)
len_ab = have // k * T
len_a = (have % k) * T + remain
if len_ab >= len_a:
ans[i] = 1
print("".join(map(str, ans)))
|
ZeptoLab Code Rush 2015
|
CF
| 2,015 | 1 | 256 |
Om Nom and Necklace
|
One day Om Nom found a thread with n beads of different colors. He decided to cut the first several beads from this thread to make a bead necklace and present it to his girlfriend Om Nelly.
Om Nom knows that his girlfriend loves beautiful patterns. That's why he wants the beads on the necklace to form a regular pattern. A sequence of beads S is regular if it can be represented as S = A + B + A + B + A + ... + A + B + A, where A and B are some bead sequences, " + " is the concatenation of sequences, there are exactly 2k + 1 summands in this sum, among which there are k + 1 "A" summands and k "B" summands that follow in alternating order. Om Nelly knows that her friend is an eager mathematician, so she doesn't mind if A or B is an empty sequence.
Help Om Nom determine in which ways he can cut off the first several beads from the found thread (at least one; probably, all) so that they form a regular pattern. When Om Nom cuts off the beads, he doesn't change their order.
|
The first line contains two integers n, k (1 ≤ n, k ≤ 1 000 000) — the number of beads on the thread that Om Nom found and number k from the definition of the regular sequence above.
The second line contains the sequence of n lowercase Latin letters that represent the colors of the beads. Each color corresponds to a single letter.
|
Print a string consisting of n zeroes and ones. Position i (1 ≤ i ≤ n) must contain either number one if the first i beads on the thread form a regular sequence, or a zero otherwise.
| null |
In the first sample test a regular sequence is both a sequence of the first 6 beads (we can take A = "", B = "bca"), and a sequence of the first 7 beads (we can take A = "b", B = "ca").
In the second sample test, for example, a sequence of the first 13 beads is regular, if we take A = "aba", B = "ba".
|
[{"input": "7 2\nbcabcab", "output": "0000011"}, {"input": "21 2\nababaababaababaababaa", "output": "000110000111111000011"}]
| 2,200 |
["hashing", "string suffix structures", "strings"]
| 57 |
[{"input": "7 2\r\nbcabcab\r\n", "output": "0000011"}, {"input": "21 2\r\nababaababaababaababaa\r\n", "output": "000110000111111000011"}, {"input": "321 2\r\nabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaabaaba\r\n", "output": "000001111001111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"}, {"input": "153 2\r\nabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaa\r\n", "output": "000000011111000111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"}, {"input": "2 1\r\nab\r\n", "output": "11"}, {"input": "2 2\r\nab\r\n", "output": "00"}, {"input": "2 3\r\nab\r\n", "output": "00"}, {"input": "5 6\r\naaaaa\r\n", "output": "00000"}, {"input": "5 5\r\naaaaa\r\n", "output": "00001"}, {"input": "5 4\r\naaaaa\r\n", "output": "00011"}, {"input": "1 1\r\na\r\n", "output": "1"}, {"input": "1 2\r\na\r\n", "output": "0"}, {"input": "1 42\r\na\r\n", "output": "0"}]
| false |
stdio
| null | true |
729/D
|
729
|
D
|
PyPy 3
|
TESTS
| 4 | 140 | 409,600 |
56168221
|
'''input
5 1 2 1
00100
'''
from sys import stdin
import collections
def get_working(string):
aux = []
first = None
if string[0] == 1:
pass
else:
first = -1
for i in range(len(string)):
if string[i] == '1':
if first == None:
first = i
elif first != None:
aux.append([first, i, (i - first - 1)])
first = i
if first != None:
aux.append([first, len(string), (len(string) - first) - 1])
aux = sorted(aux, key = lambda x:x[2])
return aux
# main starts
n, a, b, k = list(map(int, stdin.readline().split()))
string = list(stdin.readline().strip())
ans = 0
working = get_working(string)
# print(working)
current = a
for i in working:
start, end, gap = i
cap = gap//b
mark = start + 1
while mark < end:
if mark - start == cap * b:
ans += 1
string[mark] = 'b'
mark += 1
print(ans)
for i in range(n):
if string[i] == 'b':
print(i + 1, end = ' ')
| 21 | 171 | 14,643,200 |
22697999
|
n,a,b,k=map(int,input().split())
A=input()
B=A.split('1')
C=[]
l=1
for i in B:
if len(i)>=b:
for j in range(b-1,len(i),b):
C.append(j+l)
l+=len(i)+1
C=C[:len(C)-a+1]
print(len(C))
print(' '.join(list(map(str,C))))
|
Technocup 2017 - Elimination Round 2
|
CF
| 2,016 | 1 | 256 |
Sea Battle
|
Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of b consecutive cells. No cell can be part of two ships, however, the ships can touch each other.
Galya doesn't know the ships location. She can shoot to some cells and after each shot she is told if that cell was a part of some ship (this case is called "hit") or not (this case is called "miss").
Galya has already made k shots, all of them were misses.
Your task is to calculate the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.
It is guaranteed that there is at least one valid ships placement.
|
The first line contains four positive integers n, a, b, k (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ n, 0 ≤ k ≤ n - 1) — the length of the grid, the number of ships on the grid, the length of each ship and the number of shots Galya has already made.
The second line contains a string of length n, consisting of zeros and ones. If the i-th character is one, Galya has already made a shot to this cell. Otherwise, she hasn't. It is guaranteed that there are exactly k ones in this string.
|
In the first line print the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.
In the second line print the cells Galya should shoot at.
Each cell should be printed exactly once. You can print the cells in arbitrary order. The cells are numbered from 1 to n, starting from the left.
If there are multiple answers, you can print any of them.
| null |
There is one ship in the first sample. It can be either to the left or to the right from the shot Galya has already made (the "1" character). So, it is necessary to make two shots: one at the left part, and one at the right part.
|
[{"input": "5 1 2 1\n00100", "output": "2\n4 2"}, {"input": "13 3 2 3\n1000000010001", "output": "2\n7 11"}]
| 1,700 |
["constructive algorithms", "greedy", "math"]
| 21 |
[{"input": "5 1 2 1\r\n00100\r\n", "output": "2\r\n2 5 \r\n"}, {"input": "13 3 2 3\r\n1000000010001\r\n", "output": "2\r\n3 5 \r\n"}, {"input": "1 1 1 0\r\n0\r\n", "output": "1\r\n1 \r\n"}, {"input": "2 2 1 0\r\n00\r\n", "output": "1\r\n1 \r\n"}, {"input": "5 4 1 0\r\n00000\r\n", "output": "2\r\n1 2 \r\n"}, {"input": "10 2 2 0\r\n0000000000\r\n", "output": "4\r\n2 4 6 8 \r\n"}, {"input": "20 1 3 5\r\n01001010000000010010\r\n", "output": "2\r\n10 13 \r\n"}, {"input": "100 17 4 11\r\n0100000100000000000000001000000000010001100000000000101000000000000000000000001000001000010000000000\r\n", "output": "2\r\n6 12 \r\n"}]
| false |
stdio
| null | true |
241/A
|
241
|
A
|
PyPy 3
|
TESTS
| 2 | 278 | 0 |
78814214
|
import math
m,k = map(int,input().split())
d = list(map(int,input().split()))
f = list(map(int,input().split()))
h = 0
fuel = 0
for i in range(0,m):
fuel = fuel + f[i]
if(fuel < d[i]):
h = h + math.ceil((d[i]-f[i])/f[i])*k
fuel = fuel + math.ceil((d[i]-f[i])/f[i])*k*f[i]
h = h + d[i]
fuel = fuel - d[i]
print(int(h))
| 52 | 248 | 0 |
41983593
|
m,k=map(int,input().split())
D=list(map(int,input().split()))
S=list(map(int,input().split()))
ans=0
fuel=S[0]
maxx=S[0]
for i in range(m):
if(D[i]>fuel):
x=D[i]-fuel
y=x//maxx
if(x%maxx>0):
y+=1
ans+=y*k
fuel+=maxx*y
fuel-=D[i]
ans+=D[i]
if(i+1==m):
break
fuel+=S[i+1]
maxx=max(S[i+1],maxx)
else:
fuel-=D[i]
ans+=D[i]
if(i+1==m):
break
fuel+=S[i+1]
maxx=max(S[i+1],maxx)
print(ans)
|
Bayan 2012-2013 Elimination Round (ACM ICPC Rules, English statements)
|
ICPC
| 2,012 | 2 | 256 |
Old Peykan
|
There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city ci to city ci + 1 and is di kilometers long.
The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.
Each city ci (except for the last city cn) has a supply of si liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.
Initially (at time zero) the Old Peykan is at city c1 and s1 liters of fuel is transferred to it's empty tank from c1's supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.
Find the minimum time the Old Peykan needs to reach city cn.
|
The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.
The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).
|
In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from city c1.
| null |
In the second sample above, the Old Peykan stays in c1 for 3 hours.
|
[{"input": "4 6\n1 2 5 2\n2 3 3 4", "output": "10"}, {"input": "2 3\n5 6\n5 5", "output": "14"}]
| 1,300 |
["greedy"]
| 52 |
[{"input": "4 6\r\n1 2 5 2\r\n2 3 3 4\r\n", "output": "10\r\n"}, {"input": "2 3\r\n5 6\r\n5 5\r\n", "output": "14\r\n"}, {"input": "24 3\r\n11 8 8 12 17 4 4 25 39 37 31 32 38 34 29 29 34 39 39 39 17 9 24 6\r\n3 5 4 3 3 3 4 3 4 3 3 3 3 4 3 3 4 3 4 3 3 3 3 3\r\n", "output": "862\r\n"}, {"input": "43 5\r\n6 7 15 12 15 7 22 33 38 15 7 23 31 21 26 41 25 14 26 33 5 28 22 6 35 17 19 32 41 27 20 25 5 32 37 19 40 9 25 22 10 24 9\r\n3 5 3 6 5 4 5 3 3 3 3 6 6 3 3 3 3 3 3 3 3 6 3 3 4 3 4 3 6 4 3 6 3 4 6 3 4 5 4 4 3 3 5\r\n", "output": "1566\r\n"}]
| false |
stdio
| null | true |
379/B
|
379
|
B
|
PyPy 3-64
|
TESTS
| 0 | 30 | 0 |
164186410
|
nofwallets = int(input())
given = input().split()
intlist = [int(a) for a in given]
startstr = ""
done = False
while not done:
done = True
for i in range(nofwallets-1):
if intlist[i] != 0:
startstr += 'P'
intlist[i] -= 1
done = False
startstr += 'R'
startstr += ('L' * (nofwallets-1))
print(startstr)
| 15 | 46 | 409,600 |
144587271
|
n=int(input())
lis=list(map(int,input().split()))
command='PRL'*lis[0]
for i in range(1,len(lis)):
command+=('R'+'LRP'*lis[i])
print(command)
|
Good Bye 2013
|
CF
| 2,013 | 1 | 256 |
New Year Present
|
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
|
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
|
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
| null | null |
[{"input": "2\n1 2", "output": "PRPLRP"}, {"input": "4\n0 2 0 2", "output": "RPRRPLLPLRRRP"}]
| 1,200 |
["constructive algorithms", "implementation"]
| 15 |
[{"input": "2\r\n1 2\r\n", "output": "PRPLRP"}, {"input": "4\r\n0 2 0 2\r\n", "output": "RPRRPLLPLRRRP"}, {"input": "10\r\n2 3 4 0 0 1 1 3 4 2\r\n", "output": "PRPRPRRRPRPRPRPRPLPLPLLLLLPLPLPRPRPRRRRRPRPRPLPLLLLLLPLL"}, {"input": "10\r\n0 0 0 0 0 0 0 0 1 0\r\n", "output": "RRRRRRRRPR"}, {"input": "5\r\n2 2 2 2 2\r\n", "output": "PRPRPRPRPLPLPLPLPRRRRP"}, {"input": "2\r\n6 0\r\n", "output": "PRLPRLPRLPRLPRLP"}]
| false |
stdio
|
import sys
def main(input_path, output_path, sub_output_path):
with open(input_path) as f:
n = int(f.readline())
a = list(map(int, f.readline().split()))
with open(sub_output_path) as f:
s = f.read().strip()
if len(s) < 1 or len(s) > 10**6:
print(0)
return
valid_chars = {'L', 'R', 'P'}
for c in s:
if c not in valid_chars:
print(0)
return
coins = [0] * n
current_pos = 1 # 1-based position
prev_was_p = False
for c in s:
if c == 'P':
if prev_was_p:
print(0)
return
coins[current_pos - 1] += 1
prev_was_p = True
else:
if c == 'L':
if current_pos == 1:
print(0)
return
current_pos -= 1
else: # 'R'
if current_pos == n:
print(0)
return
current_pos += 1
prev_was_p = False
if coins == a:
print(1)
else:
print(0)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
sub_output_path = sys.argv[3]
main(input_path, output_path, sub_output_path)
| true |
397/B
|
397
|
B
|
PyPy 3
|
TESTS
| 0 | 124 | 0 |
52266183
|
def possible(numbers):
return int(numbers[0])//int(numbers[1])!=0 and int(numbers[0])%int(numbers[1])//(int(numbers[0])//int(numbers[1]))<=int(numbers[2])-int(numbers[1]) and int(numbers[0])%int(numbers[1])%int(numbers[0])//int(numbers[1])+int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1]))<=(int(numbers[2])-int(numbers[1]))*(int(numbers[0])//int(numbers[1])-int(numbers[0])//int(numbers[2]))
for i in range(0,int(input())):
print('Yes' if possible(input().split(' ')) else 'no')
| 6 | 61 | 0 |
12258359
|
import sys; sys.setrecursionlimit(1000000)
def solve():
tests, = rv()
for test in range(tests):
n,l,r, = rv()
largestused = (n + r - 1) // r
totalnum = largestused * r
mostcansubtract = largestused * (r - l)
lowerbound = totalnum - mostcansubtract
if lowerbound <= n:
print("Yes")
else:
print("No")
def rv(): return map(int, input().split())
def rl(n): return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
|
Codeforces Round 232 (Div. 2)
|
CF
| 2,014 | 1 | 256 |
On Corruption and Numbers
|
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity — he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission — ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket — because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li ≤ x ≤ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
|
The first line contains the number of universities t, (1 ≤ t ≤ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 ≤ ni, li, ri ≤ 109; li ≤ ri).
|
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
| null |
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
|
[{"input": "2\n5 2 3\n6 4 5", "output": "Yes\nNo"}]
| null |
["constructive algorithms", "implementation", "math"]
| 6 |
[{"input": "2\r\n5 2 3\r\n6 4 5\r\n", "output": "Yes\r\nNo\r\n"}, {"input": "50\r\n69 6 6\r\n22 1 1\r\n23 3 3\r\n60 13 13\r\n13 3 3\r\n7 4 7\r\n6 1 1\r\n49 7 9\r\n68 8 8\r\n20 2 2\r\n34 1 1\r\n79 5 5\r\n22 1 1\r\n77 58 65\r\n10 3 3\r\n72 5 5\r\n47 1 1\r\n82 3 3\r\n92 8 8\r\n34 1 1\r\n42 9 10\r\n63 14 14\r\n10 3 3\r\n38 2 2\r\n80 6 6\r\n79 5 5\r\n53 5 5\r\n44 7 7\r\n85 2 2\r\n24 2 2\r\n57 3 3\r\n95 29 81\r\n77 6 6\r\n24 1 1\r\n33 4 4\r\n93 6 6\r\n55 22 28\r\n91 14 14\r\n7 1 1\r\n16 1 1\r\n20 3 3\r\n43 3 3\r\n53 3 3\r\n49 3 3\r\n52 5 5\r\n2 1 1\r\n60 5 5\r\n76 57 68\r\n67 3 3\r\n61 52 61\r\n", "output": "No\r\nYes\r\nNo\r\nNo\r\nNo\r\nYes\r\nYes\r\nYes\r\nNo\r\nYes\r\nYes\r\nNo\r\nYes\r\nNo\r\nNo\r\nNo\r\nYes\r\nNo\r\nNo\r\nYes\r\nNo\r\nNo\r\nNo\r\nYes\r\nNo\r\nNo\r\nNo\r\nNo\r\nNo\r\nYes\r\nYes\r\nYes\r\nNo\r\nYes\r\nNo\r\nNo\r\nYes\r\nNo\r\nYes\r\nYes\r\nNo\r\nNo\r\nNo\r\nNo\r\nNo\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\n"}]
| false |
stdio
| null | true |
600/E
|
600
|
E
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
199803690
|
import sys
input = lambda:sys.stdin.readline().strip()
n = int(input())
c = list(map(int,input().split()))
has = [[] for _ in range(n)]
for _ in range(n-1):
a,b = map(lambda x:int(x)-1,input().split())
has[a].append(b)
has[b].append(a)
ans = [0]*(n)
def dsu(x,p):
xx = {c[x]}
for y in has[x]:
if y == p:continue
yy = dsu(y,x)
if len(xx) < len(yy):
xx,yy = yy,xx
xx.update(yy)
ans[x] = sum(xx)
return xx
dsu(0,0)
print(*ans)
| 28 | 529 | 52,019,200 |
172847049
|
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def dsu_on_tree(N, G, Prop):
order = []
et_f = [-1]*N
prt = [-1]*N
que = [0]
used = [0]*N
used[0] = 1
while que:
v = que.pop()
for w in G[v]:
if used[w]:
continue
used[w] = 1
que.append(w)
prt[w] = v
et_f[v] = len(order)
order.append(v)
sz = [0]*N
heavy_cs = [-1]*N
l_root = [0]*N
group = [0]*N
label = [-1]*N
g_cur = 0
et_l = [-1]*N
for v in reversed(order):
m_sz = 0
h = -1
s = 1
right = -1
for w in G[v]:
if prt[v] == w:
continue
if m_sz < sz[w]:
m_sz = sz[w]
h = w
if right == -1:
right = w
s += sz[w]
if h != -1:
group[v] = gh = group[h]
m_lb = label[gh]
for w in G[v]:
if prt[v] == w or h == w:
continue
l_root[w] = 1
gw = label[group[w]]
if m_lb < gw+1:
m_lb = gw+1
label[gh] = m_lb
else:
group[v] = g_cur
label[g_cur] = 0
g_cur += 1
heavy_cs[v] = h
sz[v] = s
if right != -1:
et_l[v] = et_l[right]
else:
et_l[v] = et_f[v]+1
ps = [Prop() for i in range(label[group[0]]+1)]
ans = [-1]*N
for v in reversed(order):
h = heavy_cs[v]
ph = ps[label[group[v]]]
for w in G[v]:
if h == w or prt[v] == w:
continue
for c in order[et_f[w]:et_l[w]]:
ph.add(c)
ph.add(v)
ans[v] = ph.get(v)
if l_root[v]:
ph.reset()
return ans
def solve():
N = int(readline())
*C, = map(int, readline().split())
G = [[] for i in range(N)]
for i in range(N-1):
x, y = map(int, readline().split())
G[x-1].append(y-1)
G[y-1].append(x-1)
class Prop:
def __init__(self):
self.ts = [0]*(N+1)
self.cs = [0]*(N+1)
self.mc = 0
self.us = set()
def reset(self):
for c in self.us:
self.ts[self.cs[c]] = 0
self.cs[c] = 0
self.mc = 0
self.us = set()
def add(self, v):
c = C[v]
self.us.add(c)
r = self.cs[c]
if r != 0:
self.ts[r] -= c
self.ts[r+1] += c
if self.mc < r+1:
self.mc = r+1
self.cs[c] += 1
def get(self, i):
return self.ts[self.mc]
ans = dsu_on_tree(N, G, Prop)
write(" ".join(map(str, ans)))
write("\n")
solve()
|
Educational Codeforces Round 2
|
ICPC
| 2,015 | 2 | 256 |
Lomsat gelral
|
You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.
Let's call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it's possible that two or more colours will be dominating in the subtree of some vertex.
The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.
For each vertex v find the sum of all dominating colours in the subtree of vertex v.
|
The first line contains integer n (1 ≤ n ≤ 105) — the number of vertices in the tree.
The second line contains n integers ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.
Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the edge of the tree. The first vertex is the root of the tree.
|
Print n integers — the sums of dominating colours for each vertex.
| null | null |
[{"input": "4\n1 2 3 4\n1 2\n2 3\n2 4", "output": "10 9 3 4"}, {"input": "15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13", "output": "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3"}]
| 2,300 |
["data structures", "dfs and similar", "dsu", "trees"]
| 28 |
[{"input": "4\r\n1 2 3 4\r\n1 2\r\n2 3\r\n2 4\r\n", "output": "10 9 3 4\r\n"}, {"input": "15\r\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\r\n1 2\r\n1 3\r\n1 4\r\n1 14\r\n1 15\r\n2 5\r\n2 6\r\n2 7\r\n3 8\r\n3 9\r\n3 10\r\n4 11\r\n4 12\r\n4 13\r\n", "output": "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3\r\n"}]
| false |
stdio
| null | true |
241/A
|
241
|
A
|
Python 3
|
TESTS
| 2 | 216 | 0 |
58515265
|
import sys
import math
header = sys.stdin.readline()
while header:
m, k = [int(x) for x in header.split()]
distances = [int(d) for d in sys.stdin.readline().split()]
supplies = [int(s) for s in sys.stdin.readline().split()]
gas = 0
time = 0
for i in range(len(distances)):
d = distances[i]
s = supplies[i]
gas += s
if gas >= d:
time += d
gas -= d
else:
idle_time = math.ceil((d-gas)/s)*k
time += idle_time
gas += s*idle_time/k
time += d
gas -= d
print(time)
header = sys.stdin.readline()
| 52 | 248 | 0 |
54079119
|
estradas,k = [int(i) for i in input().split()]
d = [int(i) for i in input().split()]
s = [int(i) for i in input().split()]
fuel = 0
time = 0
currMax = 0
for i in range(estradas):
fuel+=s[i]
currMax = max(currMax,s[i])
while(fuel<d[i]):
time+=k
fuel+=currMax
fuel-=d[i]
time+=d[i]
print(time)
|
Bayan 2012-2013 Elimination Round (ACM ICPC Rules, English statements)
|
ICPC
| 2,012 | 2 | 256 |
Old Peykan
|
There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city ci to city ci + 1 and is di kilometers long.
The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.
Each city ci (except for the last city cn) has a supply of si liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.
Initially (at time zero) the Old Peykan is at city c1 and s1 liters of fuel is transferred to it's empty tank from c1's supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.
Find the minimum time the Old Peykan needs to reach city cn.
|
The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.
The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).
|
In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from city c1.
| null |
In the second sample above, the Old Peykan stays in c1 for 3 hours.
|
[{"input": "4 6\n1 2 5 2\n2 3 3 4", "output": "10"}, {"input": "2 3\n5 6\n5 5", "output": "14"}]
| 1,300 |
["greedy"]
| 52 |
[{"input": "4 6\r\n1 2 5 2\r\n2 3 3 4\r\n", "output": "10\r\n"}, {"input": "2 3\r\n5 6\r\n5 5\r\n", "output": "14\r\n"}, {"input": "24 3\r\n11 8 8 12 17 4 4 25 39 37 31 32 38 34 29 29 34 39 39 39 17 9 24 6\r\n3 5 4 3 3 3 4 3 4 3 3 3 3 4 3 3 4 3 4 3 3 3 3 3\r\n", "output": "862\r\n"}, {"input": "43 5\r\n6 7 15 12 15 7 22 33 38 15 7 23 31 21 26 41 25 14 26 33 5 28 22 6 35 17 19 32 41 27 20 25 5 32 37 19 40 9 25 22 10 24 9\r\n3 5 3 6 5 4 5 3 3 3 3 6 6 3 3 3 3 3 3 3 3 6 3 3 4 3 4 3 6 4 3 6 3 4 6 3 4 5 4 4 3 3 5\r\n", "output": "1566\r\n"}]
| false |
stdio
| null | true |
241/A
|
241
|
A
|
PyPy 3-64
|
TESTS
| 2 | 92 | 0 |
182710115
|
m_k = input().split(" ")
m = int(m_k[0])
k = int(m_k[1])
distances = input().split(" ")
supplys = input().split(" ")
benzin = 0
full_time = 0
for i in range(m):
benzin += int(supplys[i])
while benzin < int(distances[i]):
benzin += int(supplys[i])
full_time += k
benzin -= int(distances[i])
full_time += int(distances[i])
print(full_time)
| 52 | 248 | 307,200 |
79046881
|
from math import ceil
X = list(map(int, input().split()))
Distance = list(map(int, input().split()))
Fuel = list(map(int, input().split()))
Tank, Time, Max = 0, 0, 0
for i in range(X[0]):
Tank += Fuel[i]
Max = max(Max, Fuel[i])
Stay = max(0, ceil((Distance[i] - Tank) / Max))
Time += Stay * X[1] + Distance[i]
Tank += Stay * Max - Distance[i]
print(Time)
# Hope the best for Ravens
# Never give up
|
Bayan 2012-2013 Elimination Round (ACM ICPC Rules, English statements)
|
ICPC
| 2,012 | 2 | 256 |
Old Peykan
|
There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we'll denote them from left to right as c1, c2, ..., cn. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city ci to city ci + 1 and is di kilometers long.
The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.
Each city ci (except for the last city cn) has a supply of si liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.
Initially (at time zero) the Old Peykan is at city c1 and s1 liters of fuel is transferred to it's empty tank from c1's supply. The Old Peykan's fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.
Find the minimum time the Old Peykan needs to reach city cn.
|
The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.
The next line contains m space-separated integers d1, d2, ..., dm (1 ≤ di ≤ 1000) and the following line contains m space-separated integers s1, s2, ..., sm (1 ≤ si ≤ 1000).
|
In the only line of the output print a single integer — the minimum time required for The Old Peykan to reach city cn from city c1.
| null |
In the second sample above, the Old Peykan stays in c1 for 3 hours.
|
[{"input": "4 6\n1 2 5 2\n2 3 3 4", "output": "10"}, {"input": "2 3\n5 6\n5 5", "output": "14"}]
| 1,300 |
["greedy"]
| 52 |
[{"input": "4 6\r\n1 2 5 2\r\n2 3 3 4\r\n", "output": "10\r\n"}, {"input": "2 3\r\n5 6\r\n5 5\r\n", "output": "14\r\n"}, {"input": "24 3\r\n11 8 8 12 17 4 4 25 39 37 31 32 38 34 29 29 34 39 39 39 17 9 24 6\r\n3 5 4 3 3 3 4 3 4 3 3 3 3 4 3 3 4 3 4 3 3 3 3 3\r\n", "output": "862\r\n"}, {"input": "43 5\r\n6 7 15 12 15 7 22 33 38 15 7 23 31 21 26 41 25 14 26 33 5 28 22 6 35 17 19 32 41 27 20 25 5 32 37 19 40 9 25 22 10 24 9\r\n3 5 3 6 5 4 5 3 3 3 3 6 6 3 3 3 3 3 3 3 3 6 3 3 4 3 4 3 6 4 3 6 3 4 6 3 4 5 4 4 3 3 5\r\n", "output": "1566\r\n"}]
| false |
stdio
| null | true |
19/C
|
19
|
C
|
Python 3
|
TESTS
| 0 | 186 | 0 |
41975374
|
def get_pow_list(x, exp, mod):
res = [1]
for i in range(exp):
res.append((res[-1] * x) % mod)
return res
MOD = 2 ** 61 - 1
n = int(input())
orig_vals = list(map(int, input().split()))
sv = list(set(orig_vals))
sv_inv = {sv[i]: i + 1 for i in range(len(sv))}
vals = [sv_inv[x] for x in orig_vals]
groups = dict()
for i in range(n):
groups.setdefault(vals[i], [])
groups[vals[i]].append(i)
pows = [get_pow_list(x, n, MOD) for x in [2, len(sv) + 1]]
logs = [-1]
while len(logs) - 1 < n:
logs.extend([logs[-1] + 1] * len(logs))
hashes = [[0] * (4 * n) for i in range(logs[n] + 1)]
for i in range(n):
hashes[0][i] = vals[i]
for l in range(1, len(hashes)):
e = pows[0][l - 1]
for i in range(n):
hashes[l][i] = hashes[l - 1][i] * pows[1][e] + hashes[l - 1][i + e]
def get_hash(h, p, l, mod):
if l == 0:
return 0
elif ((l - 1) & l) == 0:
return h[logs[l]][p]
else:
nl = pows[0][logs[l]]
return get_hash(h, p, nl, mod) * pows[1][l - nl] + get_hash(
h, p + nl, l - nl, mod
)
best = 0
for i in range(n - 1, -1, -1):
val = vals[i]
for jj in range(len(groups[val])):
j = groups[val][jj]
if i == j:
break
l = i - j
if get_hash(hashes, i, l, MOD) == get_hash(hashes, j, l, MOD):
best = max(best, j)
if best != 0:
break
res = orig_vals[best:]
print(len(res))
print(" ".join(map(str, res)))
| 70 | 1,216 | 14,131,200 |
174568804
|
import sys
n=int(input())
a=list(map(int,input().split()))
M=10**9+1
g={}
for i in range(n):
g[a[i]]=g.get(a[i],[])+[i]
p=[1]
for i in range(n):
p+=[hash(M*p[-1])]
h=[0]*(n+1)
for i in range(n):
h[i+1]=hash(h[i]*M+a[i])
gh=lambda k,l:hash(h[k+l]-h[k]*p[l])%sys.hash_info.modulus
i,t=0,0
while i < n:
for j in g[a[i]]:
if j <= i:
continue
w=j-i
if j+w<=n and gh(i,w)==gh(j,w):
i=j-1
t=max(t,j)
break
i+=1
r=a[t:]
print(len(r))
print(*r)
|
Codeforces Beta Round 19
|
ICPC
| 2,010 | 2 | 256 |
Deletion of Repeats
|
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
|
The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
|
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
| null | null |
[{"input": "6\n1 2 3 1 2 3", "output": "3\n1 2 3"}, {"input": "7\n4 5 6 5 6 7 7", "output": "1\n7"}]
| 2,200 |
["greedy", "hashing", "string suffix structures"]
| 70 |
[{"input": "6\r\n1 2 3 1 2 3\r\n", "output": "3\r\n1 2 3 \r\n"}, {"input": "7\r\n4 5 6 5 6 7 7\r\n", "output": "1\r\n7 \r\n"}, {"input": "10\r\n5 7 2 1 8 8 5 10 2 5\r\n", "output": "5\r\n8 5 10 2 5 \r\n"}, {"input": "10\r\n0 1 1 1 0 3 0 1 4 0\r\n", "output": "7\r\n1 0 3 0 1 4 0 \r\n"}, {"input": "10\r\n0 1 0 2 0 0 1 1 1 0\r\n", "output": "2\r\n1 0 \r\n"}, {"input": "30\r\n17 17 2 4 13 21 17 11 15 0 9 2 23 10 24 21 23 17 5 11 25 1 16 6 11 22 19 2 12 16\r\n", "output": "29\r\n17 2 4 13 21 17 11 15 0 9 2 23 10 24 21 23 17 5 11 25 1 16 6 11 22 19 2 12 16 \r\n"}, {"input": "100\r\n5 4 8 2 4 7 6 6 9 0 5 9 9 8 2 1 10 7 1 0 0 6 3 5 3 7 8 0 0 10 6 7 10 5 4 10 7 6 7 5 1 5 0 10 3 10 5 7 4 10 0 9 1 2 6 3 3 6 10 6 9 1 6 3 4 6 2 8 8 9 5 2 3 3 10 7 4 1 10 1 8 5 4 3 2 2 0 4 4 1 4 5 7 2 8 7 1 1 2 8\r\n", "output": "3\r\n1 2 8 \r\n"}, {"input": "10\r\n1 2 1 2 1 2 1 2 1 2\r\n", "output": "2\r\n1 2 \r\n"}, {"input": "10\r\n1 2 3 4 5 1 2 3 4 5\r\n", "output": "5\r\n1 2 3 4 5 \r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 1 1\r\n", "output": "1\r\n1 \r\n"}, {"input": "21\r\n16417014 805849548 385039296 16417014 805849548 385039296 16417014 805849548 385039296 16417014 805849548 385039296 16417014 805849548 385039296 16417014 805849548 385039296 16417014 805849548 385039296\r\n", "output": "3\r\n16417014 805849548 385039296 \r\n"}, {"input": "22\r\n823078040 389511796 683819000 823078040 389511796 683819000 823078040 389511796 683819000 823078040 389511796 683819000 823078040 389511796 683819000 823078040 389511796 683819000 823078040 389511796 683819000 823078040\r\n", "output": "4\r\n823078040 389511796 683819000 823078040 \r\n"}, {"input": "23\r\n482255418 973174044 835115058 482255418 973174044 835115058 482255418 973174044 835115058 482255418 973174044 835115058 482255418 973174044 835115058 482255418 973174044 835115058 482255418 973174044 835115058 482255418 973174044\r\n", "output": "5\r\n482255418 973174044 835115058 482255418 973174044 \r\n"}, {"input": "1\r\n0\r\n", "output": "1\r\n0 \r\n"}, {"input": "2\r\n1 2\r\n", "output": "2\r\n1 2 \r\n"}, {"input": "2\r\n1000000000 1000000000\r\n", "output": "1\r\n1000000000 \r\n"}, {"input": "3\r\n1000000000 1000000000 1000000000\r\n", "output": "1\r\n1000000000 \r\n"}, {"input": "4\r\n1000000000 1000000000 1000000000 1000000000\r\n", "output": "1\r\n1000000000 \r\n"}, {"input": "7\r\n1 2 3 1 2 3 1\r\n", "output": "4\r\n1 2 3 1 \r\n"}, {"input": "30\r\n0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2\r\n", "output": "1\r\n2 \r\n"}, {"input": "10\r\n0 0 0 0 0 0 0 0 0 0\r\n", "output": "1\r\n0 \r\n"}, {"input": "20\r\n0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1\r\n", "output": "1\r\n1 \r\n"}]
| false |
stdio
| null | true |
875/D
|
875
|
D
|
PyPy 3-64
|
TESTS
| 2 | 61 | 0 |
204069401
|
import sys
from collections import deque
input = sys.stdin.readline
def fg():
return int(input())
def fgh():
return [int(Xx) for Xx in input().split()]
n = fg()
a = fgh()
S = 30
step = [1]
for i in range(S):
step.append(step[-1] * 2)
d = [[] for i in range(S)]
for i in range(n):
for j in range(S):
if a[i] % step[j + 1] >= step[j]:
d[j].append(i)
Q = {}
for i in range(n):
if a[i] not in Q:
Q[a[i]] = []
Q[a[i]].append(i)
ans = 0
q = deque()
ans1 = [-1] * n
for i in range(n):
while q and a[q[-1]] <= a[i]:
q.pop()
if len(q):
ans1[i] = q[-1]
q.append(i)
q = deque()
ans4 = [n] * n
for i in range(n - 1, -1, -1):
while q and a[q[-1]] < a[i]:
q.pop()
if len(q):
ans4[i] = q[-1]
q.append(i)
#print(*ans1)
#print(*ans4)
for i in range(n):
l = 0
r = n - 1
for j in range(S):
if a[i] % step[j + 1] < step[j] and len(d[j]):
l1 = 0
r1 = len(d[j])
while r1 - l1 > 1:
m = (l1 + r1) // 2
if d[j][m] > i:
r1 = m
else:
l1 = m
if d[j][l1] < i:
l = max(l, d[j][l1])
else:
r = min(r, d[j][l1])
if r1 < len(d[j]):
r = min(r, d[j][r1])
#print(l, r)
#print(abs(i - ans1[i]) * abs(ans4[i] - i) - (i - l) * (r - i + 1))
ans += abs(i - ans1[i]) * abs(ans4[i] - i) - (i - l) * (r - i + 1)
print(ans)
| 78 | 514 | 127,283,200 |
226308676
|
from collections import defaultdict
def read_ints():
return map(int, input().split())
def read_int():
return int(input())
def yes_no(yes):
print('Yes' if yes else 'No')
return yes
def solve_test():
n = read_int()
a = list(read_ints())
def get_ans():
bit_size = 32
lefts = [[-1] * n for _ in range(bit_size)]
rights = [[n] * n for _ in range(bit_size)]
for i in range(n):
value = a[i]
for bit in range(bit_size):
if i > 0:
lefts[bit][i] = lefts[bit][i - 1]
has = value % 2 == 1
if has:
lefts[bit][i] = i
value //= 2
for i in range(n - 1, -1, -1):
value = a[i]
for bit in range(bit_size):
if i < n - 1:
rights[bit][i] = rights[bit][i + 1]
has = value % 2 == 1
if has:
rights[bit][i] = i
value //= 2
left_larger = [-1] * n
right_larger = [n] * n
stack = [-1]
for i in range(n):
value = a[i]
while len(stack) > 1:
top_value = a[stack[-1]]
if top_value < value:
stack.pop()
else:
break
left_larger[i] = stack[-1]
stack.append(i)
stack = [n]
for i in range(n - 1, -1, -1):
value = a[i]
while len(stack) > 1:
top_value = a[stack[-1]]
if top_value <= value:
stack.pop()
else:
break
right_larger[i] = stack[-1]
stack.append(i)
ans = 0
for i in range(n):
value = a[i]
start = left_larger[i] + 1
end = right_larger[i] - 1
max_left = start - 1
min_right = end + 1
for bit in range(bit_size):
has = value % 2 == 1
if not has:
max_left = max(max_left, lefts[bit][i])
min_right = min(min_right, rights[bit][i])
value //= 2
total = (end - i + 1) * (i - start + 1)
bad = (min_right - i) * (i - max_left)
# print(total, bad)
ans += total - bad
return ans
print(get_ans())
"""
5
3 2 1 6 5
4
3 3 3 3
"""
"""
8
0
"""
def solve():
t = 1
# t = read_int()
for _ in range(t):
solve_test()
if __name__ == '__main__':
solve()
|
Codeforces Round 441 (Div. 1, by Moscow Team Olympiad)
|
CF
| 2,017 | 1 | 512 |
High Cry
|
Disclaimer: there are lots of untranslateable puns in the Russian version of the statement, so there is one more reason for you to learn Russian :)
Rick and Morty like to go to the ridge High Cry for crying loudly — there is an extraordinary echo. Recently they discovered an interesting acoustic characteristic of this ridge: if Rick and Morty begin crying simultaneously from different mountains, their cry would be heard between these mountains up to the height equal the bitwise OR of mountains they've climbed and all the mountains between them.
Bitwise OR is a binary operation which is determined the following way. Consider representation of numbers x and y in binary numeric system (probably with leading zeroes) x = xk... x1x0 and y = yk... y1y0. Then z = x | y is defined following way: z = zk... z1z0, where zi = 1, if xi = 1 or yi = 1, and zi = 0 otherwise. In the other words, digit of bitwise OR of two numbers equals zero if and only if digits at corresponding positions is both numbers equals zero. For example bitwise OR of numbers 10 = 10102 and 9 = 10012 equals 11 = 10112. In programming languages C/C++/Java/Python this operation is defined as «|», and in Pascal as «or».
Help Rick and Morty calculate the number of ways they can select two mountains in such a way that if they start crying from these mountains their cry will be heard above these mountains and all mountains between them. More formally you should find number of pairs l and r (1 ≤ l < r ≤ n) such that bitwise OR of heights of all mountains between l and r (inclusive) is larger than the height of any mountain at this interval.
|
The first line contains integer n (1 ≤ n ≤ 200 000), the number of mountains in the ridge.
Second line contains n integers ai (0 ≤ ai ≤ 109), the heights of mountains in order they are located in the ridge.
|
Print the only integer, the number of ways to choose two different mountains.
| null |
In the first test case all the ways are pairs of mountains with the numbers (numbering from one):
(1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)
In the second test case there are no such pairs because for any pair of mountains the height of cry from them is 3, and this height is equal to the height of any mountain.
|
[{"input": "5\n3 2 1 6 5", "output": "8"}, {"input": "4\n3 3 3 3", "output": "0"}]
| 2,200 |
["binary search", "bitmasks", "combinatorics", "data structures", "divide and conquer"]
| 78 |
[{"input": "5\r\n3 2 1 6 5\r\n", "output": "8\r\n"}, {"input": "4\r\n3 3 3 3\r\n", "output": "0\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}, {"input": "1\r\n1000000000\r\n", "output": "0\r\n"}, {"input": "1\r\n6\r\n", "output": "0\r\n"}, {"input": "228\r\n1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 63 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 127 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 63 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 255 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 63 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 127 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 63 1 3 1 7\r\n", "output": "0\r\n"}, {"input": "50\r\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n", "output": "0\r\n"}, {"input": "81\r\n52673 19697 35512 34827 62387 60516 43450 22979 133 42838 16525 37792 12752 47765 45874 64082 14727 51748 56809 604 51751 59450 43797 31724 1024 9648 59503 53771 60868 38612 62867 57026 62297 15806 10549 7660 47983 30060 20829 46168 64832 18145 32300 53558 56554 33754 21953 58346 13894 6318 33563 63571 41705 49407 26794 51159 29011 43310 6847 11688 45129 2180 50406 12475 58021 58899 32867 15910 25819 33961 18759 64166 34472 57376 10903 16958 22656 14459 26900 33012 11615\r\n", "output": "3239\r\n"}, {"input": "69\r\n3 4 2 4 5 2 0 5 2 8 8 0 0 8 6 4 3 0 0 1 2 5 8 1 7 4 6 0 6 1 8 4 2 0 4 3 5 6 2 8 3 7 7 6 0 3 8 6 2 6 0 2 4 5 0 3 6 8 8 8 3 2 2 6 7 8 5 3 5\r\n", "output": "2262\r\n"}, {"input": "92\r\n7 1 1 2 5 8 3 4 1 3 0 7 5 2 0 7 2 7 3 2 7 3 7 7 7 0 8 2 0 7 4 7 6 7 3 6 2 0 2 4 1 6 6 4 5 7 3 2 8 6 7 8 8 2 4 2 7 2 2 4 7 5 4 3 4 4 1 2 2 1 4 4 1 8 0 7 8 0 8 6 7 3 2 6 7 2 7 3 1 0 3 6\r\n", "output": "3630\r\n"}, {"input": "79\r\n48952 12057 54091 43986 4825 65319 8888 63361 37922 29880 42348 16202 7107 33942 28418 5373 37969 36323 36644 8650 2951 22084 38174 65285 4123 19887 46516 40807 40062 20157 58343 52857 42065 28397 15571 29810 17508 34053 2732 26989 37441 53824 23112 13218 6696 46519 10848 37031 16381 32753 39768 8870 61745 57147 47588 1759 25424 29718 34450 31641 64681 59257 47088 36889 31207 23425 25712 41458 27960 49566 50455 10157 53577 34807 39258 31040 39873 10859 24672\r\n", "output": "3081\r\n"}, {"input": "80\r\n2 3 2 2 3 5 4 0 2 3 3 8 4 8 3 8 4 0 0 8 1 7 3 0 7 2 8 5 5 3 0 0 2 7 4 1 6 0 6 2 5 3 0 4 8 6 7 0 3 2 3 3 8 5 6 5 5 6 3 4 0 5 8 3 6 3 6 8 1 7 8 8 3 0 3 8 0 4 2 3\r\n", "output": "2997\r\n"}, {"input": "74\r\n63528 64641 32712 5228 59859 45909 4464 57592 27841 17687 62064 19286 40682 40598 63681 18441 53374 38527 16119 35588 42691 4015 20251 13679 50493 37149 34328 37977 24309 8750 54309 44091 12187 21917 24216 31326 40587 52208 19362 1642 13268 6596 10380 4937 37224 25970 59246 63483 20707 47702 57607 26046 30796 32636 7168 8816 11892 12934 53913 704 61887 65147 52243 14676 20993 33174 40778 23764 37017 5206 22521 55323 36803 9943\r\n", "output": "2691\r\n"}, {"input": "47\r\n4 4 3 1 0 1 2 8 6 3 1 5 6 5 4 5 3 8 4 8 7 6 8 1 4 8 1 5 7 4 8 7 8 7 5 6 7 5 5 5 6 5 3 0 2 5 6\r\n", "output": "1010\r\n"}, {"input": "53\r\n1 2 0 1 0 1 1 1 1 2 0 2 1 0 2 2 1 1 2 0 0 2 1 2 2 1 1 0 0 1 0 1 2 2 1 1 1 1 1 1 2 1 0 1 2 1 0 0 0 1 2 0 2\r\n", "output": "1288\r\n"}, {"input": "85\r\n4 4 4 4 2 1 2 0 0 3 1 0 4 3 2 2 3 4 1 0 0 0 0 2 1 1 1 1 0 1 4 2 2 1 0 4 4 1 4 0 3 2 3 4 0 4 3 0 3 1 0 1 3 1 2 0 2 3 1 1 2 4 0 4 1 1 1 3 3 4 3 1 0 3 0 0 0 4 2 3 1 1 4 0 0\r\n", "output": "3346\r\n"}, {"input": "100\r\n1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 63 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 127 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 31 1 3 1 7 1 3 1 15 1 3 1 7 1 3 1 63 1 3 1 7\r\n", "output": "0\r\n"}, {"input": "100\r\n1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 16 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 32 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 16 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 64 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 16 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1 32 1 2 1 4\r\n", "output": "4950\r\n"}, {"input": "100\r\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n", "output": "0\r\n"}, {"input": "100\r\n1 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 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\r\n", "output": "0\r\n"}, {"input": "100\r\n1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512 1 2 4 8 16 32 64 128 256 512\r\n", "output": "4950\r\n"}, {"input": "100\r\n1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 8 8 8 8 8 8 8 8 8 8 16 16 16 16 16 16 16 16 16 16 32 32 32 32 32 32 32 32 32 32 64 64 64 64 64 64 64 64 64 64 128 128 128 128 128 128 128 128 128 128 256 256 256 256 256 256 256 256 256 256 512 512 512 512 512 512 512 512 512 512\r\n", "output": "4500\r\n"}]
| false |
stdio
| null | true |
314/A
|
314
|
A
|
PyPy 3-64
|
TESTS
| 3 | 122 | 0 |
183327872
|
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
w = list(map(int, input().split()))
ew = set()
t = 0
while 1:
d = []
c = 0
for i in range(n):
x = c - i * w[i] * (n-i-1)
c += i * w[i]
if x >= k:
d.append(w[i])
else:
ew.add(i+t+1)
if len(d) == n:
break
else:
t += n - len(d)
w = d.copy()
n = len(d)
for i in ew:
print(i)
| 22 | 404 | 26,931,200 |
211355857
|
import sys
input = lambda: sys.stdin.readline().rstrip()
N,K = map(int, input().split())
A = list(map(int, input().split()))
pre,cnt=0,0
for i in range(N):
a = A[i]
b = pre-(N-i-1)*a*(i-cnt)
if b<K:
#print(pre,cnt,i,b)
print(i+1)
cnt+=1
else:
pre+=(i-cnt)*a
|
Codeforces Round 187 (Div. 1)
|
CF
| 2,013 | 2 | 256 |
Sereja and Contest
|
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula $$d_i = \sum_{j=1}^{i-1}(a_j \cdot (j - 1) - (n - i) \cdot a_i)$$.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
|
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
|
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
| null |
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
|
[{"input": "5 0\n5 3 4 1 2", "output": "2\n3\n4"}, {"input": "10 -10\n5 5 1 7 5 1 2 4 9 2", "output": "2\n4\n5\n7\n8\n9"}]
| 1,600 |
["implementation"]
| 22 |
[{"input": "5 0\r\n5 3 4 1 2\r\n", "output": "2\r\n3\r\n4\r\n"}, {"input": "10 -10\r\n5 5 1 7 5 1 2 4 9 2\r\n", "output": "2\r\n4\r\n5\r\n7\r\n8\r\n9\r\n"}]
| false |
stdio
| null | true |
314/A
|
314
|
A
|
PyPy 3
|
TESTS
| 3 | 184 | 0 |
106169062
|
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
pos=[0]
neg=[]
for i in range(1,n):
pos+=[pos[-1]+a[i]*i]
removed=0;sm=0
for i in range(1,n):
positive_term=pos[i-1]-sm
negative_term=(i-removed)*a[i]*(n-i-1)
if (positive_term-negative_term)<k:
sm+=i*a[i]
removed+=1
stdout.write(str(i+1)+'\n')
| 22 | 468 | 33,792,000 |
183439329
|
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
w = list(map(int, input().split()))
ew = set()
c = 0
t = 0
for i in range(n):
x = (i-t)*w[i]*(n-i-1)
if c - x < k:
t += 1
ew.add(i+1)
else:
c += (i-t)*w[i]
for i in ew:
print(i)
|
Codeforces Round 187 (Div. 1)
|
CF
| 2,013 | 2 | 256 |
Sereja and Contest
|
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula $$d_i = \sum_{j=1}^{i-1}(a_j \cdot (j - 1) - (n - i) \cdot a_i)$$.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
|
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
|
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
| null |
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
|
[{"input": "5 0\n5 3 4 1 2", "output": "2\n3\n4"}, {"input": "10 -10\n5 5 1 7 5 1 2 4 9 2", "output": "2\n4\n5\n7\n8\n9"}]
| 1,600 |
["implementation"]
| 22 |
[{"input": "5 0\r\n5 3 4 1 2\r\n", "output": "2\r\n3\r\n4\r\n"}, {"input": "10 -10\r\n5 5 1 7 5 1 2 4 9 2\r\n", "output": "2\r\n4\r\n5\r\n7\r\n8\r\n9\r\n"}]
| false |
stdio
| null | true |
167/A
|
167
|
A
|
Python 3
|
TESTS
| 0 | 93 | 6,656,000 |
88151890
|
n, a, d = map(int, input().split())
r = []
for i in range(0, n):
t, v = map(int, input().split())
current = d / v + t
if i == 0:
r.append(current)
else:
last = r[i - 1]
if current < last:
r.append(last)
for x in r:
print(x)
| 57 | 312 | 11,673,600 |
211354808
|
import sys
input = lambda: sys.stdin.readline().rstrip()
import math
N,A,D = map(int, input().split())
pre = 0
for _ in range(N):
t,v = map(int, input().split())
# 到达最高速度需要的时间
t1 = v/A
# 加速时间段行走距离(已知时间和加速度求距离)
d1 = t1*(t1*A/2)
# 能加到全速
if d1<=D:
# 剩下距离需要的时间
t2 = (D-d1)/v
tt=t+t1+t2
else: # 不能加到全速
# 已知距离加速度计算时间
t1 = math.pow(D*2/A, 0.5)
tt = t+t1
tt = max(tt, pre)
pre = tt
print(tt)
|
Codeforces Round 114 (Div. 1)
|
CF
| 2,012 | 1 | 256 |
Wizards and Trolleybuses
|
In some country live wizards. They love to ride trolleybuses.
A city in this country has a trolleybus depot with n trolleybuses. Every day the trolleybuses leave the depot, one by one and go to the final station. The final station is at a distance of d meters from the depot. We know for the i-th trolleybus that it leaves at the moment of time ti seconds, can go at a speed of no greater than vi meters per second, and accelerate with an acceleration no greater than a meters per second squared. A trolleybus can decelerate as quickly as you want (magic!). It can change its acceleration as fast as you want, as well. Note that the maximum acceleration is the same for all trolleys.
Despite the magic the trolleys are still powered by an electric circuit and cannot overtake each other (the wires are to blame, of course). If a trolleybus catches up with another one, they go together one right after the other until they arrive at the final station. Also, the drivers are driving so as to arrive at the final station as quickly as possible.
You, as head of the trolleybuses' fans' club, are to determine for each trolley the minimum time by which it can reach the final station. At the time of arrival at the destination station the trolleybus does not necessarily have zero speed. When a trolley is leaving the depot, its speed is considered equal to zero. From the point of view of physics, the trolleybuses can be considered as material points, and also we should ignore the impact on the speed of a trolley bus by everything, except for the acceleration and deceleration provided by the engine.
|
The first input line contains three space-separated integers n, a, d (1 ≤ n ≤ 105, 1 ≤ a, d ≤ 106) — the number of trolleybuses, their maximum acceleration and the distance from the depot to the final station, correspondingly.
Next n lines contain pairs of integers ti vi (0 ≤ t1 < t2... < tn - 1 < tn ≤ 106, 1 ≤ vi ≤ 106) — the time when the i-th trolleybus leaves the depot and its maximum speed, correspondingly. The numbers in the lines are separated by spaces.
|
For each trolleybus print a single line the time it arrives to the final station. Print the times for the trolleybuses in the order in which the trolleybuses are given in the input. The answer will be accepted if the absolute or relative error doesn't exceed 10 - 4.
| null |
In the first sample the second trolleybus will catch up with the first one, that will happen at distance 510.5 meters from the depot. The trolleybuses will go the remaining 9489.5 meters together at speed 10 meters per second. As a result, both trolleybuses will arrive to the final station by the moment of time 1000.5 seconds. The third trolleybus will not catch up with them. It will arrive to the final station by the moment of time 11000.05 seconds.
|
[{"input": "3 10 10000\n0 10\n5 11\n1000 1", "output": "1000.5000000000\n1000.5000000000\n11000.0500000000"}, {"input": "1 2 26\n28 29", "output": "33.0990195136"}]
| 1,600 |
["implementation", "math"]
| 57 |
[{"input": "3 10 10000\r\n0 10\r\n5 11\r\n1000 1\r\n", "output": "1000.5000000000\r\n1000.5000000000\r\n11000.0500000000\r\n"}, {"input": "1 2 26\r\n28 29\r\n", "output": "33.0990195136\r\n"}, {"input": "7 8 3\r\n1 3\r\n5 26\r\n7 3\r\n10 15\r\n18 7\r\n21 17\r\n23 21\r\n", "output": "2.1875000000\r\n5.8660254038\r\n8.1875000000\r\n10.8660254038\r\n18.8660254038\r\n21.8660254038\r\n23.8660254038\r\n"}, {"input": "3 6 6\r\n2 10\r\n14 19\r\n18 14\r\n", "output": "3.4142135624\r\n15.4142135624\r\n19.4142135624\r\n"}, {"input": "10 7 8\r\n2 4\r\n3 13\r\n4 7\r\n5 1\r\n9 16\r\n10 9\r\n12 18\r\n16 4\r\n17 16\r\n20 6\r\n", "output": "4.2857142857\r\n4.5118578920\r\n5.6428571429\r\n13.0714285714\r\n13.0714285714\r\n13.0714285714\r\n13.5118578920\r\n18.2857142857\r\n18.5118578920\r\n21.7619047619\r\n"}, {"input": "8 4 13\r\n0 18\r\n6 24\r\n10 25\r\n11 5\r\n12 18\r\n20 22\r\n21 8\r\n22 12\r\n", "output": "2.5495097568\r\n8.5495097568\r\n12.5495097568\r\n14.2250000000\r\n14.5495097568\r\n22.5495097568\r\n23.6250000000\r\n24.5495097568\r\n"}, {"input": "1 2 7\r\n20 13\r\n", "output": "22.6457513111\r\n"}, {"input": "3 3 3\r\n13 1\r\n18 12\r\n19 2\r\n", "output": "16.1666666667\r\n19.4142135624\r\n20.8333333333\r\n"}, {"input": "8 7 21\r\n2 11\r\n3 4\r\n4 3\r\n9 23\r\n15 9\r\n16 5\r\n22 17\r\n24 10\r\n", "output": "4.6948051948\r\n8.5357142857\r\n11.2142857143\r\n11.4494897428\r\n17.9761904762\r\n20.5571428571\r\n24.4495798319\r\n26.8142857143\r\n"}, {"input": "3 6 19\r\n12 3\r\n20 24\r\n30 2\r\n", "output": "18.5833333333\r\n22.5166114784\r\n39.6666666667\r\n"}, {"input": "4 5 14\r\n11 1\r\n16 20\r\n17 15\r\n21 7\r\n", "output": "25.1000000000\r\n25.1000000000\r\n25.1000000000\r\n25.1000000000\r\n"}, {"input": "1 1 722397\r\n556297 454495\r\n", "output": "557498.9958402590\r\n"}, {"input": "1 100000 363166\r\n560443 753304\r\n", "output": "560445.6950547304\r\n"}, {"input": "1 124232 477338\r\n899117 898233\r\n", "output": "899119.7721151346\r\n"}, {"input": "1 1000000 1000000\r\n0 1000000\r\n", "output": "1.5000000000\r\n"}, {"input": "1 1 1\r\n0 1000000\r\n", "output": "1.4142135624\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, a, d = map(int, f.readline().split())
# Read correct output
correct = []
with open(output_path) as f:
for line in f:
line = line.strip()
if line:
correct.append(float(line))
if len(correct) != n:
print(0)
return
# Read submission output
submission = []
try:
with open(submission_path) as f:
submission_lines = f.readlines()
except:
print(0)
return
if len(submission_lines) != n:
print(0)
return
for line in submission_lines:
stripped = line.strip()
if not stripped:
print(0)
return
try:
t = float(stripped)
except ValueError:
print(0)
return
submission.append(t)
# Check each value
eps = 1e-4
for c, s in zip(correct, submission):
abs_err = abs(c - s)
if abs_err <= eps:
continue
rel_err = abs_err / abs(c)
if rel_err <= eps:
continue
print(0)
return
print(1)
if __name__ == "__main__":
main()
| true |
314/A
|
314
|
A
|
Python 3
|
TESTS
| 3 | 92 | 5,529,600 |
31660973
|
f = lambda: map(int, input().split())
n, k = f()
s = j = 0
for i, q in enumerate(f(), 1):
if s - j * (n - i) * q < k: print(i)
else:
s += q * (i - 1)
j += 1
| 22 | 778 | 23,961,600 |
197616117
|
n, k = map(int, input().split())
S = D = i = 0
ans = []
for val in map(int, input().split()):
f = (n - i - 1) * val * (i - D)
if S - f < k:
D += 1
ans.append(i + 1)
else: S += val * (i - D)
i += 1
print(' '.join(map(str, ans)))
|
Codeforces Round 187 (Div. 1)
|
CF
| 2,013 | 2 | 256 |
Sereja and Contest
|
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula $$d_i = \sum_{j=1}^{i-1}(a_j \cdot (j - 1) - (n - i) \cdot a_i)$$.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
|
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
|
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
| null |
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
|
[{"input": "5 0\n5 3 4 1 2", "output": "2\n3\n4"}, {"input": "10 -10\n5 5 1 7 5 1 2 4 9 2", "output": "2\n4\n5\n7\n8\n9"}]
| 1,600 |
["implementation"]
| 22 |
[{"input": "5 0\r\n5 3 4 1 2\r\n", "output": "2\r\n3\r\n4\r\n"}, {"input": "10 -10\r\n5 5 1 7 5 1 2 4 9 2\r\n", "output": "2\r\n4\r\n5\r\n7\r\n8\r\n9\r\n"}]
| false |
stdio
| null | true |
545/B
|
545
|
B
|
Python 3
|
TESTS
| 0 | 15 | 0 |
178522689
|
n = int(input())
tree = [] # empty list
for i in range(n):
tree.append(list(map(int,input().split())))
s = 2
t = sorted(tree,reverse = True)
t = sorted(tree) # arrange in ascending order
for j in range(1,-1):
a,b = tree[i]
if x - tree[i-1][0] > b:
s += 1
elif tree[i-1][0] -x >b:
s += 1
tree[i][0] += b
print(s if n>1 else 1)
| 54 | 109 | 409,600 |
11151633
|
s=input()
s2=input()
s3=''
f = True
for i in range(len(s)):
if s[i] == s2[i]:
s3 +='1'
elif f:
s3 += s[i]
f = not(f)
else:
s3 += s2[i]
f = not(f)
if f:
print(s3)
else:
print('impossible')
|
Codeforces Round 303 (Div. 2)
|
CF
| 2,015 | 1 | 256 |
Equidistant String
|
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
|
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
|
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
| null |
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
|
[{"input": "0001\n1011", "output": "0011"}, {"input": "000\n111", "output": "impossible"}]
| 1,100 |
["greedy"]
| 54 |
[{"input": "0001\r\n1011\r\n", "output": "0011\r\n"}, {"input": "000\r\n111\r\n", "output": "impossible\r\n"}, {"input": "1010101011111110111111001111111111111111111111101101110111111111111110110110101011111110110111111101\r\n0101111111000100010100001100010101000000011000000000011011000001100100001110111011111000001110011111\r\n", "output": "1111101111101100110110001110110111010101011101001001010011101011101100100110111011111100100110111111\r\n"}, {"input": "0000000001000000000000100000100001000000\r\n1111111011111101111111111111111111111111\r\n", "output": "0101010011010100101010110101101011010101\r\n"}, {"input": "10101000101001001101010010000101100011010011000011001001001111110010100110000001111111\r\n01001011110111111101111011011111110000000111111001000011010101001010000111101010000101\r\n", "output": "11101010111101101101110011001101110010010111010001001011000111011010100111001000101101\r\n"}, {"input": "1111111111111111111111111110111111111111111111111111111111111110111111101111111111111111111111111111\r\n1111111111111111111001111111110010111111111111111111001111111111111111111111111111111111111111111111\r\n", "output": "1111111111111111111101111110110110111111111111111111101111111110111111111111111111111111111111111111\r\n"}, {"input": "0000000000000000000000000000111111111111111111111111111111111111111111111111111111111111111111111111\r\n1111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000\r\n", "output": "0101010101010101010101000000101010101010101010101010101010101010101010101010101010101010101010101010\r\n"}, {"input": "00001111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\r\n11111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\r\n", "output": "01011110101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010\r\n"}, {"input": "0\r\n0\r\n", "output": "0\r\n"}, {"input": "0\r\n1\r\n", "output": "impossible\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n0\r\n", "output": "impossible\r\n"}, {"input": "1111\r\n0000\r\n", "output": "1010\r\n"}, {"input": "1111\r\n1001\r\n", "output": "1101\r\n"}, {"input": "0000\r\n1111\r\n", "output": "0101\r\n"}, {"input": "1010\r\n0101\r\n", "output": "1111\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, 'r') as f:
s = f.readline().strip()
t = f.readline().strip()
n = len(s)
assert len(t) == n
with open(submission_path, 'r') as f:
submission_lines = f.readlines()
if len(submission_lines) == 0:
submission = ''
elif len(submission_lines) == 1:
submission = submission_lines[0].rstrip('\n')
else:
print(0)
return
if submission == 'impossible':
d = sum(c1 != c2 for c1, c2 in zip(s, t))
if d % 2 == 1:
print(1)
else:
print(0)
return
else:
if len(submission) != n:
print(0)
return
if any(c not in {'0', '1'} for c in submission):
print(0)
return
d_s = sum(c != pc for c, pc in zip(s, submission))
d_t = sum(c != pc for c, pc in zip(t, submission))
if d_s == d_t:
print(1)
else:
print(0)
return
if __name__ == '__main__':
main()
| true |
545/A
|
545
|
A
|
PyPy 3-64
|
TESTS
| 0 | 30 | 0 |
195250157
|
def solve():
n=int(input());a=[];ans=[]
for i in range(n):
r=list(map(int,input().split()));a.append(r)
for index,i in enumerate(a):
for j in i:
if j==1 or j==3:break
else:ans.append(index+1)
print(*ans) if len(ans)>0 else print(0)
solve()
| 35 | 31 | 0 |
205323724
|
n = int(input())
collisions = []
for i in range(n):
row = list(map(int, input().split()))
collisions.append(row)
good_cars = set(range(n)) # initially all cars are good
for i in range(n):
for j in range(i+1, n):
if collisions[i][j] == 1:
good_cars.discard(i) # i-th car turned over, so it's not good
elif collisions[i][j] == 2:
good_cars.discard(j) # j-th car turned over, so it's not good
elif collisions[i][j] == 3:
good_cars.discard(i) # both cars turned over, so neither is good
good_cars.discard(j)
good_cars = sorted(list(good_cars))
print(len(good_cars))
print(' '.join(str(car+1) for car in good_cars))
|
Codeforces Round 303 (Div. 2)
|
CF
| 2,015 | 1 | 256 |
Toy Cars
|
Little Susie, thanks to her older brother, likes to play with cars. Today she decided to set up a tournament between them. The process of a tournament is described in the next paragraph.
There are n toy cars. Each pair collides. The result of a collision can be one of the following: no car turned over, one car turned over, both cars turned over. A car is good if it turned over in no collision. The results of the collisions are determined by an n × n matrix А: there is a number on the intersection of the і-th row and j-th column that describes the result of the collision of the і-th and the j-th car:
- - 1: if this pair of cars never collided. - 1 occurs only on the main diagonal of the matrix.
- 0: if no car turned over during the collision.
- 1: if only the i-th car turned over during the collision.
- 2: if only the j-th car turned over during the collision.
- 3: if both cars turned over during the collision.
Susie wants to find all the good cars. She quickly determined which cars are good. Can you cope with the task?
|
The first line contains integer n (1 ≤ n ≤ 100) — the number of cars.
Each of the next n lines contains n space-separated integers that determine matrix A.
It is guaranteed that on the main diagonal there are - 1, and - 1 doesn't appear anywhere else in the matrix.
It is guaranteed that the input is correct, that is, if Aij = 1, then Aji = 2, if Aij = 3, then Aji = 3, and if Aij = 0, then Aji = 0.
|
Print the number of good cars and in the next line print their space-separated indices in the increasing order.
| null | null |
[{"input": "3\n-1 0 0\n0 -1 1\n0 2 -1", "output": "2\n1 3"}, {"input": "4\n-1 3 3 3\n3 -1 3 3\n3 3 -1 3\n3 3 3 -1", "output": "0"}]
| 900 |
["implementation"]
| 35 |
[{"input": "3\r\n-1 0 0\r\n0 -1 1\r\n0 2 -1\r\n", "output": "2\r\n1 3 "}, {"input": "4\r\n-1 3 3 3\r\n3 -1 3 3\r\n3 3 -1 3\r\n3 3 3 -1\r\n", "output": "0\r\n"}, {"input": "1\r\n-1\r\n", "output": "1\r\n1 "}, {"input": "2\r\n-1 0\r\n0 -1\r\n", "output": "2\r\n1 2 "}, {"input": "2\r\n-1 1\r\n2 -1\r\n", "output": "1\r\n2 "}, {"input": "2\r\n-1 2\r\n1 -1\r\n", "output": "1\r\n1 "}, {"input": "2\r\n-1 3\r\n3 -1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
600/E
|
600
|
E
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
230239871
|
def cin(r=1):
if not r:
return int(input())
if r == 1:
return list(map(int, input().split()))
return [list(map(int, input().split())), cin(r - 1)]
n = cin(0)
c = [0] + cin()
e = [[] for i in range(n + 1)]
res = [0] * (n + 1)
def dfs(u):
cnt = {c[u]: 1}
for i in e[u]:
x = dfs(i)
for k in x.keys():
cnt[k] = cnt.get(k, 0) + x[k]
mx, sum = 0, 0
for k in cnt.keys():
if cnt[k] > mx:
mx = cnt[k]
sum = k
elif cnt[k] == mx:
sum += k
res[u] = sum
return cnt
if __name__ == "__main__":
for i in range(n - 1):
u, v = cin()
e[u].append(v)
dfs(1)
for i in res[1:]:
print(i, end=" ")
| 28 | 560 | 72,499,200 |
219697319
|
import sys
input = sys.stdin.buffer.readline
def merge(u, v):
if len(colors[u]) < len(colors[v]):
colors[v], colors[u] = colors[u], colors[v]
_sum[v], _sum[u] = _sum[u], _sum[v]
mx[v], mx[u] = mx[u], mx[v]
for c, value in colors[v].items():
if c not in colors[u]:
colors[u][c] = 0
colors[u][c] += value
if colors[u][c] > mx[u]:
mx[u] = colors[u][c]
_sum[u] = c
elif colors[u][c] == mx[u]:
_sum[u] += c
n = int(input())
_sum = list(map(int,input().split()))
mx = [1] * n
colors = [{c:1} for c in _sum]
adj = [[] for _ in range(n)]
for _ in range(n - 1):
u, v = [int(x) - 1 for x in input().split()]
adj[u].append(v)
adj[v].append(u)
ans = [0] * n
prev = [0] * n
time_seen = [-1] * n
time_seen[0] = 0
Q = [0]
while Q:
u = Q[-1]
if time_seen[u] == len(adj[u]):
ans[u] = _sum[u]
merge(prev[u], u)
Q.pop()
else:
v = adj[u][time_seen[u]]
if v == prev[u]:
time_seen[u] += 1
if time_seen[u] == len(adj[u]):
continue
v = adj[u][time_seen[u]]
prev[v] = u
time_seen[u] += 1
time_seen[v] = 0
Q.append(v)
print(*ans)
|
Educational Codeforces Round 2
|
ICPC
| 2,015 | 2 | 256 |
Lomsat gelral
|
You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.
Let's call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it's possible that two or more colours will be dominating in the subtree of some vertex.
The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.
For each vertex v find the sum of all dominating colours in the subtree of vertex v.
|
The first line contains integer n (1 ≤ n ≤ 105) — the number of vertices in the tree.
The second line contains n integers ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.
Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the edge of the tree. The first vertex is the root of the tree.
|
Print n integers — the sums of dominating colours for each vertex.
| null | null |
[{"input": "4\n1 2 3 4\n1 2\n2 3\n2 4", "output": "10 9 3 4"}, {"input": "15\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\n1 2\n1 3\n1 4\n1 14\n1 15\n2 5\n2 6\n2 7\n3 8\n3 9\n3 10\n4 11\n4 12\n4 13", "output": "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3"}]
| 2,300 |
["data structures", "dfs and similar", "dsu", "trees"]
| 28 |
[{"input": "4\r\n1 2 3 4\r\n1 2\r\n2 3\r\n2 4\r\n", "output": "10 9 3 4\r\n"}, {"input": "15\r\n1 2 3 1 2 3 3 1 1 3 2 2 1 2 3\r\n1 2\r\n1 3\r\n1 4\r\n1 14\r\n1 15\r\n2 5\r\n2 6\r\n2 7\r\n3 8\r\n3 9\r\n3 10\r\n4 11\r\n4 12\r\n4 13\r\n", "output": "6 5 4 3 2 3 3 1 1 3 2 2 1 2 3\r\n"}]
| false |
stdio
| null | true |
314/A
|
314
|
A
|
PyPy 3
|
TESTS
| 3 | 248 | 0 |
64365517
|
def IO():
import sys
sys.stdout = open('output.txt', 'w')
sys.stdin = open('input.txt', 'r')
###################### MAIN PROGRAM #####################
def main():
#IO()
n, k = map(int, input().split())
a = list(map(int, input().split()))
ans = list()
run = int(0)
deleted = 0
for i in range(n):
ratingCh = run - a[i] * (n - i - 1) * (i - deleted)
if (ratingCh < k):
deleted += 1
ans.append(i)
else:
run += a[i] * i
for e in ans:
print(f"{e+1} ")
##################### END OF PROGRAM ####################
if __name__ == "__main__":
main()
| 22 | 840 | 16,896,000 |
14704422
|
n, k = map(int, input().split())
c, v = 0, []
for i, a in enumerate(map(int, input().split())):
j = i + 1 - len(v)
d = c - (j - 1) * (n - j) * a
if d < k:
v.append(i + 1)
n -= 1
else:
c += a * (j - 1)
print('\n'.join(map(str, v)))
|
Codeforces Round 187 (Div. 1)
|
CF
| 2,013 | 2 | 256 |
Sereja and Contest
|
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula $$d_i = \sum_{j=1}^{i-1}(a_j \cdot (j - 1) - (n - i) \cdot a_i)$$.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
|
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
|
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
| null |
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
|
[{"input": "5 0\n5 3 4 1 2", "output": "2\n3\n4"}, {"input": "10 -10\n5 5 1 7 5 1 2 4 9 2", "output": "2\n4\n5\n7\n8\n9"}]
| 1,600 |
["implementation"]
| 22 |
[{"input": "5 0\r\n5 3 4 1 2\r\n", "output": "2\r\n3\r\n4\r\n"}, {"input": "10 -10\r\n5 5 1 7 5 1 2 4 9 2\r\n", "output": "2\r\n4\r\n5\r\n7\r\n8\r\n9\r\n"}]
| false |
stdio
| null | true |
314/A
|
314
|
A
|
PyPy 3
|
TESTS
| 3 | 248 | 0 |
92461680
|
n, k = input().split()
n = int(n)
k = int(k)
v = []
v.append(0)
for aux in input().split():
aux = int(aux)
v.append(aux)
s = []
s.append(0)
ant = 0
for i in range(1, n+1):
if i==1:
s.append(0)
else:
ant+=(v[i-1]*(i-2))
s.append(ant-((n-i)*v[i]*(i-1)))
ans = []
cn = 0
for i in range(2, n+1):
tam = len(ans)
if s[i]+cn+tam*(n-i)*v[i] < k:
ans.append(i)
cn-=v[i]*(i-1)
for l in ans:
print(l)
| 22 | 872 | 17,305,600 |
106177048
|
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for _ in range(1):#nmbr()):
n,k=lst()
a=lst()
removed=0
positive_term=0
for i in range(1,n):
negative_term=(i-removed)*a[i]*(n-i-1)
if (positive_term-negative_term)<k:
removed+=1
stdout.write(str(i+1)+'\n')
else:positive_term += (i - removed) * a[i]
|
Codeforces Round 187 (Div. 1)
|
CF
| 2,013 | 2 | 256 |
Sereja and Contest
|
During the last Sereja's Codesecrof round the server crashed many times, so the round was decided to be made unrated for some participants.
Let's assume that n people took part in the contest. Let's assume that the participant who got the first place has rating a1, the second place participant has rating a2, ..., the n-th place participant has rating an. Then changing the rating on the Codesecrof site is calculated by the formula $$d_i = \sum_{j=1}^{i-1}(a_j \cdot (j - 1) - (n - i) \cdot a_i)$$.
After the round was over, the Codesecrof management published the participants' results table. They decided that if for a participant di < k, then the round can be considered unrated for him. But imagine the management's surprise when they found out that the participants' rating table is dynamic. In other words, when some participant is removed from the rating, he is removed from the results' table and the rating is recalculated according to the new table. And of course, all applications for exclusion from the rating are considered in view of the current table.
We know that among all the applications for exclusion from the rating the first application to consider is from the participant with the best rank (the rank with the minimum number), for who di < k. We also know that the applications for exclusion from rating were submitted by all participants.
Now Sereja wonders, what is the number of participants to be excluded from the contest rating, and the numbers of the participants in the original table in the order of their exclusion from the rating. Pay attention to the analysis of the first test case for a better understanding of the statement.
|
The first line contains two integers n, k (1 ≤ n ≤ 2·105, - 109 ≤ k ≤ 0). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — ratings of the participants in the initial table.
|
Print the numbers of participants in the order in which they were removed from the table. Print the initial numbers of the participants, that is, the numbers that the participants had in the initial table.
| null |
Consider the first test sample.
1. Initially the sequence of the contest participants' ratings equals [5, 3, 4, 1, 2]. You can use this sequence to calculate the sequence of rating changes: [0, -9, -13, 8, 14]. According to the problem statement, the application of the participant who won the second place will be considered first.
2. As soon as the second place winner is out from the ratings, the participants' rating sequence will equal [5, 4, 1, 2]. By this sequence you can count the new sequence of rating changes: [0, -8, 2, 6]. According to the problem statement, the application of the participant who won the second place will be considered. Initially this participant won third place.
3. The new rating sequence equals [5, 1, 2], the new sequence of rating changes equals [0, -1, 1]. The second place participant's application is taken into consideration, initially this participant won the fourth place.
4. The new rating sequence equals [5, 2], the new sequence of rating changes equals [0, 0]. No more applications will be considered.
Thus, you should print 2, 3, 4.
|
[{"input": "5 0\n5 3 4 1 2", "output": "2\n3\n4"}, {"input": "10 -10\n5 5 1 7 5 1 2 4 9 2", "output": "2\n4\n5\n7\n8\n9"}]
| 1,600 |
["implementation"]
| 22 |
[{"input": "5 0\r\n5 3 4 1 2\r\n", "output": "2\r\n3\r\n4\r\n"}, {"input": "10 -10\r\n5 5 1 7 5 1 2 4 9 2\r\n", "output": "2\r\n4\r\n5\r\n7\r\n8\r\n9\r\n"}]
| false |
stdio
| null | true |
280/C
|
280
|
C
|
Python 3
|
TESTS
| 1 | 60 | 0 |
193944668
|
# LUOGU_RID: 102285147
import sys
sys.setrecursionlimit(10**6)
(n, ), *a = [[*map(int, s.split())] for s in open(0)]
g = [[] for _ in range(n)]
for u, v in a:
u -= 1
v -= 1
g[u] += v,
g[v] += u,
d = [1] * n
def dfs(u):
for v in g[u]:
if d[v] == 1 and v != 1:
d[v] = d[u] + 1
dfs(v)
dfs(1)
s = 0
for d in d:
s += 1 / d
print(s)
| 52 | 280 | 16,076,800 |
198355800
|
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
n = int(input())
adj = [[] for _ in range(n)]
for _ in range(n-1):
a, b = map(int, input().split())
a -= 1
b -= 1
adj[a].append(b)
adj[b].append(a)
d = [0] * n
d[0] = 1
q = [0]
while q:
v = q.pop()
for nei in adj[v]:
if d[nei] == 0:
d[nei] = d[v] + 1
q.append(nei)
ans = sum(1 / x for x in d)
print(ans)
t = 1
for _ in range(t):
solve()
|
Codeforces Round 172 (Div. 1)
|
CF
| 2,013 | 1 | 256 |
Game on Tree
|
Momiji has got a rooted tree, consisting of n nodes. The tree nodes are numbered by integers from 1 to n. The root has number 1. Momiji decided to play a game on this tree.
The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let's denote it by v) and removes all the subtree nodes with the root in node v from the tree. Node v gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number 1.
Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.
|
The first line contains integer n (1 ≤ n ≤ 105) — the number of nodes in the tree. The next n - 1 lines contain the tree edges. The i-th line contains integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the numbers of the nodes that are connected by the i-th edge.
It is guaranteed that the given graph is a tree.
|
Print a single real number — the expectation of the number of steps in the described game.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
| null |
In the first sample, there are two cases. One is directly remove the root and another is remove the root after one step. Thus the expected steps are:
1 × (1 / 2) + 2 × (1 / 2) = 1.5
In the second sample, things get more complex. There are two cases that reduce to the first sample, and one case cleaned at once. Thus the expected steps are:
1 × (1 / 3) + (1 + 1.5) × (2 / 3) = (1 / 3) + (5 / 3) = 2
|
[{"input": "2\n1 2", "output": "1.50000000000000000000"}, {"input": "3\n1 2\n1 3", "output": "2.00000000000000000000"}]
| 2,200 |
["implementation", "math", "probabilities", "trees"]
| 52 |
[{"input": "2\r\n1 2\r\n", "output": "1.50000000000000000000\r\n"}, {"input": "3\r\n1 2\r\n1 3\r\n", "output": "2.00000000000000000000\r\n"}, {"input": "10\r\n1 2\r\n2 3\r\n3 4\r\n1 5\r\n2 6\r\n6 7\r\n4 8\r\n6 9\r\n9 10\r\n", "output": "3.81666666666666690000\r\n"}, {"input": "6\r\n1 3\r\n2 4\r\n5 6\r\n3 6\r\n5 4\r\n", "output": "2.45000000000000020000\r\n"}]
| false |
stdio
|
import sys
def read_number(path):
with open(path, 'r') as f:
lines = f.readlines()
if len(lines) != 1:
return None
line = lines[0].strip()
try:
return float(line)
except:
return None
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
expected = read_number(output_path)
submission = read_number(submission_path)
if expected is None or submission is None:
print(0)
return
abs_err = abs(expected - submission)
if abs_err <= 1e-6:
print(1)
return
max_val = max(abs(expected), abs(submission))
if abs_err <= 1e-6 * max_val:
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
962/D
|
962
|
D
|
Python 3
|
TESTS
| 10 | 576 | 13,004,800 |
38591601
|
def appendDict(dict, key, value):
if key in dict:
dict[key] += [value]
else:
dict[key] = [value]
def main():
input()
numbers = list(map(int, input().split()))
dict = {}
idx = 0
for number in numbers:
appendDict(dict, number, idx)
idx += 1
while len(dict.keys()) != 0:
minkey = min(dict)
arr = dict[minkey]
if len(arr) > 1:
for i in range(0, len(arr) - 1, 2):
numbers[arr[i]] = 0
numbers[arr[i + 1]] = minkey * 2
idx = arr[i + 1]
appendDict(dict, minkey * 2, idx)
del dict[minkey]
numbers = [str(x) for x in numbers if x > 0]
print(len(numbers))
print(' '.join(numbers))
if __name__ == "__main__":
main()
| 26 | 140 | 27,648,000 |
229880624
|
input()
a = {}
for i in list(map(int,input().split())):
while i in a:
del a[i]
i *= 2
a[i] = 1
print(len(a))
print(*a)# 1698329597.4687905
|
Educational Codeforces Round 42 (Rated for Div. 2)
|
ICPC
| 2,018 | 2 | 256 |
Merge Equals
|
You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $$$x$$$ that occurs in the array $$$2$$$ or more times. Take the first two occurrences of $$$x$$$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $$$2 \cdot x$$$).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like $$$[3, 4, 1, 2, 2, 1, 1]$$$. It will be changed in the following way: $$$[3, 4, 1, 2, 2, 1, 1]~\rightarrow~[3, 4, 2, 2, 2, 1]~\rightarrow~[3, 4, 4, 2, 1]~\rightarrow~[3, 8, 2, 1]$$$.
If the given array is look like $$$[1, 1, 3, 1, 1]$$$ it will be changed in the following way: $$$[1, 1, 3, 1, 1]~\rightarrow~[2, 3, 1, 1]~\rightarrow~[2, 3, 2]~\rightarrow~[3, 4]$$$.
|
The first line contains a single integer $$$n$$$ ($$$2 \le n \le 150\,000$$$) — the number of elements in the array.
The second line contains a sequence from $$$n$$$ elements $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$) — the elements of the array.
|
In the first line print an integer $$$k$$$ — the number of elements in the array after all the performed operations. In the second line print $$$k$$$ integers — the elements of the array after all the performed operations.
| null |
The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change.
|
[{"input": "7\n3 4 1 2 2 1 1", "output": "4\n3 8 2 1"}, {"input": "5\n1 1 3 1 1", "output": "2\n3 4"}, {"input": "5\n10 40 20 50 30", "output": "5\n10 40 20 50 30"}]
| 1,600 |
["data structures", "implementation"]
| 26 |
[{"input": "7\r\n3 4 1 2 2 1 1\r\n", "output": "4\r\n3 8 2 1 \r\n"}, {"input": "5\r\n1 1 3 1 1\r\n", "output": "2\r\n3 4 \r\n"}, {"input": "5\r\n10 40 20 50 30\r\n", "output": "5\r\n10 40 20 50 30 \r\n"}, {"input": "100\r\n10 10 15 12 15 13 15 12 10 10 15 11 13 14 13 14 10 13 12 10 14 12 13 11 14 15 12 11 11 15 12 12 11 14 14 14 15 10 10 15 15 13 13 15 10 12 14 10 12 13 11 15 11 13 14 12 10 12 11 14 13 15 13 15 13 14 14 11 12 13 11 14 10 10 15 10 15 12 15 12 13 10 11 13 15 11 10 12 10 12 14 14 13 12 14 10 12 13 11 13\r\n", "output": "12\r\n88 240 15 44 160 192 208 224 20 24 11 26 \r\n"}, {"input": "2\r\n1000000000 1000000000\r\n", "output": "1\r\n2000000000 \r\n"}, {"input": "3\r\n500000000 500000000 1000000000\r\n", "output": "1\r\n2000000000 \r\n"}, {"input": "9\r\n8 536870913 536870913 536870913 536870913 536870913 536870913 536870913 536870913\r\n", "output": "2\r\n8 4294967304 \r\n"}, {"input": "34\r\n967614464 967614464 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\r\n", "output": "2\r\n1935228928 32000000000 \r\n"}]
| false |
stdio
| null | true |
545/A
|
545
|
A
|
Python 3
|
TESTS
| 0 | 30 | 0 |
219937961
|
n = int(input())
s = 1
m = []
for i in range(n):
row = list(map(int, input().split()))
m.append(row)
ch = ''
for i in range(n):
contains_1_or_3 = False
for j in range(n):
if m[i][j] == 1 or m[i][j] == 3:
contains_1_or_3 = True
break
if not contains_1_or_3:
ch = ch + str(i + 1) + ' '
if ch == '':
print(0)
else:
print(len(ch.strip().split()) - 1)
print(ch.strip())
| 35 | 46 | 0 |
187908142
|
n,a=int(input()),[]
for i in range(n):
x = list(map(int, input().split(" ")))
if 1 not in x and 3 not in x:
a.append(i+1)
print(len(a))
print(*sorted(a))
|
Codeforces Round 303 (Div. 2)
|
CF
| 2,015 | 1 | 256 |
Toy Cars
|
Little Susie, thanks to her older brother, likes to play with cars. Today she decided to set up a tournament between them. The process of a tournament is described in the next paragraph.
There are n toy cars. Each pair collides. The result of a collision can be one of the following: no car turned over, one car turned over, both cars turned over. A car is good if it turned over in no collision. The results of the collisions are determined by an n × n matrix А: there is a number on the intersection of the і-th row and j-th column that describes the result of the collision of the і-th and the j-th car:
- - 1: if this pair of cars never collided. - 1 occurs only on the main diagonal of the matrix.
- 0: if no car turned over during the collision.
- 1: if only the i-th car turned over during the collision.
- 2: if only the j-th car turned over during the collision.
- 3: if both cars turned over during the collision.
Susie wants to find all the good cars. She quickly determined which cars are good. Can you cope with the task?
|
The first line contains integer n (1 ≤ n ≤ 100) — the number of cars.
Each of the next n lines contains n space-separated integers that determine matrix A.
It is guaranteed that on the main diagonal there are - 1, and - 1 doesn't appear anywhere else in the matrix.
It is guaranteed that the input is correct, that is, if Aij = 1, then Aji = 2, if Aij = 3, then Aji = 3, and if Aij = 0, then Aji = 0.
|
Print the number of good cars and in the next line print their space-separated indices in the increasing order.
| null | null |
[{"input": "3\n-1 0 0\n0 -1 1\n0 2 -1", "output": "2\n1 3"}, {"input": "4\n-1 3 3 3\n3 -1 3 3\n3 3 -1 3\n3 3 3 -1", "output": "0"}]
| 900 |
["implementation"]
| 35 |
[{"input": "3\r\n-1 0 0\r\n0 -1 1\r\n0 2 -1\r\n", "output": "2\r\n1 3 "}, {"input": "4\r\n-1 3 3 3\r\n3 -1 3 3\r\n3 3 -1 3\r\n3 3 3 -1\r\n", "output": "0\r\n"}, {"input": "1\r\n-1\r\n", "output": "1\r\n1 "}, {"input": "2\r\n-1 0\r\n0 -1\r\n", "output": "2\r\n1 2 "}, {"input": "2\r\n-1 1\r\n2 -1\r\n", "output": "1\r\n2 "}, {"input": "2\r\n-1 2\r\n1 -1\r\n", "output": "1\r\n1 "}, {"input": "2\r\n-1 3\r\n3 -1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
962/D
|
962
|
D
|
Python 3
|
TESTS
| 10 | 358 | 20,992,000 |
37257591
|
from collections import defaultdict
from heapq import heappush , heappop
def process() :
n = int(input())
l = [int(x) for x in input().split(' ')]
sorted_list = []
d = defaultdict(list)
for index,i in enumerate(l) :
heappush(d[i],index)
for i in d.keys() :
heappush(sorted_list,i)
zero = 0
while sorted_list :
min_num = heappop(sorted_list)
length = len(d[min_num])
if length < 2 :
continue
if length % 2 :
d[min_num] = d[min_num][: length-1]
for i,index in enumerate(d[min_num]):
if i % 2 :
if min_num*2 not in d :
heappush(sorted_list,min_num*2)
heappush(d[min_num*2],index)
l[index] = min_num*2
zero += 1
else :
l[index] = 0
print(n-zero)
for i in l :
if i != 0 :
print(i ,end=' ')
process()
| 26 | 155 | 29,696,000 |
193672717
|
n=int(input())
a=list(map(int,input().split()))
d={}
for i in range(n):
while a[i] in d:
del d[a[i]]
a[i]*=2
d[a[i]]=i
print(len(d))
print(*d,sep=" ")
|
Educational Codeforces Round 42 (Rated for Div. 2)
|
ICPC
| 2,018 | 2 | 256 |
Merge Equals
|
You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $$$x$$$ that occurs in the array $$$2$$$ or more times. Take the first two occurrences of $$$x$$$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $$$2 \cdot x$$$).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like $$$[3, 4, 1, 2, 2, 1, 1]$$$. It will be changed in the following way: $$$[3, 4, 1, 2, 2, 1, 1]~\rightarrow~[3, 4, 2, 2, 2, 1]~\rightarrow~[3, 4, 4, 2, 1]~\rightarrow~[3, 8, 2, 1]$$$.
If the given array is look like $$$[1, 1, 3, 1, 1]$$$ it will be changed in the following way: $$$[1, 1, 3, 1, 1]~\rightarrow~[2, 3, 1, 1]~\rightarrow~[2, 3, 2]~\rightarrow~[3, 4]$$$.
|
The first line contains a single integer $$$n$$$ ($$$2 \le n \le 150\,000$$$) — the number of elements in the array.
The second line contains a sequence from $$$n$$$ elements $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^{9}$$$) — the elements of the array.
|
In the first line print an integer $$$k$$$ — the number of elements in the array after all the performed operations. In the second line print $$$k$$$ integers — the elements of the array after all the performed operations.
| null |
The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change.
|
[{"input": "7\n3 4 1 2 2 1 1", "output": "4\n3 8 2 1"}, {"input": "5\n1 1 3 1 1", "output": "2\n3 4"}, {"input": "5\n10 40 20 50 30", "output": "5\n10 40 20 50 30"}]
| 1,600 |
["data structures", "implementation"]
| 26 |
[{"input": "7\r\n3 4 1 2 2 1 1\r\n", "output": "4\r\n3 8 2 1 \r\n"}, {"input": "5\r\n1 1 3 1 1\r\n", "output": "2\r\n3 4 \r\n"}, {"input": "5\r\n10 40 20 50 30\r\n", "output": "5\r\n10 40 20 50 30 \r\n"}, {"input": "100\r\n10 10 15 12 15 13 15 12 10 10 15 11 13 14 13 14 10 13 12 10 14 12 13 11 14 15 12 11 11 15 12 12 11 14 14 14 15 10 10 15 15 13 13 15 10 12 14 10 12 13 11 15 11 13 14 12 10 12 11 14 13 15 13 15 13 14 14 11 12 13 11 14 10 10 15 10 15 12 15 12 13 10 11 13 15 11 10 12 10 12 14 14 13 12 14 10 12 13 11 13\r\n", "output": "12\r\n88 240 15 44 160 192 208 224 20 24 11 26 \r\n"}, {"input": "2\r\n1000000000 1000000000\r\n", "output": "1\r\n2000000000 \r\n"}, {"input": "3\r\n500000000 500000000 1000000000\r\n", "output": "1\r\n2000000000 \r\n"}, {"input": "9\r\n8 536870913 536870913 536870913 536870913 536870913 536870913 536870913 536870913\r\n", "output": "2\r\n8 4294967304 \r\n"}, {"input": "34\r\n967614464 967614464 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\r\n", "output": "2\r\n1935228928 32000000000 \r\n"}]
| false |
stdio
| null | true |
717/D
|
717
|
D
|
PyPy 3
|
TESTS
| 1 | 249 | 4,915,200 |
52463904
|
from sys import stdin,stdout
from decimal import Decimal
from functools import reduce
n,x = map(int,stdin.readline().split())
print(1 - reduce(lambda x,y: x +y*y,map(Decimal,stdin.readline().split()),Decimal(0)))
| 11 | 249 | 4,915,200 |
29285114
|
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
n,x=map(int,input().split())
def mult(a,b):
# compute a*b
c=[0]*128
for i in range(128):
for j in range(128):
c[i^j]+=a[i]*b[j]
return c
def quickpow(a,b):
# compute a**b
if b==1:
return a
if b&1:
return mult(quickpow(mult(a,a),b//2),a)
return quickpow(mult(a,a),b//2)
prob=list(map(float,input().split()))
prob+=[0.0]*(128-len(prob))
print("%.9f"%(1-quickpow(prob,n)[0]))
|
Bubble Cup 9 - Finals [Online Mirror]
|
ICPC
| 2,016 | 1 | 256 |
Dexterina’s Lab
|
Dexterina and Womandark have been arch-rivals since they’ve known each other. Since both are super-intelligent teenage girls, they’ve always been trying to solve their disputes in a peaceful and nonviolent way. After god knows how many different challenges they’ve given to one another, their score is equal and they’re both desperately trying to best the other in various games of wits. This time, Dexterina challenged Womandark to a game of Nim.
Nim is a two-player game in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects from a single heap. The player who can't make a turn loses. By their agreement, the sizes of piles are selected randomly from the range [0, x]. Each pile's size is taken independently from the same probability distribution that is known before the start of the game.
Womandark is coming up with a brand new and evil idea on how to thwart Dexterina’s plans, so she hasn’t got much spare time. She, however, offered you some tips on looking fabulous in exchange for helping her win in Nim. Your task is to tell her what is the probability that the first player to play wins, given the rules as above.
|
The first line of the input contains two integers n (1 ≤ n ≤ 109) and x (1 ≤ x ≤ 100) — the number of heaps and the maximum number of objects in a heap, respectively. The second line contains x + 1 real numbers, given with up to 6 decimal places each: P(0), P(1), ... , P(X). Here, P(i) is the probability of a heap having exactly i objects in start of a game. It's guaranteed that the sum of all P(i) is equal to 1.
|
Output a single real number, the probability that the first player wins. The answer will be judged as correct if it differs from the correct answer by at most 10 - 6.
| null | null |
[{"input": "2 2\n0.500000 0.250000 0.250000", "output": "0.62500000"}]
| 1,900 |
["games", "matrices", "probabilities"]
| 11 |
[{"input": "2 2\r\n0.500000 0.250000 0.250000\r\n", "output": "0.62500000\r\n"}, {"input": "9 9\r\n0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000 0.100000\r\n", "output": "0.93687014\r\n"}, {"input": "1000001 5\r\n0.000000 0.300000 0.000000 0.500000 0.000000 0.200000\r\n", "output": "1.00000000\r\n"}, {"input": "1000000 5\r\n0.000000 0.000000 1.000000 0.000000 0.000000 0.000000\r\n", "output": "0.00000000\r\n"}, {"input": "100 20\r\n0.065682 0.015963 0.024840 0.025856 0.000807 0.004366 0.020977 0.151841 0.050870 0.009918 0.044345 0.009672 0.077615 0.019520 0.166791 0.028122 0.010873 0.018634 0.172648 0.037614 0.043046\r\n", "output": "0.96875000\r\n"}, {"input": "1000000 50\r\n0.034403 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.033928 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.071853 0.000000 0.000000 0.016440 0.000000 0.000000 0.000000 0.000000 0.023167 0.067938 0.000000 0.000000 0.000000 0.000000 0.176349 0.030154 0.180266 0.272662 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.014462 0.000000 0.000000 0.037916 0.040462\r\n", "output": "0.98437500\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]
try:
with open(correct_output_path, 'r') as f:
correct_line = f.read().strip()
correct_val = float(correct_line)
except:
print(0)
return
try:
with open(submission_output_path, 'r') as f:
submission_line = f.read().strip()
submission_val = float(submission_line)
except:
print(0)
return
diff = abs(submission_val - correct_val)
print(1 if diff <= 1e-6 else 0)
if __name__ == "__main__":
main()
| true |
177/D1
|
177
|
D2
|
Python 3
|
TESTS2
| 2 | 124 | 307,200 |
109287019
|
n,m,c=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
sum=0
for i in range(n):
if i<m:
sum=(sum+b[i])%c
if i>=n-m+1:
sum=(sum+b[i-(n-m+1)])%c
a[i]=(a[i]+sum)%c
print(' '.join(map(str,a)))
| 12 | 124 | 307,200 |
4855563
|
n, m, c = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
s, k = 0, n - m
for i in range(n):
if i < m: s += b[i]
a[i] = (a[i] + s) % c
if i >= k: s -= b[i - k]
print(' '.join(str(i) for i in a))
|
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 |
773/B
|
773
|
B
|
Python 3
|
TESTS
| 0 | 77 | 0 |
48309925
|
n = int(input())
Mines =[[int(x) for x in input().split(' ')] for y in range(n)]
diff =[]
for i in range(n):
diff.append(abs( Mines[i][0] - Mines[i][1] ))
print(diff.index(max(diff))+1)
| 40 | 62 | 307,200 |
26938082
|
from sys import stdin, stdout
fractionBoundaries = [1/2,1/4,1/8,1/16,1/32,0]
scoreList = [500,1000,1500,2000,2500,3000]
def assessScores(n,Vsol,Psol,solvers):
vScore=0
pScore=0
for i in range(5):
for j in range(6):
if solvers[i]/n>fractionBoundaries[j]:
qnScore=scoreList[j]
break
if Vsol[i]>=0:
vScore+=qnScore-(qnScore*Vsol[i]//250)
if Psol[i]>=0:
pScore+=qnScore-(qnScore*Psol[i]//250)
return vScore>pScore
def main():
n = int(stdin.readline().rstrip())
Vsol = [int(x) for x in stdin.readline().rstrip().split()]
Psol = [int(x) for x in stdin.readline().rstrip().split()]
solvers = [0]*5
for _ in range(n-2):
a = [int(x) for x in stdin.readline().rstrip().split()]
for i in range(5):
if a[i]>=0:
solvers[i]+=1
for i in range(5):
if Vsol[i]>=0:
solvers[i]+=1
if Psol[i]>=0:
solvers[i]+=1
newSolvers=0
vWins=[]
pWins=[]
for i in range(5):
if (Vsol[i]<Psol[i] and Vsol[i]>=0) or (Vsol[i]>=0 and Psol[i]<0):
vWins.append(i)
elif (Psol[i]<Vsol[i] and Psol[i]>=0 and Vsol[i]>=0):
pWins.append(i)
if len(vWins)==0:
print(-1)
else:
while not assessScores(n+newSolvers,Vsol,Psol,solvers) and newSolvers<=100000007:
solversNeeded=9999999999999
for i in range(5):
if i in vWins:
currentRatio = solvers[i]/(newSolvers+n)
for j in range(6):
if solvers[i]/(newSolvers+n)>fractionBoundaries[j]:
nextBoundary = fractionBoundaries[j]
break
if nextBoundary!=0:
if solvers[i]%nextBoundary==0:
solversNeeded = min([solvers[i]//nextBoundary - (newSolvers+n),solversNeeded])
else:
solversNeeded = min([solvers[i]//nextBoundary - (newSolvers+n)+1,solversNeeded])
elif i in pWins and Vsol[i]>0:
currentRatio = solvers[i]/(newSolvers+n)
for j in range(6):
if solvers[i]/(newSolvers+n)>fractionBoundaries[j]:
if j>0:
nextBoundary = fractionBoundaries[j-1]
solversNeeded = min([(nextBoundary*(newSolvers+n)-solvers[i])//(1-nextBoundary)+1,solversNeeded])
break
newSolvers+=solversNeeded
for x in pWins:
solvers[x]+=solversNeeded
if newSolvers>1000000007:
print(-1)
else:
print(int(newSolvers))
main()
|
VK Cup 2017 - Round 3
|
CF
| 2,017 | 2 | 256 |
Dynamic Problem Scoring
|
Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.
For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.
$$\begin{array}{|c|c|}
\hline
\text{Solvers fraction} & \text{Maximum point value} \\
\hline
(1/2, 1] & 500 \\
(1/4, 1/2] & 1000 \\
(1/8, 1/4] & 1500 \\
(1/16, 1/8] & 2000 \\
(1/32, 1/16] & 2500 \\
[0, 1/32] & 3000 \\
\hline
\end{array}$$
Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.
If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000·(1 - 40 / 250) = 1680 points for this problem.
There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.
With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.
Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.
Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.
Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.
|
The first line contains a single integer n (2 ≤ n ≤ 120) — the number of round participants, including Vasya and Petya.
Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 ≤ ai, j ≤ 119) — the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.
It is guaranteed that each participant has made at least one successful submission.
Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.
|
Output a single integer — the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.
| null |
In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.
In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.
In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.
|
[{"input": "2\n5 15 40 70 115\n50 45 40 30 15", "output": "2"}, {"input": "3\n55 80 10 -1 -1\n15 -1 79 60 -1\n42 -1 13 -1 -1", "output": "3"}, {"input": "5\n119 119 119 119 119\n0 0 0 0 -1\n20 65 12 73 77\n78 112 22 23 11\n1 78 60 111 62", "output": "27"}, {"input": "4\n-1 20 40 77 119\n30 10 73 50 107\n21 29 -1 64 98\n117 65 -1 -1 -1", "output": "-1"}]
| 2,000 |
["brute force", "greedy"]
| 40 |
[{"input": "2\r\n5 15 40 70 115\r\n50 45 40 30 15\r\n", "output": "2\r\n"}, {"input": "3\r\n55 80 10 -1 -1\r\n15 -1 79 60 -1\r\n42 -1 13 -1 -1\r\n", "output": "3\r\n"}, {"input": "5\r\n119 119 119 119 119\r\n0 0 0 0 -1\r\n20 65 12 73 77\r\n78 112 22 23 11\r\n1 78 60 111 62\r\n", "output": "27\r\n"}, {"input": "4\r\n-1 20 40 77 119\r\n30 10 73 50 107\r\n21 29 -1 64 98\r\n117 65 -1 -1 -1\r\n", "output": "-1\r\n"}, {"input": "2\r\n33 15 51 7 101\r\n41 80 40 13 46\r\n", "output": "0\r\n"}, {"input": "9\r\n57 52 60 56 91\r\n32 40 107 89 36\r\n80 0 45 92 119\r\n62 9 107 24 61\r\n43 28 4 26 113\r\n31 91 86 13 95\r\n4 2 88 38 68\r\n83 35 57 101 28\r\n12 40 37 56 73\r\n", "output": "9\r\n"}, {"input": "19\r\n78 100 74 31 2\r\n27 45 72 63 0\r\n42 114 31 106 79\r\n88 119 118 69 90\r\n68 14 90 104 70\r\n106 21 96 15 73\r\n75 66 54 46 107\r\n108 49 17 34 90\r\n76 112 49 56 76\r\n34 43 5 57 67\r\n47 43 114 73 109\r\n79 118 69 22 19\r\n31 74 21 84 79\r\n1 64 88 97 79\r\n115 14 119 101 28\r\n55 9 43 67 10\r\n33 40 26 10 11\r\n92 0 60 14 48\r\n58 57 8 12 118\r\n", "output": "133\r\n"}, {"input": "17\r\n66 15 -1 42 90\r\n67 108 104 16 110\r\n76 -1 -1 -1 96\r\n108 32 100 91 17\r\n87 -1 85 10 -1\r\n70 55 102 15 23\r\n-1 33 111 105 63\r\n-1 56 104 68 116\r\n56 111 102 89 63\r\n63 -1 68 80 -1\r\n80 61 -1 81 19\r\n101 -1 87 -1 89\r\n92 82 4 105 83\r\n19 30 114 77 104\r\n100 99 29 68 82\r\n98 -1 62 52 -1\r\n108 -1 -1 50 -1\r\n", "output": "5\r\n"}, {"input": "3\r\n20 65 12 73 77\r\n78 112 22 23 11\r\n1 78 60 111 62\r\n", "output": "3\r\n"}, {"input": "4\r\n66 55 95 78 114\r\n70 98 8 95 95\r\n17 47 88 71 18\r\n23 22 9 104 38\r\n", "output": "4\r\n"}, {"input": "10\r\n-1 18 44 61 115\r\n-1 34 12 40 114\r\n-1 86 100 119 58\r\n-1 4 36 8 91\r\n1 58 85 13 82\r\n-1 9 85 109 -1\r\n13 75 0 71 42\r\n116 75 42 79 88\r\n62 -1 98 114 -1\r\n68 96 44 61 35\r\n", "output": "62\r\n"}, {"input": "26\r\n3 -1 71 -1 42\r\n85 72 48 38 -1\r\n-1 -1 66 24 -1\r\n46 -1 60 99 107\r\n53 106 51 -1 104\r\n-1 17 98 54 -1\r\n44 107 66 65 102\r\n47 40 62 34 5\r\n-1 10 -1 98 -1\r\n-1 69 47 85 75\r\n12 62 -1 15 -1\r\n48 63 72 32 99\r\n91 104 111 -1 -1\r\n92 -1 52 -1 11\r\n118 25 97 1 108\r\n-1 61 97 37 -1\r\n87 47 -1 -1 21\r\n79 87 73 82 70\r\n90 108 19 25 57\r\n37 -1 51 8 119\r\n64 -1 -1 38 82\r\n42 61 63 25 27\r\n82 -1 15 82 15\r\n-1 89 73 95 -1\r\n4 8 -1 70 116\r\n89 21 65 -1 88\r\n", "output": "10\r\n"}, {"input": "2\r\n0 0 0 0 1\r\n0 0 0 1 0\r\n", "output": "2\r\n"}]
| false |
stdio
| null | true |
965/E
|
965
|
E
|
Python 3
|
TESTS
| 0 | 15 | 0 |
158708882
|
n = int(input())
c="ode"
d="aacada"
v="resistance"
for i in range(n):
m = str(input())
if m == c:
print (6)
if m == d:
print(11)
if m == v:
print(3)
| 24 | 156 | 22,937,600 |
177672519
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def trie():
G, cnt = [[-1] * 26], [0]
return G, cnt
def insert(s):
j = 0
for i in s:
if not G[j][i - 97] ^ -1:
G[j][i - 97] = len(cnt)
G.append([-1] * 26)
cnt.append(0)
j = len(cnt) - 1
else:
j = G[j][i - 97]
cnt[j] += 1
return
n = int(input())
G, cnt = trie()
for _ in range(n):
s = list(input().rstrip())
insert(s)
m = len(cnt)
inf = pow(10, 9) + 1
dist = [inf] * m
dist[0] = 0
parent = [-1] * m
q, k = [0], 0
while len(q) ^ k:
i = q[k]
di = dist[i]
for j in G[i]:
if j ^ -1:
q.append(j)
dist[j] = di + 1
parent[j] = i
k += 1
ans = 0
for i in range(m):
if cnt[i]:
ans += dist[i]
for i in q[::-1]:
if not cnt[i]:
continue
u = i
while parent[u]:
u = parent[u]
if not cnt[u]:
break
if not cnt[u]:
cnt[u] = 1
ans -= dist[i] - dist[u]
print(ans)
|
Codeforces Round 476 (Div. 2) [Thanks, Telegram!]
|
CF
| 2,018 | 1 | 256 |
Short Code
|
Arkady's code contains $$$n$$$ variables. Each variable has a unique name consisting of lowercase English letters only. One day Arkady decided to shorten his code.
He wants to replace each variable name with its non-empty prefix so that these new names are still unique (however, a new name of some variable can coincide with some old name of another or same variable). Among such possibilities he wants to find the way with the smallest possible total length of the new names.
A string $$$a$$$ is a prefix of a string $$$b$$$ if you can delete some (possibly none) characters from the end of $$$b$$$ and obtain $$$a$$$.
Please find this minimum possible total length of new names.
|
The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of variables.
The next $$$n$$$ lines contain variable names, one per line. Each name is non-empty and contains only lowercase English letters. The total length of these strings is not greater than $$$10^5$$$. The variable names are distinct.
|
Print a single integer — the minimum possible total length of new variable names.
| null |
In the first example one of the best options is to shorten the names in the given order as "cod", "co", "c".
In the second example we can shorten the last name to "aac" and the first name to "a" without changing the other names.
|
[{"input": "3\ncodeforces\ncodehorses\ncode", "output": "6"}, {"input": "5\nabba\nabb\nab\naa\naacada", "output": "11"}, {"input": "3\ntelegram\ndigital\nresistance", "output": "3"}]
| 2,200 |
["data structures", "dp", "greedy", "strings", "trees"]
| 24 |
[{"input": "3\r\ncodeforces\r\ncodehorses\r\ncode\r\n", "output": "6\r\n"}, {"input": "5\r\nabba\r\nabb\r\nab\r\naa\r\naacada\r\n", "output": "11\r\n"}, {"input": "3\r\ntelegram\r\ndigital\r\nresistance\r\n", "output": "3\r\n"}, {"input": "1\r\na\r\n", "output": "1\r\n"}, {"input": "10\r\naaaba\r\nbabba\r\nbbba\r\naaabb\r\nabba\r\na\r\nbbb\r\nbaa\r\naaba\r\naa\r\n", "output": "23\r\n"}, {"input": "10\r\naaaaaaaaaaaaaaa\r\naaaaaaaaaaaaaa\r\naaaaaaaa\r\naaa\r\naaaaaaaaaaaa\r\naa\r\naaaaaaa\r\naaaaaaaaa\r\naaaaaaaaaaaaa\r\naaaaaaaaaa\r\n", "output": "55\r\n"}, {"input": "26\r\ni\r\nm\r\nz\r\na\r\nv\r\nu\r\nq\r\nt\r\nj\r\nn\r\nr\r\nk\r\nd\r\ng\r\ns\r\nc\r\no\r\ne\r\np\r\nl\r\ny\r\nw\r\nh\r\nf\r\nb\r\nx\r\n", "output": "26\r\n"}]
| false |
stdio
| null | true |
610/D
|
610
|
D
|
Python 3
|
TESTS
| 0 | 15 | 0 |
215716364
|
# LUOGU_RID: 117562212
n = int(input())
rows = set()
cols = set()
for _ in range(n):
x1, y1, x2, y2 = map(int, input().split())
# 将起点和终点所在的行号添加到 rows 集合中
rows.add(y1)
rows.add(y2)
# 将起点和终点所在的列号添加到 cols 集合中
cols.add(x1)
cols.add(x2)
# 计算涂色的方格总数
painted_cells = len(rows) * len(cols)
print(painted_cells)
| 47 | 1,278 | 54,681,600 |
216911246
|
# i'm from jasnah, do not ban me
from sys import stdin
from itertools import repeat
from collections import defaultdict
num_rectangles = int(stdin.readline())
horizontal_segments = defaultdict(list)
vertical_segments = defaultdict(list)
unique_y_values = set()
events = defaultdict(list)
for _ in repeat(None, num_rectangles):
x1, y1, x2, y2 = map(int, input().split())
if x1 > x2:
x1, x2 = x2, x1
if y1 > y2:
y1, y2 = y2, y1
if x1 == x2:
horizontal_segments[x1].append((y1, y2))
else:
vertical_segments[y1].append((x1, x2))
total_area = 0
for x, segments in horizontal_segments.items():
segments.sort()
segments.append((1001001001, 1001001001))
new_segments = []
start = end = segments[0][0]
for y1, y2 in segments:
if y1 > end:
new_segments.append((start, end))
total_area += end - start + 1
events[start].append((1, x))
events[end + 1].append((-1, x))
start = end = y1
if end < y2:
end = y2
segments = new_segments
unique_y_values.add(x)
for y, segments in vertical_segments.items():
segments.sort()
segments.append((1001001001, 1001001001))
new_segments = []
start = end = segments[0][0]
for x1, x2 in segments:
if x1 > end:
new_segments.append((start, end))
unique_y_values.add(start)
unique_y_values.add(end)
events[y].append((2, start, end))
total_area += end - start + 1
start = end = x1
if end < x2:
end = x2
segments = new_segments
unique_y_values = [-1001001001] + sorted(unique_y_values)
unique_y_values_mapping = {y: i for i, y in enumerate(unique_y_values)}
length = len(unique_y_values)
binary_indexed_tree = [0] * length
for x_value in sorted(events.keys()):
for event in sorted(events[x_value]):
if event[0] <= 1:
operation, y_value = event[0], unique_y_values_mapping[event[1]]
while y_value < length:
binary_indexed_tree[y_value] += operation
y_value += y_value - (y_value & (y_value - 1))
else:
operation, start, end = event[0], unique_y_values_mapping[event[1]] - 1, unique_y_values_mapping[event[2]]
while end > 0:
total_area -= binary_indexed_tree[end]
end = end & (end - 1)
while start > 0:
total_area += binary_indexed_tree[start]
start = start & (start - 1)
print(total_area)
|
Codeforces Round 337 (Div. 2)
|
CF
| 2,015 | 2 | 256 |
Vika and Segments
|
Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black horizontal and vertical segments parallel to the coordinate axes. All segments have width equal to 1 square, that means every segment occupy some set of neighbouring squares situated in one row or one column.
Your task is to calculate the number of painted cells. If a cell was painted more than once, it should be calculated exactly once.
|
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of segments drawn by Vika.
Each of the next n lines contains four integers x1, y1, x2 and y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — the coordinates of the endpoints of the segments drawn by Vika. It is guaranteed that all the segments are parallel to coordinate axes. Segments may touch, overlap and even completely coincide.
|
Print the number of cells painted by Vika. If a cell was painted more than once, it should be calculated exactly once in the answer.
| null |
In the first sample Vika will paint squares (0, 1), (1, 1), (2, 1), (1, 2), (1, 3), (1, 4), (0, 3) and (2, 3).
|
[{"input": "3\n0 1 2 1\n1 4 1 2\n0 3 2 3", "output": "8"}, {"input": "4\n-2 -1 2 -1\n2 1 -2 1\n-1 -2 -1 2\n1 2 1 -2", "output": "16"}]
| 2,300 |
["constructive algorithms", "data structures", "geometry", "two pointers"]
| 47 |
[{"input": "3\r\n0 1 2 1\r\n1 4 1 2\r\n0 3 2 3\r\n", "output": "8\r\n"}, {"input": "4\r\n-2 -1 2 -1\r\n2 1 -2 1\r\n-1 -2 -1 2\r\n1 2 1 -2\r\n", "output": "16\r\n"}, {"input": "1\r\n1 1 1 1\r\n", "output": "1\r\n"}, {"input": "10\r\n-357884841 -999999905 -357884841 999999943\r\n-130177221 999999983 -130177221 -999999974\r\n627454332 999999936 627454332 -999999900\r\n999999986 366591992 -999999919 366591992\r\n488824292 999999952 488824292 -999999979\r\n-261575319 999999910 -261575319 -999999995\r\n837827059 -999999983 837827059 999999984\r\n-999999947 543634048 999999977 543634048\r\n512878899 -999999968 512878899 999999926\r\n239286254 -999999975 239286254 999999937\r\n", "output": "19999999073\r\n"}, {"input": "4\r\n553245544 -999999997 553245544 -918743333\r\n999999988 -668043590 264717840 -668043590\r\n-999999961 121002405 999999920 121002405\r\n999999985 121822043 -796706494 121822043\r\n", "output": "4613245176\r\n"}]
| false |
stdio
| null | true |
852/G
|
852
|
G
|
Python 3
|
TESTS
| 2 | 77 | 716,800 |
105063852
|
import re,sys
r=sys.stdin.readline
N,M=map(int,r().split())
words=[]
for _ in range(N):
words.append(r().strip())
for _ in range(M):
pattern=r().strip()
p=re.compile(pattern.replace("?",".?"))
cnt=0
for word in words:
if re.match(p,word):
cnt+=1
print(cnt)
| 15 | 872 | 31,846,400 |
105100981
|
import sys
from itertools import product
from collections import defaultdict
r=sys.stdin.readline
N,M=map(int,r().split())
words=defaultdict(int)
tb=['a','b','c','d','e']
st=set()
cnt=0
res=""
def dfs(u):
global res,cnt
if u==l:
if res in st:
return
if words[res]>0: cnt+=words[res]
st.add(res)
return
if pattern[u]=='?':
for i in range(6):
if i!=5:res+=tb[i]
dfs(u+1)
if i!=5:res=res[:-1]
else:
res+=pattern[u]
dfs(u+1)
res=res[:-1]
for _ in range(N):
word=r().strip()
words[word]+=1
for _ in range(M):
cnt=0
st.clear()
pattern=r().strip()
l=len(pattern)
res=""
dfs(0)
print(cnt)
|
Bubble Cup X - Finals [Online Mirror]
|
ICPC
| 2,017 | 2 | 256 |
Bathroom terminal
|
Smith wakes up at the side of a dirty, disused bathroom, his ankle chained to pipes. Next to him is tape-player with a hand-written message "Play Me". He finds a tape in his own back pocket. After putting the tape in the tape-player, he sees a key hanging from a ceiling, chained to some kind of a machine, which is connected to the terminal next to him. After pressing a Play button a rough voice starts playing from the tape:
"Listen up Smith. As you can see, you are in pretty tough situation and in order to escape, you have to solve a puzzle.
You are given N strings which represent words. Each word is of the maximum length L and consists of characters 'a'-'e'. You are also given M strings which represent patterns. Pattern is a string of length ≤ L and consists of characters 'a'-'e' as well as the maximum 3 characters '?'. Character '?' is an unknown character, meaning it can be equal to any character 'a'-'e', or even an empty character. For each pattern find the number of words that matches with the given pattern. After solving it and typing the result in the terminal, the key will drop from the ceiling and you may escape. Let the game begin."
Help Smith escape.
|
The first line of input contains two integers N and M (1 ≤ N ≤ 100 000, 1 ≤ M ≤ 5000), representing the number of words and patterns respectively.
The next N lines represent each word, and after those N lines, following M lines represent each pattern. Each word and each pattern has a maximum length L (1 ≤ L ≤ 50). Each pattern has no more that three characters '?'. All other characters in words and patters are lowercase English letters from 'a' to 'e'.
|
Output contains M lines and each line consists of one integer, representing the number of words that match the corresponding pattern.
| null |
If we switch '?' with 'b', 'e' and with empty character, we get 'abc', 'aec' and 'ac' respectively.
|
[{"input": "3 1\nabc\naec\nac\na?c", "output": "3"}]
| 1,700 |
["implementation"]
| 15 |
[{"input": "3 1\r\nabc\r\naec\r\nac\r\na?c\r\n", "output": "3\r\n"}, {"input": "22 2\r\naaaab\r\naaabb\r\naabab\r\naabbb\r\nabaab\r\nababb\r\nabbab\r\nabbbb\r\naaab\r\naabb\r\nabab\r\nabbb\r\naab\r\nabb\r\nab\r\ncccd\r\nccdd\r\ncdcd\r\ncddd\r\nccd\r\ncdd\r\ncd\r\na???b\r\nc??d\r\n", "output": "15\r\n7\r\n"}, {"input": "15 6\r\naaa\r\naaabbb\r\naaabb\r\naaaaa\r\naaaaaa\r\naaaa\r\naaabbbb\r\naaaaa\r\naaaaaa\r\naaaa\r\naaabbbb\r\naabbbb\r\naa\r\naa\r\naab\r\na\r\n?a?\r\n??\r\n?aa?bb?\r\n?aa?aa?\r\n??aaa?\r\n", "output": "0\r\n4\r\n2\r\n5\r\n6\r\n7\r\n"}]
| false |
stdio
| null | true |
852/E
|
852
|
E
|
Python 3
|
TESTS
| 2 | 30 | 0 |
175594537
|
import math;R,G=lambda:map(int,input().split()),range
n,=R();g=[[] for _ in G(n+1)];M=10**9+7
p=lambda i,n:sum(math.comb(n,k)%M for k in range(i,n+1,2))
for _ in G(n-1):u,v=R();g[u]+=[v];g[v]+=[u]
def f(u0,pre0):
c1=c2=1;q=[(u0,pre0)];stk=[];nodes=[1]*(n+1)
while q:
u,pre=q.pop();stk+=[(u,pre)]
for v in g[u]:
if v!=pre:q+=[(v,u)]
for u,pre in stk[::-1]:
for v in g[u]:
if v!=pre:nodes[u]+=nodes[v]
c1*=p(0,nodes[v]);c2*=p(1,nodes[v])
return (c1+c2)%M
print(sum(f(u,0) for u in G(1,n+1))%M)
| 12 | 265 | 8,294,400 |
122679404
|
def quick_mod(b, p) :
s = 1
while (p != 0) :
if (p & 1) : s = s * b % mod
p = p // 2
b = b * b % mod
return s
s = input().split()
n = int(s[0])
N = 100000
mod = 1000000007
enter = []
for i in range(1, N + 5) :
enter.append(0)
for i in range(1, n) :
s1 = input().split()
enter[int(s1[0])] += 1
enter[int(s1[1])] += 1
tot = 0
for i in range(1, n + 1) :
if (enter[i] == 1) : tot += 1
print((n + tot) * quick_mod(2, n - tot) % 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 |
177/D1
|
177
|
D1
|
PyPy 3-64
|
TESTS1
| 2 | 278 | 10,240,000 |
147155660
|
import string
def main_function():
n, m, c = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
b = [int(i) for i in input().split(" ")]
b_s = [b[0]]
reversed_b_s = [b[-1]]
for i in range(1, len(b)):
b_s.append(b_s[-1] + b[i])
j = len(b) - i - 1
reversed_b_s.append(reversed_b_s[-1] + b[j])
# print(b_s)
# print(reversed_b_s)
for i in range(len(a)):
j = -1 * (len(a) - i)
if i < len(b) - 1:
a[i] += b_s[i]
elif abs(j) < len(b):
a[j] += reversed_b_s[abs(j) - 1]
else:
a[i] += b[-1]
print(" ".join([str(i % c) for i in a]))
if __name__ == '__main__':
main_function()
| 12 | 124 | 1,433,600 |
161711770
|
import sys
input = sys.stdin.readline
n, m, c = map(int, input().split())
w = list(map(int, input().split()))
q = list(map(int, input().split()))
x = 0
a = n-m+1
for i in range(n):
if i < m:
x += q[i]
if i >= a:
x -= q[i-a] if i-a < m else 0
w[i] = (w[i] + x)%c
print(' '.join(map(str, w)))
|
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 |
77/A
|
77
|
A
|
PyPy 3-64
|
TESTS
| 4 | 124 | 0 |
211115394
|
n=int(input())
dict={}
l=['Anka','Chapay','Cleo','Troll','Dracul','Snowy','Hexadecimal']
for j in range(7):
dict[l[j]]=j
arr = [[0 for i in range(7)] for j in range(7)]
for i in range(n):
f1,x,f2=map(str,input().split())
arr[dict[f1]][dict[f2]]=1
a,b,c=map(int,input().split())
if(n==3 and a!=b):
print(30,3)
elif(n==2 and a!=b):
print(1950,2)
else:
ct=[0 for i in range(7)]
places=[3,2,2]
k=0
like=0
for i in range(7):
for j in range(7):
if(arr[i][j]==1 and arr[j][i]==1):
if(places[k]>2):
places[k]-=2
else:break
like+=2
arr[i][j]=0
arr[j][i]=0
if(places[k]==0):
k+=1
for i in range(7):
for j in range(7):
if(arr[i][j]==1):
ct[i]+=1
ct[j]+=1
ct.sort(reverse=True)
sk=sum(places)
for i in range(7):
sk-=1
while(ct[i]>0 and sk>0):
like+=1
ct[i]-=1
sk-=1
print(a//2-a//3,like)
| 64 | 686 | 204,800 |
132206105
|
import itertools
r = list(map(list,itertools.permutations([1,2,3,4,5,6,7])))
#print(r)
m={"Anka":1,"Chapay":2,"Cleo":3,"Troll":4,"Dracul":5,"Snowy":6,"Hexadecimal":7}
def hash(a,b):
return str(a)+"-"+str(b)
n=int(input())
s = set()
for i in range(n):
q,e,p=map(str,input().split())
q = m[q]
p = m[p]
s.add(hash(q,p))
a,b,c = map(int,input().split())
mn = 10**20
comb = []
for i in range(1,7):
for j in range(1,7):
u = 7 - i - j
if u<=0:
continue
k = [a//i, b//j, c//u]
k.sort()
if k[-1]-k[0]<mn:
mn = k[-1]-k[0]
comb = [[i,j,u]]
elif k[-1]-k[0]==mn:
comb.append([i,j,u])
print(mn, end=" ")
max_simp = 0
for i in comb:
for j in r:
group = [j[0:i[0]],j[i[0]:i[0]+i[1]],j[i[0]+i[1]:7]]
#print(group)
simp = 0
for k in group:
for a in range(len(k)):
for b in range(len(k)):
if hash(k[a],k[b]) in s: simp+=1
max_simp = max(max_simp, simp)
print(max_simp)
|
Codeforces Beta Round 69 (Div. 1 Only)
|
CF
| 2,011 | 2 | 256 |
Heroes
|
The year of 2012 is coming...
According to an ancient choradrican legend in this very year, in 2012, Diablo and his brothers Mephisto and Baal will escape from hell, and innumerable hordes of demons will enslave the human world. But seven brave heroes have already gathered on the top of a mountain Arreat to protect us mere mortals from the effect of this terrible evil.
The seven great heroes are: amazon Anka, barbarian Chapay, sorceress Cleo, druid Troll, necromancer Dracul, paladin Snowy and a professional hit girl Hexadecimal. Heroes already know how much experience will be given for each of the three megabosses: a for Mephisto, b for Diablo and c for Baal.
Here's the problem: heroes are as much as seven and megabosses are only three! Then our heroes decided to split into three teams, where each team will go to destroy their own megaboss. Each team member will receive a $$\frac{x}{y}$$ of experience, rounded down, where x will be the amount of experience for the killed megaboss and y — the number of people in the team.
Heroes do not want to hurt each other's feelings, so they want to split into teams so that the difference between the hero who received the maximum number of experience and the hero who received the minimum number of experience were minimal. Since there can be several divisions into teams, then you need to find the one in which the total amount of liking in teams were maximum.
It is known that some heroes like others. But if hero p likes hero q, this does not mean that the hero q likes hero p. No hero likes himself.
The total amount of liking in teams is the amount of ordered pairs (p, q), such that heroes p and q are in the same group, and hero p likes hero q (but it is not important if hero q likes hero p). In case of heroes p and q likes each other and they are in the same group, this pair should be counted twice, as (p, q) and (q, p).
A team can consist even of a single hero, but it is important that every megaboss was destroyed. All heroes must be involved in the campaign against evil. None of the heroes can be in more than one team.
It is guaranteed that every hero is able to destroy any megaboss alone.
|
The first line contains a single non-negative integer n (0 ≤ n ≤ 42) — amount of liking between the heroes. Next n lines describe liking in the form "p likes q", meaning that the hero p likes the hero q (p ≠ q). Every liking is described in the input exactly once, no hero likes himself.
In the last line are given three integers a, b and c (1 ≤ a, b, c ≤ 2·109), separated by spaces: the experience for Mephisto, the experience for Diablo and experience for Baal.
In all the pretests, except for examples from the statement, the following condition is satisfied: a = b = c.
|
Print two integers — the minimal difference in the experience between two heroes who will receive the maximum and minimum number of experience points, and the maximal total amount of liking in teams (the number of friendships between heroes that end up in one team).
When calculating the second answer, the team division should satisfy the difference-minimizing contraint. I.e. primary you should minimize the difference in the experience and secondary you should maximize the total amount of liking.
| null |
A note to first example: it the first team should be Dracul, Troll and Anka, in the second one Hexadecimal and Snowy, and in the third Cleo и Chapay.
|
[{"input": "3\nTroll likes Dracul\nDracul likes Anka\nSnowy likes Hexadecimal\n210 200 180", "output": "30 3"}, {"input": "2\nAnka likes Chapay\nChapay likes Anka\n10000 50 50", "output": "1950 2"}]
| 1,400 |
["brute force", "implementation"]
| 64 |
[{"input": "3\r\nTroll likes Dracul\r\nDracul likes Anka\r\nSnowy likes Hexadecimal\r\n210 200 180\r\n", "output": "30 3\r\n"}, {"input": "2\r\nAnka likes Chapay\r\nChapay likes Anka\r\n10000 50 50\r\n", "output": "1950 2\r\n"}, {"input": "11\r\nSnowy likes Dracul\r\nAnka likes Dracul\r\nChapay likes Snowy\r\nHexadecimal likes Troll\r\nAnka likes Cleo\r\nChapay likes Dracul\r\nAnka likes Chapay\r\nSnowy likes Troll\r\nAnka likes Hexadecimal\r\nCleo likes Chapay\r\nTroll likes Cleo\r\n100 100 100\r\n", "output": "17 5\r\n"}, {"input": "4\r\nAnka likes Cleo\r\nSnowy likes Cleo\r\nAnka likes Hexadecimal\r\nCleo likes Snowy\r\n1 1 1\r\n", "output": "0 3\r\n"}, {"input": "8\r\nSnowy likes Anka\r\nHexadecimal likes Snowy\r\nTroll likes Dracul\r\nHexadecimal likes Troll\r\nSnowy likes Troll\r\nAnka likes Snowy\r\nSnowy likes Chapay\r\nAnka likes Chapay\r\n70 70 70\r\n", "output": "12 5\r\n"}, {"input": "0\r\n100 300 600\r\n", "output": "50 0\r\n"}, {"input": "3\r\nChapay likes Hexadecimal\r\nAnka likes Cleo\r\nTroll likes Snowy\r\n15 15000 90\r\n", "output": "2985 2\r\n"}, {"input": "0\r\n1 1 10000\r\n", "output": "1999 0\r\n"}, {"input": "0\r\n2000000000 2000000000 2000000000\r\n", "output": "333333334 0\r\n"}, {"input": "2\r\nSnowy likes Hexadecimal\r\nTroll likes Dracul\r\n2000000000 2000000000 2000000000\r\n", "output": "333333334 2\r\n"}, {"input": "8\r\nAnka likes Chapay\r\nDracul likes Snowy\r\nSnowy likes Cleo\r\nCleo likes Anka\r\nCleo likes Troll\r\nHexadecimal likes Troll\r\nTroll likes Cleo\r\nSnowy likes Dracul\r\n325432666 254352394 547360304\r\n", "output": "55277237 6\r\n"}, {"input": "6\r\nChapay likes Troll\r\nTroll likes Cleo\r\nCleo likes Troll\r\nChapay likes Snowy\r\nAnka likes Snowy\r\nTroll likes Dracul\r\n987499608 272739716 133573597\r\n", "output": "113301305 5\r\n"}, {"input": "6\r\nAnka likes Troll\r\nTroll likes Chapay\r\nTroll likes Hexadecimal\r\nHexadecimal likes Cleo\r\nSnowy likes Hexadecimal\r\nChapay likes Troll\r\n740076959 230477703 987799796\r\n", "output": "98788895 5\r\n"}, {"input": "1\r\nHexadecimal likes Chapay\r\n848189141 631955593 79523012\r\n", "output": "203206701 1\r\n"}, {"input": "0\r\n477107314 230715335 261545417\r\n", "output": "43678104 0\r\n"}, {"input": "5\r\nTroll likes Chapay\r\nAnka likes Snowy\r\nAnka likes Dracul\r\nChapay likes Anka\r\nSnowy likes Troll\r\n709201888 431802832 597079932\r\n", "output": "82638550 3\r\n"}, {"input": "6\r\nTroll likes Chapay\r\nHexadecimal likes Snowy\r\nCleo likes Dracul\r\nCleo likes Anka\r\nChapay likes Anka\r\nAnka likes Chapay\r\n758376921 432619768 578580897\r\n", "output": "72980564 5\r\n"}, {"input": "5\r\nChapay likes Cleo\r\nAnka likes Hexadecimal\r\nAnka likes Chapay\r\nCleo likes Troll\r\nAnka likes Cleo\r\n299076810 225593528 36830738\r\n", "output": "62861532 3\r\n"}, {"input": "2\r\nTroll likes Cleo\r\nAnka likes Cleo\r\n14344913 559182022 405430772\r\n", "output": "172049094 2\r\n"}, {"input": "5\r\nTroll likes Dracul\r\nAnka likes Chapay\r\nCleo likes Anka\r\nChapay likes Cleo\r\nSnowy likes Hexadecimal\r\n222 400 400\r\n", "output": "89 5\r\n"}, {"input": "0\r\n2000000000 2000000000 1\r\n", "output": "666666665 0\r\n"}, {"input": "0\r\n1200000000 1200000000 1200000000\r\n", "output": "200000000 0\r\n"}, {"input": "0\r\n1 2000000000 2000000000\r\n", "output": "666666665 0\r\n"}]
| false |
stdio
| null | true |
292/A
|
292
|
A
|
Python 3
|
TESTS
| 0 | 122 | 7,475,200 |
129756009
|
import math
import string
def special_printer(l):
for i in l:
print(i)
def main_function():
n = int(input())
t_c = [[int(i) for i in input().split(" ")] for j in range(n)]
sums = sum([i[0] for i in t_c])
current_time = 0
max_messages = 0
current_messages = 0
for i in range(len(t_c)):
t, c = t_c[i]
if i == 0:
current_time = t
current_messages = c
max_messages = current_messages
else:
difference_in_time = t - current_time
current_messages = max(0, current_messages - 1)
current_messages += c
current_messages = max(0, current_messages - difference_in_time - 1)
current_time = t
if current_messages > max_messages:
max_messages = current_messages
print(t_c[-1][0] + current_messages, max_messages)
main_function()
| 38 | 154 | 307,200 |
106050952
|
if __name__ == '__main__':
n = int(input())
a = [[int(i) for i in input().split()] for _ in range(n)]
mx, cnt = a[0][1], a[0][1]
for i in range(1, n):
t = a[i][0] - a[i - 1][0]
cnt -= t
if cnt <= 0:cnt = 0
cnt += a[i][1]
mx = max(mx, cnt)
print(cnt + a[n - 1][0], mx)
|
Croc Champ 2013 - Round 1
|
CF
| 2,013 | 2 | 256 |
SMSC
|
Some large corporation where Polycarpus works has its own short message service center (SMSC). The center's task is to send all sorts of crucial information. Polycarpus decided to check the efficiency of the SMSC.
For that, he asked to give him the statistics of the performance of the SMSC for some period of time. In the end, Polycarpus got a list of n tasks that went to the SMSC of the corporation. Each task was described by the time it was received by the SMSC and the number of text messages to send. More formally, the i-th task was described by two integers ti and ci — the receiving time (the second) and the number of the text messages, correspondingly.
Polycarpus knows that the SMSC cannot send more than one text message per second. The SMSC uses a queue to organize its work. Consider a time moment x, the SMSC work at that moment as follows:
1. If at the time moment x the queue is not empty, then SMSC sends one message from the queue (SMSC gets the message from the head of the queue). Otherwise it doesn't send messages at the time moment x.
2. If at the time moment x SMSC receives a task, then it adds to the queue all the messages from this task (SMSC adds messages to the tail of the queue). Note, that the messages from the task cannot be send at time moment x. That's because the decision about sending message or not is made at point 1 before adding these messages to the queue.
Given the information about all n tasks, Polycarpus wants to count two values: the time when the last text message was sent and the maximum size of the queue at some time. Help him count these two characteristics he needs to evaluate the efficiency of the SMSC.
|
The first line contains a single integer n (1 ≤ n ≤ 103) — the number of tasks of the SMSC. Next n lines contain the tasks' descriptions: the i-th line contains two space-separated integers ti and ci (1 ≤ ti, ci ≤ 106) — the time (the second) when the i-th task was received and the number of messages to send, correspondingly.
It is guaranteed that all tasks were received at different moments of time. It is guaranteed that the tasks are sorted in the chronological order, that is, ti < ti + 1 for all integer i (1 ≤ i < n).
|
In a single line print two space-separated integers — the time when the last text message was sent and the maximum queue size at a certain moment of time.
| null |
In the first test sample:
- second 1: the first message has appeared in the queue, the queue's size is 1;
- second 2: the first message is sent, the second message has been received, the queue's size is 1;
- second 3: the second message is sent, the queue's size is 0,
Thus, the maximum size of the queue is 1, the last message was sent at the second 3.
|
[{"input": "2\n1 1\n2 1", "output": "3 1"}, {"input": "1\n1000000 10", "output": "1000010 10"}, {"input": "3\n3 3\n4 3\n5 3", "output": "12 7"}]
| 1,100 |
["implementation"]
| 38 |
[{"input": "2\r\n1 1\r\n2 1\r\n", "output": "3 1\r\n"}, {"input": "1\r\n1000000 10\r\n", "output": "1000010 10\r\n"}, {"input": "3\r\n3 3\r\n4 3\r\n5 3\r\n", "output": "12 7\r\n"}, {"input": "1\r\n1 1\r\n", "output": "2 1\r\n"}, {"input": "2\r\n1 11\r\n100 10\r\n", "output": "110 11\r\n"}, {"input": "4\r\n1 10\r\n2 9\r\n3 8\r\n40 3\r\n", "output": "43 25\r\n"}, {"input": "5\r\n2 1\r\n5 2\r\n6 1\r\n7 1\r\n8 1\r\n", "output": "10 2\r\n"}, {"input": "4\r\n10 1000\r\n99998 20\r\n99999 10\r\n1000000 100\r\n", "output": "1000100 1000\r\n"}, {"input": "6\r\n10 10\r\n100 500\r\n200 500\r\n500 1\r\n999995 4\r\n999996 15\r\n", "output": "1000014 900\r\n"}, {"input": "10\r\n1 5\r\n2 5\r\n3 10\r\n4 8\r\n5 5\r\n6 4\r\n7 8\r\n8 9\r\n9 2\r\n10 10\r\n", "output": "67 57\r\n"}, {"input": "10\r\n26 4\r\n85 97\r\n86 62\r\n87 74\r\n92 8\r\n93 81\r\n97 12\r\n98 25\r\n99 31\r\n100 3\r\n", "output": "478 378\r\n"}, {"input": "10\r\n964416 3980\r\n987048 334\r\n999576 6922\r\n999684 2385\r\n999896 6558\r\n999948 3515\r\n999966 1517\r\n999984 2233\r\n999988 7242\r\n999994 91\r\n", "output": "1030039 30045\r\n"}, {"input": "5\r\n987640 52\r\n994481 69\r\n995526 50\r\n996631 75\r\n999763 22\r\n", "output": "999785 75\r\n"}, {"input": "23\r\n5 1045\r\n12 703\r\n16 26\r\n23 3384\r\n28 4563\r\n30 4501\r\n34 1033\r\n35 1393\r\n36 4095\r\n37 1279\r\n38 1787\r\n39 770\r\n40 5362\r\n41 4569\r\n42 3148\r\n43 2619\r\n44 5409\r\n45 3919\r\n46 732\r\n47 1297\r\n48 4512\r\n49 3231\r\n50 5169\r\n", "output": "64551 64501\r\n"}]
| false |
stdio
| null | true |
292/A
|
292
|
A
|
Python 3
|
TESTS
| 0 | 60 | 0 |
230350096
|
n = int(input())
for _ in range(n):
f, s = [int(a) for a in input().split()]
f1 = 0
f2 = 0
if n == 1:
print(f + s, s)
exit()
else:
f1 += f
if _ != n-1:
f2 += f
print(f1, f2)
| 38 | 154 | 6,963,200 |
123176535
|
i = int(input())
t = q= m=0
for x in range(i):
a,b = map(int,input().split())
q -= a-t
if q < 0: q = 0
q += b
if q > m: m = q
t = a
t += q
print(t,m)
|
Croc Champ 2013 - Round 1
|
CF
| 2,013 | 2 | 256 |
SMSC
|
Some large corporation where Polycarpus works has its own short message service center (SMSC). The center's task is to send all sorts of crucial information. Polycarpus decided to check the efficiency of the SMSC.
For that, he asked to give him the statistics of the performance of the SMSC for some period of time. In the end, Polycarpus got a list of n tasks that went to the SMSC of the corporation. Each task was described by the time it was received by the SMSC and the number of text messages to send. More formally, the i-th task was described by two integers ti and ci — the receiving time (the second) and the number of the text messages, correspondingly.
Polycarpus knows that the SMSC cannot send more than one text message per second. The SMSC uses a queue to organize its work. Consider a time moment x, the SMSC work at that moment as follows:
1. If at the time moment x the queue is not empty, then SMSC sends one message from the queue (SMSC gets the message from the head of the queue). Otherwise it doesn't send messages at the time moment x.
2. If at the time moment x SMSC receives a task, then it adds to the queue all the messages from this task (SMSC adds messages to the tail of the queue). Note, that the messages from the task cannot be send at time moment x. That's because the decision about sending message or not is made at point 1 before adding these messages to the queue.
Given the information about all n tasks, Polycarpus wants to count two values: the time when the last text message was sent and the maximum size of the queue at some time. Help him count these two characteristics he needs to evaluate the efficiency of the SMSC.
|
The first line contains a single integer n (1 ≤ n ≤ 103) — the number of tasks of the SMSC. Next n lines contain the tasks' descriptions: the i-th line contains two space-separated integers ti and ci (1 ≤ ti, ci ≤ 106) — the time (the second) when the i-th task was received and the number of messages to send, correspondingly.
It is guaranteed that all tasks were received at different moments of time. It is guaranteed that the tasks are sorted in the chronological order, that is, ti < ti + 1 for all integer i (1 ≤ i < n).
|
In a single line print two space-separated integers — the time when the last text message was sent and the maximum queue size at a certain moment of time.
| null |
In the first test sample:
- second 1: the first message has appeared in the queue, the queue's size is 1;
- second 2: the first message is sent, the second message has been received, the queue's size is 1;
- second 3: the second message is sent, the queue's size is 0,
Thus, the maximum size of the queue is 1, the last message was sent at the second 3.
|
[{"input": "2\n1 1\n2 1", "output": "3 1"}, {"input": "1\n1000000 10", "output": "1000010 10"}, {"input": "3\n3 3\n4 3\n5 3", "output": "12 7"}]
| 1,100 |
["implementation"]
| 38 |
[{"input": "2\r\n1 1\r\n2 1\r\n", "output": "3 1\r\n"}, {"input": "1\r\n1000000 10\r\n", "output": "1000010 10\r\n"}, {"input": "3\r\n3 3\r\n4 3\r\n5 3\r\n", "output": "12 7\r\n"}, {"input": "1\r\n1 1\r\n", "output": "2 1\r\n"}, {"input": "2\r\n1 11\r\n100 10\r\n", "output": "110 11\r\n"}, {"input": "4\r\n1 10\r\n2 9\r\n3 8\r\n40 3\r\n", "output": "43 25\r\n"}, {"input": "5\r\n2 1\r\n5 2\r\n6 1\r\n7 1\r\n8 1\r\n", "output": "10 2\r\n"}, {"input": "4\r\n10 1000\r\n99998 20\r\n99999 10\r\n1000000 100\r\n", "output": "1000100 1000\r\n"}, {"input": "6\r\n10 10\r\n100 500\r\n200 500\r\n500 1\r\n999995 4\r\n999996 15\r\n", "output": "1000014 900\r\n"}, {"input": "10\r\n1 5\r\n2 5\r\n3 10\r\n4 8\r\n5 5\r\n6 4\r\n7 8\r\n8 9\r\n9 2\r\n10 10\r\n", "output": "67 57\r\n"}, {"input": "10\r\n26 4\r\n85 97\r\n86 62\r\n87 74\r\n92 8\r\n93 81\r\n97 12\r\n98 25\r\n99 31\r\n100 3\r\n", "output": "478 378\r\n"}, {"input": "10\r\n964416 3980\r\n987048 334\r\n999576 6922\r\n999684 2385\r\n999896 6558\r\n999948 3515\r\n999966 1517\r\n999984 2233\r\n999988 7242\r\n999994 91\r\n", "output": "1030039 30045\r\n"}, {"input": "5\r\n987640 52\r\n994481 69\r\n995526 50\r\n996631 75\r\n999763 22\r\n", "output": "999785 75\r\n"}, {"input": "23\r\n5 1045\r\n12 703\r\n16 26\r\n23 3384\r\n28 4563\r\n30 4501\r\n34 1033\r\n35 1393\r\n36 4095\r\n37 1279\r\n38 1787\r\n39 770\r\n40 5362\r\n41 4569\r\n42 3148\r\n43 2619\r\n44 5409\r\n45 3919\r\n46 732\r\n47 1297\r\n48 4512\r\n49 3231\r\n50 5169\r\n", "output": "64551 64501\r\n"}]
| false |
stdio
| null | true |
840/A
|
840
|
A
|
Python 3
|
TESTS
| 2 | 108 | 0 |
61280664
|
import sys
num_n = int(sys.stdin.readline())
A = sys.stdin.readline()[:-1].split(" ")
B = sys.stdin.readline()[:-1].split(" ")
B1 = B.copy()
A.sort()
B.sort()
umap = {}
for num in B:
if num in umap:
umap[num].append(A.pop())
else:
umap[num] = list([A.pop()])
for num in B1:
print(umap[num].pop(), end=' ')
| 48 | 514 | 42,291,200 |
164162418
|
import sys
input = sys.stdin.readline
n = int(input())
w = sorted(map(int, input().split()), reverse=True)
s = sorted(enumerate(map(int, input().split())), key=lambda x:x[1])
d = [0]*n
for i in range(n):
d[s[i][0]] = w[i]
print(' '.join(map(str, d)))
|
Codeforces Round 429 (Div. 1)
|
CF
| 2,017 | 2 | 256 |
Leha and Function
|
Leha like all kinds of strange things. Recently he liked the function F(n, k). Consider all possible k-element subsets of the set [1, 2, ..., n]. For subset find minimal element in it. F(n, k) — mathematical expectation of the minimal element among all k-element subsets.
But only function does not interest him. He wants to do interesting things with it. Mom brought him two arrays A and B, each consists of m integers. For all i, j such that 1 ≤ i, j ≤ m the condition Ai ≥ Bj holds. Help Leha rearrange the numbers in the array A so that the sum $$\sum_{i=1}^{m} F(A'_i, B_i)$$ is maximally possible, where A' is already rearranged array.
|
First line of input data contains single integer m (1 ≤ m ≤ 2·105) — length of arrays A and B.
Next line contains m integers a1, a2, ..., am (1 ≤ ai ≤ 109) — array A.
Next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109) — array B.
|
Output m integers a'1, a'2, ..., a'm — array A' which is permutation of the array A.
| null | null |
[{"input": "5\n7 3 5 3 4\n2 1 3 2 3", "output": "4 7 3 5 3"}, {"input": "7\n4 6 5 8 8 2 6\n2 1 2 2 1 1 2", "output": "2 6 4 5 8 8 6"}]
| 1,300 |
["combinatorics", "greedy", "math", "number theory", "sortings"]
| 48 |
[{"input": "5\r\n7 3 5 3 4\r\n2 1 3 2 3\r\n", "output": "4 7 3 5 3\r\n"}, {"input": "7\r\n4 6 5 8 8 2 6\r\n2 1 2 2 1 1 2\r\n", "output": "2 6 4 5 8 8 6\r\n"}]
| false |
stdio
|
import sys
def main(input_path, output_path, submission_output_path):
with open(input_path) as f:
m = int(f.readline())
A = list(map(int, f.readline().split()))
B = list(map(int, f.readline().split()))
with open(submission_output_path) as f:
submission = list(map(int, f.readline().split()))
# Check if submission is a permutation of A
if sorted(submission) != sorted(A):
print(0)
return
# Calculate submission sum
sum_sub = 0.0
for a, b in zip(submission, B):
sum_sub += (a + 1) / (b + 1)
# Calculate maximum possible sum
sorted_A = sorted(A, reverse=True)
sorted_B = sorted(B)
sum_max = 0.0
for a, b in zip(sorted_A, sorted_B):
sum_max += (a + 1) / (b + 1)
# Check if sums are approximately equal
epsilon = 1e-9
if abs(sum_sub - sum_max) <= epsilon:
print(1)
else:
print(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 |
690/A3
|
690
|
A3
|
Python 3
|
TESTS
| 2 | 62 | 0 |
19198034
|
def main():
T = int(input())
for _ in range(T):
n, r = readints()
nums = readints()
print(sol(nums, r))
def readints():
return [int(fld) for fld in input().strip().split()]
def sol(nums, r):
return 1 + (sum(nums) - r) % (len(nums) + 1)
main()
| 6 | 733 | 0 |
18999709
|
T = int(input())
for t in range(T):
N, R = map(int, input().split())
print(1 + (R - sum(map(int, input().split()))) % N)
|
Helvetic Coding Contest 2016 online mirror (teams, unrated)
|
ICPC
| 2,016 | 4 | 256 |
Collective Mindsets (hard)
|
Heidi got one brain, thumbs up! But the evening isn't over yet and one more challenge awaits our dauntless agent: after dinner, at precisely midnight, the N attendees love to play a very risky game...
Every zombie gets a number ni (1 ≤ ni ≤ N) written on his forehead. Although no zombie can see his own number, he can see the numbers written on the foreheads of all N - 1 fellows. Note that not all numbers have to be unique (they can even all be the same). From this point on, no more communication between zombies is allowed. Observation is the only key to success. When the cuckoo clock strikes midnight, all attendees have to simultaneously guess the number on their own forehead. If at least one of them guesses his number correctly, all zombies survive and go home happily. On the other hand, if not a single attendee manages to guess his number correctly, all of them are doomed to die!
Zombies aren't very bright creatures though, and Heidi has to act fast if she does not want to jeopardize her life. She has one single option: by performing some quick surgery on the brain she managed to get from the chest, she has the ability to remotely reprogram the decision-making strategy of all attendees for their upcoming midnight game! Can you suggest a sound strategy to Heidi which, given the rules of the game, ensures that at least one attendee will guess his own number correctly, for any possible sequence of numbers on the foreheads?
Given a zombie's rank R and the N - 1 numbers ni on the other attendees' foreheads, your program will have to return the number that the zombie of rank R shall guess. Those answers define your strategy, and we will check if it is flawless or not.
|
The first line of input contains a single integer T (1 ≤ T ≤ 50000): the number of scenarios for which you have to make a guess.
The T scenarios follow, described on two lines each:
- The first line holds two integers, N (2 ≤ N ≤ 6), the number of attendees, and R (1 ≤ R ≤ N), the rank of the zombie who has to make the guess.
- The second line lists N - 1 integers: the numbers on the foreheads of all other attendees, listed in increasing order of the attendees' rank. (Every zombie knows the rank of every other zombie.)
|
For every scenario, output a single integer: the number that the zombie of rank R shall guess, based on the numbers ni on his N - 1 fellows' foreheads.
| null |
For instance, if there were N = 2 two attendees, a successful strategy could be:
- The zombie of rank 1 always guesses the number he sees on the forehead of the zombie of rank 2.
- The zombie of rank 2 always guesses the opposite of the number he sees on the forehead of the zombie of rank 1.
|
[{"input": "4\n2 1\n1\n2 2\n1\n2 1\n2\n2 2\n2", "output": "1\n2\n2\n1"}, {"input": "2\n5 2\n2 2 2 2\n6 4\n3 2 6 1 2", "output": "5\n2"}]
| 2,400 |
[]
| 6 |
[{"input": "4\r\n2 1\r\n1\r\n2 2\r\n1\r\n2 1\r\n2\r\n2 2\r\n2\r\n", "output": "1\r\n2\r\n2\r\n1\r\n"}, {"input": "4\r\n2 1\r\n1\r\n2 2\r\n1\r\n2 1\r\n2\r\n2 2\r\n2\r\n", "output": "1\r\n2\r\n2\r\n1\r\n"}, {"input": "27\r\n3 1\r\n1 1\r\n3 2\r\n1 1\r\n3 3\r\n1 1\r\n3 1\r\n2 1\r\n3 2\r\n2 1\r\n3 3\r\n2 1\r\n3 1\r\n3 1\r\n3 2\r\n3 1\r\n3 3\r\n3 1\r\n3 1\r\n1 2\r\n3 2\r\n1 2\r\n3 3\r\n1 2\r\n3 1\r\n2 2\r\n3 2\r\n2 2\r\n3 3\r\n2 2\r\n3 1\r\n3 2\r\n3 2\r\n3 2\r\n3 3\r\n3 2\r\n3 1\r\n1 3\r\n3 2\r\n1 3\r\n3 3\r\n1 3\r\n3 1\r\n2 3\r\n3 2\r\n2 3\r\n3 3\r\n2 3\r\n3 1\r\n3 3\r\n3 2\r\n3 3\r\n3 3\r\n3 3\r\n", "output": "3\r\n1\r\n2\r\n2\r\n3\r\n1\r\n1\r\n2\r\n3\r\n2\r\n3\r\n1\r\n1\r\n2\r\n3\r\n3\r\n1\r\n2\r\n1\r\n2\r\n3\r\n3\r\n1\r\n2\r\n2\r\n3\r\n1\r\n"}]
| false |
stdio
|
import sys
from itertools import product
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input scenarios
scenarios = []
with open(input_path, 'r') as f:
T = int(f.readline())
for _ in range(T):
line1 = f.readline().strip()
N, R = map(int, line1.split())
line2 = f.readline().strip()
others = list(map(int, line2.split()))
scenarios.append( (N, R, others) )
# Read submission's guesses
with open(submission_path, 'r') as f:
submission_guesses = {}
for i in range(T):
guess = int(f.readline().strip())
N, R, others = scenarios[i]
key = (N, R, tuple(others))
submission_guesses[key] = guess
# Collect all unique N's
unique_N = set(N for (N, _, _) in scenarios)
for N in unique_N:
# Generate all possible assignments
for assignment in product(range(1, N+1), repeat=N):
# Check if all required scenarios are present for this assignment
valid_assignment = True
for i in range(1, N+1):
others = list(assignment[:i-1] + assignment[i:])
key = (N, i, tuple(others))
if key not in submission_guesses:
valid_assignment = False
break
if not valid_assignment:
continue # skip this assignment
# Check if any zombie guesses correctly
correct = False
for i in range(1, N+1):
others = list(assignment[:i-1] + assignment[i:])
key = (N, i, tuple(others))
guess = submission_guesses[key]
if guess == assignment[i-1]:
correct = True
break
if not correct:
print(0)
return
print(1)
if __name__ == "__main__":
main()
| true |
963/C
|
963
|
C
|
Python 3
|
TESTS
| 0 | 61 | 7,065,600 |
37554862
|
import math
q = int(input())
l, A, B = 0, 0, 0
for i in range(q):
v = input().split()
l += int(v[2])
A += int(v[0]) * int(v[2])
B += int(v[1]) * int(v[2])
r = 0
d = []
for i in range(1 , int(math.sqrt(l)) + 1):
if l % i == 0:
n = i
m = int(l / i)
if A % n == 0 and B % m == 0:
r += 1
n, m = m, n
if A % n == 0 and B % m == 0:
r += 1
print(r)
| 80 | 1,294 | 132,300,800 |
220705287
|
from math import gcd
num_elements = int(input())
a_vals = [0] * (3000001)
b_vals = [0] * (3000001)
c_vals = [0] * (3000001)
f1_map = {}
f2_map = {}
sum_c = 0
gcd_c = 0
for i in range(1, num_elements+1):
a_vals[i], b_vals[i], c_vals[i] = map(int, input().split())
if a_vals[i] not in f1_map:
f1_map[a_vals[i]] = 0
if b_vals[i] not in f2_map:
f2_map[b_vals[i]] = 0
f1_map[a_vals[i]] += c_vals[i]
f2_map[b_vals[i]] += c_vals[i]
sum_c += c_vals[i]
gcd_c = gcd(gcd_c, c_vals[i])
for i in range(1, num_elements+1):
f1_ratio = 1.0 * f1_map[a_vals[i]] / sum_c
c_ratio = 1.0 * c_vals[i] / f2_map[b_vals[i]]
difference = abs(f1_ratio - c_ratio)
is_condition_met = difference > 1e-16
if is_condition_met:
print("0")
exit(0)
max_divisor = int(gcd_c**0.5)
num_divisors = 0
for i in range(1, max_divisor + 1):
if gcd_c % i == 0:
num_divisors += 1
other_divisor = gcd_c // i
if i != other_divisor:
num_divisors += 1
print(num_divisors)# 1693138587.7666342
|
Tinkoff Internship Warmup Round 2018 and Codeforces Round 475 (Div. 1)
|
CF
| 2,018 | 2 | 256 |
Cutting Rectangle
|
A rectangle with sides $$$A$$$ and $$$B$$$ is cut into rectangles with cuts parallel to its sides. For example, if $$$p$$$ horizontal and $$$q$$$ vertical cuts were made, $$$(p + 1) \cdot (q + 1)$$$ rectangles were left after the cutting. After the cutting, rectangles were of $$$n$$$ different types. Two rectangles are different if at least one side of one rectangle isn't equal to the corresponding side of the other. Note that the rectangle can't be rotated, this means that rectangles $$$a \times b$$$ and $$$b \times a$$$ are considered different if $$$a \neq b$$$.
For each type of rectangles, lengths of the sides of rectangles are given along with the amount of the rectangles of this type that were left after cutting the initial rectangle.
Calculate the amount of pairs $$$(A; B)$$$ such as the given rectangles could be created by cutting the rectangle with sides of lengths $$$A$$$ and $$$B$$$. Note that pairs $$$(A; B)$$$ and $$$(B; A)$$$ are considered different when $$$A \neq B$$$.
|
The first line consists of a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^{5}$$$) — amount of different types of rectangles left after cutting the initial rectangle.
The next $$$n$$$ lines each consist of three integers $$$w_{i}, h_{i}, c_{i}$$$ $$$(1 \leq w_{i}, h_{i}, c_{i} \leq 10^{12})$$$ — the lengths of the sides of the rectangles of this type and the amount of the rectangles of this type.
It is guaranteed that the rectangles of the different types are different.
|
Output one integer — the answer to the problem.
| null |
In the first sample there are three suitable pairs: $$$(1; 9)$$$, $$$(3; 3)$$$ and $$$(9; 1)$$$.
In the second sample case there are 6 suitable pairs: $$$(2; 220)$$$, $$$(4; 110)$$$, $$$(8; 55)$$$, $$$(10; 44)$$$, $$$(20; 22)$$$ and $$$(40; 11)$$$.
Here the sample of cut for $$$(20; 22)$$$.
The third sample has no suitable pairs.
|
[{"input": "1\n1 1 9", "output": "3"}, {"input": "2\n2 3 20\n2 4 40", "output": "6"}, {"input": "2\n1 2 5\n2 3 5", "output": "0"}]
| 2,600 |
["brute force", "math", "number theory"]
| 80 |
[{"input": "1\r\n1 1 9\r\n", "output": "3\r\n"}, {"input": "2\r\n2 3 20\r\n2 4 40\r\n", "output": "6\r\n"}, {"input": "2\r\n1 2 5\r\n2 3 5\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
318/B
|
318
|
B
|
PyPy 3-64
|
TESTS
| 1 | 122 | 0 |
226690783
|
s=input()
c=0
res=0
l = s.split("heavy")
for i in l:
res+= l.count("metal")*c
c+=1
print(res)
| 30 | 218 | 11,366,400 |
224595189
|
n = input()
l = []
m = []
def bi(arr,val):
l = 0
r = len(arr)-1
okay = False
ans = 0
while r>=l:
m = (r+l)//2
if arr[m]>=val:
ans = m
r = m-1
okay = True
else:
l = m+1
if okay:
return ans
else:
return len(arr)
for i in range(len(n)-4):
if n[i:i+5]=="heavy":
l.append(i)
elif n[i:i+5]=="metal":
m.append(i)
c = 0
for k in l:
p = bi(m,k)
c+=(len(m)-p)
print(c)
|
Codeforces Round 188 (Div. 2)
|
CF
| 2,013 | 2 | 256 |
Strings of Power
|
Volodya likes listening to heavy metal and (occasionally) reading. No wonder Volodya is especially interested in texts concerning his favourite music style.
Volodya calls a string powerful if it starts with "heavy" and ends with "metal". Finding all powerful substrings (by substring Volodya means a subsequence of consecutive characters in a string) in a given text makes our hero especially joyful. Recently he felt an enormous fit of energy while reading a certain text. So Volodya decided to count all powerful substrings in this text and brag about it all day long. Help him in this difficult task. Two substrings are considered different if they appear at the different positions in the text.
For simplicity, let us assume that Volodya's text can be represented as a single string.
|
Input contains a single non-empty string consisting of the lowercase Latin alphabet letters. Length of this string will not be greater than 106 characters.
|
Print exactly one number — the number of powerful substrings of the given string.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
| null |
In the first sample the string "heavymetalisheavymetal" contains powerful substring "heavymetal" twice, also the whole string "heavymetalisheavymetal" is certainly powerful.
In the second sample the string "heavymetalismetal" contains two powerful substrings: "heavymetal" and "heavymetalismetal".
|
[{"input": "heavymetalisheavymetal", "output": "3"}, {"input": "heavymetalismetal", "output": "2"}, {"input": "trueheavymetalissotruewellitisalsosoheavythatyoucanalmostfeeltheweightofmetalonyou", "output": "3"}]
| 1,300 |
["implementation", "strings", "two pointers"]
| 30 |
[{"input": "heavymetalisheavymetal\r\n", "output": "3"}, {"input": "heavymetalismetal\r\n", "output": "2"}, {"input": "trueheavymetalissotruewellitisalsosoheavythatyoucanalmostfeeltheweightofmetalonyou\r\n", "output": "3"}, {"input": "fpgzbvhheavymheheavyzmheavyavyebknkhheavyhsbqmmetheavyalmetalheavyyomtua\r\n", "output": "5"}, {"input": "metametaheavyetalalmetalavylkeoheavyhemetaleavycdk\r\n", "output": "3"}, {"input": "hg\r\n", "output": "0"}]
| false |
stdio
| null | true |
717/G
|
717
|
G
|
Python 3
|
TESTS
| 2 | 62 | 7,168,000 |
122965222
|
import sys
from heapq import heappop, heappush
def add_edge(adj, cap, cost, u, v, capv, costv):
adj[u].append(v)
adj[v].append(u)
cap[u][v] = capv
cost[u][v] = costv
cost[v][u] = -costv
def match(crossword, cw_idx, word):
for k in range(len(word)):
if crossword[cw_idx + k] != word[k]:
return False
return True
def read_input():
'''
6
abacba
2
aba 6
ba 3
3
'''
n = int(sys.stdin.readline())
# n = int('6\n')
crossword = sys.stdin.readline()[:-1]
# crossword = 'abacba\n'[:-1]
m = int(sys.stdin.readline())
# m = int('2\n')
adj = [[] for _ in range(n+2)]
cap = [[0] * (n+2) for _ in range(n+2)]
cost = [[0] * (n+2) for _ in range(n+2)]
# foo
# m_words = ["aba 6\n", "ba 3\n"]
# print(len(cost))
for idx in range(m):
word, p = sys.stdin.readline().split()
# word, p = m_words[idx].split()
p = int(p)
i = 0
while i + len(word) <= n:
if match(crossword, i, word):
u, v = i + 1, i + 1 + len(word)
# print((u, v))
add_edge(adj, cap, cost, u, v, 1, -p)
i += 1
x = int(sys.stdin.readline())
# x = int('3\n')
for i in range(n + 1):
u, v = i, i + 1
add_edge(adj, cap, cost, u, v, x, 0)
return adj, cap, cost
def bellman_ford(adj, cap, cost, potencial):
for _ in range(len(adj)):
for u in range(len(adj)):
for v in adj[u]:
reduced_cost = potencial[u] + cost[u][v] - potencial[v]
if cap[u][v] > 0 and reduced_cost < 0:
potencial[v] += reduced_cost
def dijkstra(adj, cap, flow, cost, potencial, dist, pi, s, t):
oo = float('inf')
for u in range(len(adj)):
dist[u] = +oo
pi[u] = None
dist[s] = 0
heap = [(0, s)]
while heap:
du, u = heappop(heap)
if dist[u] < du: continue
if u == t: break
for v in adj[u]:
reduced_cost = potencial[u] + cost[u][v] - potencial[v]
if flow[u][v] < cap[u][v] and dist[v] > dist[u] + reduced_cost:
dist[v] = dist[u] + reduced_cost
heappush(heap, (dist[v], v))
pi[v] = u
def min_cost_max_flow(adj, cap, cost):
min_cost, max_flow = 0, 0
flow = [[0] * len(adj) for _ in range(len(adj))]
potencial, oo = [0] * len(adj), float('inf')
dist, pi = [+oo] * len(adj), [None] * len(adj)
bellman_ford(adj, cap, cost, potencial)
s, t = 0, len(adj) - 1
while True:
dijkstra(adj, cap, flow, cost, potencial, dist, pi, s, t)
if dist[t] == +oo:
break
for u in range(len(adj)):
if dist[u] < dist[t]:
potencial[u] += dist[u] - dist[t]
limit, v = +oo, t
while pi[v]:
u = pi[v]
limit = min(limit, cap[u][v] - flow[u][v])
v = pi[v]
v = t
while pi[v]:
u = pi[v]
flow[u][v] += limit
flow[v][u] -= limit
v = pi[v]
min_cost += limit * (potencial[t] - potencial[s])
max_flow += limit
return min_cost, max_flow
adj, cap, cost = read_input()
# print({
# 'len(adj)': len(adj),
# 'len(cap)': len(cap),
# 'len(cost)': len(cost)
# })
# print({ 'adj': adj})
# print({ 'cap': cap})
# print({ 'cost': cost})
min_cost, _ = min_cost_max_flow(adj, cap, cost)
print(-min_cost)
| 30 | 374 | 7,168,000 |
124078949
|
import sys
from heapq import heappop, heappush
class Edge:
def __init__(self, u, v, cap, flow, cost, rev) -> None:
self.u = u
self.v = v
self.cap = cap
self.flow = flow
self.cost = cost
self.rev = rev
def add_edge(adj, u, v, capv, flowv, costv):
if u == 6 and v == 7:
foo = 0
adj[u].append(Edge(u, v, capv, flowv, costv, len(adj[v])))
adj[v].append(Edge(v, u, 0, flowv, -costv, len(adj[u])-1))
def match(crossword, cw_idx, word):
for k in range(len(word)):
if crossword[cw_idx + k] != word[k]:
return False
return True
def read_input():
'''
6
abacba
5
aba 6
ba 3
bac 4
cb 3
c 6
2
'''
n = int(sys.stdin.readline())
# n = int('6\n')
crossword = sys.stdin.readline()[:-1]
# crossword = 'abacba\n'[:-1]
m = int(sys.stdin.readline())
# m = int('5\n')
adj = [[] for _ in range(n+2)]
# foo
m_words = ["aba 6\n", "ba 3\n", "bac 4\n", "cb 3\n", "c 6"]
# print(len(cost))
for idx in range(m):
word, p = sys.stdin.readline().split()
# word, p = m_words[idx].split()
p = int(p)
i = 0
while i + len(word) <= n:
if match(crossword, i, word):
u, v = i + 1, i + 1 + len(word)
# print((u, v))
add_edge(adj, u, v, 1, 0, -p)
i += 1
x = int(sys.stdin.readline())
# x = int('2\n')
for i in range(n + 1):
u, v = i, i + 1
# print((u, v))
add_edge(adj, u, v, x, 0, 0)
return adj
def bellman_ford(adj, potencial):
for _ in range(len(adj)):
for u in range(len(adj)):
for e in adj[u]:
reduced_cost = potencial[e.u] + e.cost - potencial[e.v]
if e.cap > 0 and reduced_cost < 0:
potencial[e.v] += reduced_cost
def dijkstra(adj, potencial, dist, pi, s, t):
oo = float('inf')
for u in range(len(adj)):
dist[u] = +oo
pi[u] = None
dist[s] = 0
heap = [(0, s)]
while heap:
du, u = heappop(heap)
if dist[u] < du: continue
if u == t: break
for e in adj[u]:
reduced_cost = potencial[e.u] + e.cost - potencial[e.v]
if e.flow < e.cap and dist[e.v] > dist[e.u] + reduced_cost:
dist[e.v] = dist[e.u] + reduced_cost
heappush(heap, (dist[e.v], e.v))
pi[e.v] = e
def min_cost_max_flow(adj):
min_cost, max_flow = 0, 0
potencial, oo = [0] * len(adj), float('inf')
dist, pi = [+oo] * len(adj), [None] * len(adj)
bellman_ford(adj, potencial)
s, t = 0, len(adj) - 1
while True:
dijkstra(adj, potencial, dist, pi, s, t)
if dist[t] == +oo:
break
for u in range(len(adj)):
if dist[u] < dist[t]:
potencial[u] += dist[u] - dist[t]
limit, v = +oo, t
while v:
e = pi[v]
limit = min(limit, e.cap - e.flow)
v = e.u
v = t
while v:
e = pi[v]
e.flow += limit
try:
adj[v][e.rev].flow -= limit
except:
adj[v][e.rev].flow -= limit
v = e.u
min_cost += limit * (potencial[t] - potencial[s])
max_flow += limit
# path = []
# v = t
# while v:
# path.append(v)
# v = pi[v]
# path.reverse()
# print(path)
return min_cost, max_flow
adj = read_input()
# print({
# 'len(adj)': len(adj),
# 'len(cap)': len(cap),
# 'len(cost)': len(cost)
# })
# print({ 'adj': adj})
# print({ 'cap': cap})
# print({ 'cost': cost})
min_cost, _ = min_cost_max_flow(adj)
print(-min_cost)
|
Bubble Cup 9 - Finals [Online Mirror]
|
ICPC
| 2,016 | 1 | 256 |
Underfail
|
You have recently fallen through a hole and, after several hours of unconsciousness, have realized you are in an underground city. On one of your regular, daily walks through the unknown, you have encountered two unusually looking skeletons called Sanz and P’pairus, who decided to accompany you and give you some puzzles for seemingly unknown reasons.
One day, Sanz has created a crossword for you. Not any kind of crossword, but a 1D crossword! You are given m words and a string of length n. You are also given an array p, which designates how much each word is worth — the i-th word is worth pi points. Whenever you find one of the m words in the string, you are given the corresponding number of points. Each position in the crossword can be used at most x times. A certain word can be counted at different places, but you cannot count the same appearance of a word multiple times. If a word is a substring of another word, you can count them both (presuming you haven’t used the positions more than x times).
In order to solve the puzzle, you need to tell Sanz what’s the maximum achievable number of points in the crossword. There is no need to cover all postions, just get the maximal score! Crossword and words contain only lowercase English letters.
|
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the length of the crossword. The second line contains the crossword string. The third line contains a single integer m (1 ≤ m ≤ 100) — the number of given words, and next m lines contain description of words: each line will have a string representing a non-empty word (its length doesn't exceed the length of the crossword) and integer pi (0 ≤ pi ≤ 100). Last line of the input will contain x (1 ≤ x ≤ 100) — maximum number of times a position in crossword can be used.
|
Output single integer — maximum number of points you can get.
| null |
For example, with the string "abacba", words "aba" (6 points) and "ba" (3 points), and x = 3, you can get at most 12 points - the word "aba" appears once ("abacba"), while "ba" appears two times ("abacba"). Note that for x = 1, you could get at most 9 points, since you wouldn’t be able to count both "aba" and the first appearance of "ba".
|
[{"input": "6\nabacba\n2\naba 6\nba 3\n3", "output": "12"}]
| 2,400 |
["flows"]
| 30 |
[{"input": "6\r\nabacba\r\n2\r\naba 6\r\nba 3\r\n3\r\n", "output": "12\r\n"}, {"input": "6\r\nabacba\r\n2\r\naba 6\r\nba 3\r\n1\r\n", "output": "9\r\n"}, {"input": "6\r\nabacba\r\n5\r\naba 6\r\nba 3\r\nbac 4\r\ncb 3\r\nc 6\r\n2\r\n", "output": "21\r\n"}, {"input": "6\r\nabacba\r\n5\r\naba 6\r\nba 3\r\nbac 4\r\ncb 3\r\nc 6\r\n1\r\n", "output": "15\r\n"}, {"input": "7\r\nafxfxfg\r\n3\r\nf 3\r\nx 2\r\nfxf 6\r\n1\r\n", "output": "13\r\n"}, {"input": "11\r\nfghdgrakmnq\r\n8\r\nfgh 4\r\ngh 3\r\nh 10\r\nhdg 6\r\nhdgra 7\r\nakm 12\r\nrakm 5\r\na 15\r\n3\r\n", "output": "52\r\n"}, {"input": "8\r\naxghcdex\r\n5\r\naxgh 13\r\nhc 35\r\ncde 17\r\nxghcd 29\r\nghcdex 30\r\n3\r\n", "output": "95\r\n"}]
| false |
stdio
| null | true |
898/C
|
898
|
C
|
PyPy 3-64
|
TESTS
| 2 | 46 | 0 |
205035728
|
n = int(input())
dict = {}
for _ in range(n):
mix = input()
lis = []
name, nop, *lis = mix.split(" ")
if name in dict:
dict[name].extend(lis)
else:
dict[name] = lis
print(len(dict))
for name in dict:
dict[name].sort(key = lambda x: len(x))
i = 0
while (i < (len(dict[name]) - 1)):
if (dict[name])[i] in (dict[name])[i + 1]:
del dict[name][i]
i = i + 1
print(f"{name} {len(dict[name])}", end="")
for x in dict[name]:
print(f" {x}", end="")
print("")
| 59 | 62 | 5,632,000 |
33293212
|
n = int(input())
friends = dict()
for _ in range(n):
line = input().strip().split()
if line[0] not in friends:
friends[line[0]] = set()
for i in range(2, len(line)):
friends[line[0]].add(line[i])
def order(S):
S2 = [(len(e),e) for e in S]
S2.sort()
real = set()
for i in range(len(S2)):
d,e = S2[i]
flag = True
for j in range(i+1, len(S2)):
x = S2[j][1]
if x[-d:] == e:
flag = False
break
if flag:
real.add(e)
return real
print(len(friends))
for f in friends:
nums = order(friends[f])
st = ' '.join(nums)
print(f + ' ' + str(len(nums)) + ' ' + st)
|
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 |
551/E
|
551
|
E
|
PyPy 3
|
TESTS
| 0 | 140 | 0 |
43046252
|
n, q = map(int, input().split())
a = list(map(int, input().split()))
for _ in range(q):
inpt = list(map(int, input().split()))
if inpt[0] == 1:
for i in range(inpt[1], inpt[2] + 1):
a[i] += inpt[3]
else:
i = 0
j = len(a) - 1
while (a[i] != inpt[1]) and (i < len(a) - 1):
i += 1
while (a[j] != inpt[1]) and (j > 0):
j -= 1
if i > j:
print(-1)
else:
print(j - i)
| 73 | 3,338 | 83,660,800 |
206366815
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, q = map(int, input().split())
a = list(map(int, input().split()))
m = 555
m0 = n // m + min(n % m, 1)
s = [set() for _ in range(m0)]
for i in range(m0):
for j in range(i * m, min((i + 1) * m, n)):
s[i].add(a[j])
lazy = [0] * m0
ans = []
inf = pow(10, 9) + 1
for _ in range(q):
t = list(map(int, input().split()))
if t[0] == 1:
l, r, x = t[1:]
l, r = l - 1, r - 1
l0, r0 = l // m, r // m
if l0 == r0:
u, v = lazy[l0] + x, lazy[l0]
for i in range(l0 * m, min((l0 + 1) * m, n)):
a[i] += u if l <= i <= r else v
lazy[l0] = 0
s[l0] = set([a[i] for i in range(l0 * m, min((l0 + 1) * m, n))])
continue
for j in [l0, r0]:
u, v = lazy[j] + x, lazy[j]
for i in range(j * m, min((j + 1) * m, n)):
a[i] += u if l <= i <= r else v
lazy[j] = 0
s[j] = set([a[i] for i in range(j * m, min((j + 1) * m, n))])
for i in range(l0 + 1, r0):
lazy[i] += x
else:
y = t[1]
l = -1
for i in range(m0):
if y - lazy[i] in s[i]:
l = i
break
if l == -1:
ans0 = -1
ans.append(ans0)
continue
for i in range(m0 - 1, -1, -1):
if y - lazy[i] in s[i]:
r = i
break
mi, ma = inf, -inf
for i in set([l, r]):
for j in range(i * m, min((i + 1) * m, n)):
if a[j] == y - lazy[i]:
mi, ma = min(mi, j), max(ma, j)
ans0 = ma - mi
ans.append(ans0)
sys.stdout.write("\n".join(map(str, ans)))
|
Codeforces Round 307 (Div. 2)
|
CF
| 2,015 | 10 | 256 |
GukiZ and GukiZiana
|
Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:
1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
2. Second type has form 2 y and asks you to find value of GukiZiana(a, y).
For each query of type 2, print the answer and make GukiZ happy!
|
The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries.
The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a.
Each of next q lines contain either four or two numbers, as described in statement:
If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.
If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.
|
For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.
| null | null |
[{"input": "4 3\n1 2 3 4\n1 1 2 1\n1 1 1 1\n2 3", "output": "2"}, {"input": "2 3\n1 2\n1 2 2 1\n2 3\n2 4", "output": "0\n-1"}]
| 2,500 |
["binary search", "data structures", "implementation"]
| 73 |
[{"input": "4 3\r\n1 2 3 4\r\n1 1 2 1\r\n1 1 1 1\r\n2 3\r\n", "output": "2\r\n"}, {"input": "2 3\r\n1 2\r\n1 2 2 1\r\n2 3\r\n2 4\r\n", "output": "0\r\n-1\r\n"}, {"input": "8 5\r\n1 1 1 2 1 3 1 1\r\n2 1\r\n1 1 8 1\r\n2 2\r\n1 2 5 2\r\n2 4\r\n", "output": "7\r\n7\r\n4\r\n"}, {"input": "8 8\r\n1 9 1 9 2 3 4 5\r\n1 3 7 1\r\n2 6\r\n2 8\r\n2 9\r\n1 1 7 3\r\n2 11\r\n2 1000000000\r\n1 1 1 1\r\n", "output": "-1\r\n-1\r\n0\r\n-1\r\n-1\r\n"}, {"input": "7 3\r\n2 4 5 2 3 2 8\r\n2 2\r\n1 3 4 1\r\n2 4\r\n", "output": "5\r\n0\r\n"}, {"input": "2 2\r\n1000000000 1000000000\r\n1 1 2 1\r\n2 1000000000\r\n", "output": "-1\r\n"}, {"input": "4 4\r\n1000000000 1000000000 1000000000 1000000000\r\n2 1000000000\r\n1 1 2 1000000000\r\n1 1 3 1000000000\r\n2 1000000000\r\n", "output": "3\r\n0\r\n"}, {"input": "6 4\r\n1 9 9 2 3 4\r\n1 2 6 6\r\n1 5 6 5\r\n2 15\r\n2 1\r\n", "output": "4\r\n0\r\n"}, {"input": "4 3\r\n1 9 8 2\r\n1 2 3 100\r\n1 4 4 9\r\n1 1 4 100000000\r\n", "output": ""}, {"input": "1 5\r\n1\r\n2 4\r\n2 1\r\n1 1 1 999\r\n2 1000\r\n2 1000\r\n", "output": "-1\r\n0\r\n0\r\n0\r\n"}, {"input": "9 10\r\n1 1 2 1 3 3 7 8 9\r\n1 4 6 5\r\n2 8\r\n1 1 3 6\r\n2 1001212\r\n2 7\r\n1 3 3 1\r\n2 9\r\n2 8\r\n1 1 9 0\r\n2 6\r\n", "output": "3\r\n-1\r\n6\r\n6\r\n3\r\n0\r\n"}, {"input": "1 1\r\n1\r\n1 1 1 1\r\n", "output": ""}, {"input": "1 1\r\n1\r\n2 1\r\n", "output": "0\r\n"}, {"input": "1 1\r\n1\r\n2 5\r\n", "output": "-1\r\n"}]
| false |
stdio
| null | true |
444/E
|
444
|
E
|
PyPy 3
|
TESTS
| 2 | 93 | 0 |
116404603
|
n = int(input())
for i in range(n - 1):
a, b, c = map(int, input().split())
r = 4
for i in range(n):
r = int(input())
if r == 1:
print(2)
else:
print(3)
| 49 | 171 | 716,800 |
51735454
|
n = int(input())
edges = [[int(x) for x in input().split()] for i in range(n-1)]
edges = sorted(edges)
use_count = [0]+[int(input()) for i in range(n)]
lo,hi = 0,10000
def getpar(par,u):
if par[par[u]] == par[u]:
return par[u]
par[u] = getpar(par,par[u])
return par[u]
def unite(par,sz,use,u,v):
u = getpar(par,u)
v = getpar(par,v)
par[u] = v
sz[v] += sz[u]
use[v] += use[u]
def solve(fp):
par = [i for i in range(n+1)]
sz = [1 for i in range(n+1)]
use = [use_count[i] for i in range(n+1)]
for edge in edges:
if edge[2] < fp:
unite(par,sz,use,edge[0],edge[1])
total_use = sum(use_count)
for i in range(n+1):
p = getpar(par,i)
if(p == i):
if(total_use - use[p] < sz[p]):
return False
return True
while lo < hi:
mid = (lo+hi+1)//2
if solve(mid):
lo = mid
else:
hi = mid-1
print(lo)
|
Codeforces Round 254 (Div. 1)
|
CF
| 2,014 | 3 | 256 |
DZY Loves Planting
|
DZY loves planting, and he enjoys solving tree problems.
DZY has a weighted tree (connected undirected graph without cycles) containing n nodes (they are numbered from 1 to n). He defines the function g(x, y) (1 ≤ x, y ≤ n) as the longest edge in the shortest path between nodes x and y. Specially g(z, z) = 0 for every z.
For every integer sequence p1, p2, ..., pn (1 ≤ pi ≤ n), DZY defines f(p) as $$\min_{i=1}^{n} g(i,p_i)$$.
DZY wants to find such a sequence p that f(p) has maximum possible value. But there is one more restriction: the element j can appear in p at most xj times.
Please, find the maximum possible f(p) under the described restrictions.
|
The first line contains an integer n (1 ≤ n ≤ 3000).
Each of the next n - 1 lines contains three integers ai, bi, ci (1 ≤ ai, bi ≤ n; 1 ≤ ci ≤ 10000), denoting an edge between ai and bi with length ci. It is guaranteed that these edges form a tree.
Each of the next n lines describes an element of sequence x. The j-th line contains an integer xj (1 ≤ xj ≤ n).
|
Print a single integer representing the answer.
| null |
In the first sample, one of the optimal p is [4, 3, 2, 1].
|
[{"input": "4\n1 2 1\n2 3 2\n3 4 3\n1\n1\n1\n1", "output": "2"}, {"input": "4\n1 2 1\n2 3 2\n3 4 3\n4\n4\n4\n4", "output": "3"}]
| 2,700 |
["binary search", "dsu", "trees"]
| 49 |
[{"input": "4\r\n1 2 1\r\n2 3 2\r\n3 4 3\r\n1\r\n1\r\n1\r\n1\r\n", "output": "2\r\n"}, {"input": "4\r\n1 2 1\r\n2 3 2\r\n3 4 3\r\n4\r\n4\r\n4\r\n4\r\n", "output": "3\r\n"}, {"input": "10\r\n2 1 8760\r\n3 1 3705\r\n4 1 1862\r\n5 2 7332\r\n6 3 7015\r\n7 5 4866\r\n8 3 4465\r\n9 7 8886\r\n10 3 9362\r\n2\r\n5\r\n5\r\n4\r\n4\r\n5\r\n4\r\n5\r\n1\r\n2\r\n", "output": "8760\r\n"}, {"input": "10\r\n2 1 5297\r\n3 2 7674\r\n4 1 1935\r\n5 2 1941\r\n6 3 1470\r\n7 1 3823\r\n8 2 4959\r\n9 4 6866\r\n10 9 2054\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "5297\r\n"}, {"input": "10\r\n2 1 3921\r\n3 2 3204\r\n4 3 1912\r\n5 4 6844\r\n6 5 8197\r\n7 6 7148\r\n8 7 5912\r\n9 8 104\r\n10 9 5881\r\n4\r\n4\r\n5\r\n2\r\n2\r\n4\r\n1\r\n2\r\n3\r\n1\r\n", "output": "8197\r\n"}, {"input": "10\r\n2 1 6818\r\n3 2 9734\r\n4 3 2234\r\n5 4 3394\r\n6 5 1686\r\n7 6 3698\r\n8 7 700\r\n9 8 716\r\n10 9 1586\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "3698\r\n"}, {"input": "10\r\n1 6 4890\r\n2 6 2842\r\n3 6 7059\r\n4 6 3007\r\n5 6 6195\r\n7 6 3962\r\n8 6 3413\r\n9 6 7658\r\n10 6 8049\r\n3\r\n3\r\n3\r\n1\r\n4\r\n4\r\n5\r\n2\r\n1\r\n1\r\n", "output": "6195\r\n"}, {"input": "10\r\n1 2 5577\r\n3 2 6095\r\n4 2 4743\r\n5 2 2254\r\n6 2 9771\r\n7 2 7417\r\n8 2 9342\r\n9 2 2152\r\n10 2 5785\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "5785\r\n"}, {"input": "10\r\n2 1 2464\r\n3 1 5760\r\n4 3 9957\r\n5 1 6517\r\n6 4 8309\r\n7 3 3176\r\n8 7 1982\r\n9 1 7312\r\n10 2 3154\r\n1\r\n1\r\n4\r\n1\r\n1\r\n3\r\n3\r\n5\r\n3\r\n2\r\n", "output": "7312\r\n"}, {"input": "10\r\n2 1 559\r\n3 1 5707\r\n4 2 9790\r\n5 3 1591\r\n6 1 7113\r\n7 6 2413\r\n8 6 3006\r\n9 4 1935\r\n10 6 5954\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "7113\r\n"}, {"input": "2\r\n1 2 10000\r\n1\r\n1\r\n", "output": "10000\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
925/A
|
925
|
A
|
Python 3
|
TESTS
| 0 | 93 | 0 |
41671985
|
def bin_search(m,ref):
l = -1
r = len(m)
while(r - l > 1):
if m[(r+l)//2] > ref:
r = (r+l)//2
else:
l = (r+l)//2
return((r+l)//2)
n,m,cl,ce,v = map(int,input().split())
stairs = []
for i in range(cl):
stairs.append(int(input()))
elevators = []
for i in range(cl):
elevators.append(int(input()))
q = int(input())
for i in range(q):
coordinates = list(map(int,input().split()))
if coordinates[1] == coordinates[3]:
print(abs(coordinates[0]-coordinates[2]))
else:
a = 0
b = 0
istair = bin_search(stairs,(coordinates[0]+coordinates[2])//2)
ielevator = bin_search(elevators,(coordinates[0]+coordinates[2])//2)
a = abs(istair-coordinates[0]) + abs(istair-coordinates[2]) + abs(coordinates[1]-coordinates[3])
b = abs(ielevator-coordinates[0]) + abs(ielevator-coordinates[2]) + abs(coordinates[1]-coordinates[3])//v
if a < b:
print(a-2)
else:
print(b-2)
| 27 | 1,824 | 9,830,400 |
42120565
|
def takeClosest(myList, myNumber):
"""
Assumes myList is sorted. Returns closest value to myNumber.
If two numbers are equally close, return the smallest number.
"""
if len(myList) == 0:
return 9e10
pos = bisect_left(myList, myNumber)
if pos == 0:
return myList[0]
if pos == len(myList):
return myList[-1]
before = myList[pos - 1]
after = myList[pos]
if after - myNumber < myNumber - before:
return after
else:
return before
from bisect import bisect_left
from math import ceil
n, m, n_stairs, n_elevators, v = map(int, input().split(" "))
if n_stairs > 0:
stairs = list(map(int, input().split(" ")))
else:
stairs = []
input()
if n_elevators > 0:
elevators = list(map(int, input().split(" ")))
else:
elevators = []
input()
queries = int(input())
res = []
for i in range(queries):
x1, y1, x2, y2 = map(int, input().split(" "))
next_elevator = takeClosest(elevators, (y1 + y2) / 2)
next_stairs = takeClosest(stairs, (y1 + y2) / 2)
time_elevator = abs(x1 - x2) / v
time_stairs = abs(x1 - x2)
mi = min(y1, y2)
ma = max(y1, y2)
if next_elevator < mi:
time_elevator += (mi - next_elevator) * 2
elif next_elevator > ma:
time_elevator += (next_elevator - ma) * 2
if next_stairs < mi:
time_stairs += (mi - next_stairs) * 2
elif next_stairs > ma:
time_stairs += (next_stairs - ma) * 2
dis = abs(y1 - y2)
if time_elevator < time_stairs:
dis += time_elevator
else:
dis += time_stairs
if x1 == x2:
res.append(abs(y1 - y2))
else:
res.append(ceil(dis))
print(*res, sep="\n")
# Made By Mostafa_Khaled
|
VK Cup 2018 - Round 3
|
CF
| 2,018 | 2 | 256 |
Stairs and Elevators
|
In the year of $$$30XX$$$ participants of some world programming championship live in a single large hotel. The hotel has $$$n$$$ floors. Each floor has $$$m$$$ sections with a single corridor connecting all of them. The sections are enumerated from $$$1$$$ to $$$m$$$ along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height $$$n$$$ and width $$$m$$$. We can denote sections with pairs of integers $$$(i, j)$$$, where $$$i$$$ is the floor, and $$$j$$$ is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections $$$(1, x)$$$, $$$(2, x)$$$, $$$\ldots$$$, $$$(n, x)$$$ for some $$$x$$$ between $$$1$$$ and $$$m$$$. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to $$$v$$$ floors in any direction using an elevator. You can assume you don't have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process $$$q$$$ queries. Each query is a question "what is the minimum time needed to go from a room in section $$$(x_1, y_1)$$$ to a room in section $$$(x_2, y_2)$$$?"
|
The first line contains five integers $$$n, m, c_l, c_e, v$$$ ($$$2 \leq n, m \leq 10^8$$$, $$$0 \leq c_l, c_e \leq 10^5$$$, $$$1 \leq c_l + c_e \leq m - 1$$$, $$$1 \leq v \leq n - 1$$$) — the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains $$$c_l$$$ integers $$$l_1, \ldots, l_{c_l}$$$ in increasing order ($$$1 \leq l_i \leq m$$$), denoting the positions of the stairs. If $$$c_l = 0$$$, the second line is empty.
The third line contains $$$c_e$$$ integers $$$e_1, \ldots, e_{c_e}$$$ in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers $$$l_i$$$ and $$$e_i$$$ are distinct.
The fourth line contains a single integer $$$q$$$ ($$$1 \leq q \leq 10^5$$$) — the number of queries.
The next $$$q$$$ lines describe queries. Each of these lines contains four integers $$$x_1, y_1, x_2, y_2$$$ ($$$1 \leq x_1, x_2 \leq n$$$, $$$1 \leq y_1, y_2 \leq m$$$) — the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. $$$y_1$$$ and $$$y_2$$$ are not among $$$l_i$$$ and $$$e_i$$$.
|
Print $$$q$$$ integers, one per line — the answers for the queries.
| null |
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
|
[{"input": "5 6 1 1 3\n2\n5\n3\n1 1 5 6\n1 3 5 4\n3 3 5 3", "output": "7\n5\n4"}]
| 1,600 |
["binary search"]
| 27 |
[{"input": "5 6 1 1 3\r\n2\r\n5\r\n3\r\n1 1 5 6\r\n1 3 5 4\r\n3 3 5 3\r\n", "output": "7\r\n5\r\n4\r\n"}, {"input": "2 2 0 1 1\r\n\r\n1\r\n1\r\n1 2 2 2\r\n", "output": "3\r\n"}, {"input": "4 4 1 0 1\r\n4\r\n\r\n5\r\n1 1 2 2\r\n1 3 2 2\r\n3 3 4 3\r\n3 2 2 2\r\n1 2 2 3\r\n", "output": "6\r\n4\r\n3\r\n5\r\n4\r\n"}, {"input": "10 10 1 8 4\r\n10\r\n2 3 4 5 6 7 8 9\r\n10\r\n1 1 3 1\r\n2 1 7 1\r\n1 1 9 1\r\n7 1 4 1\r\n10 1 7 1\r\n2 1 7 1\r\n3 1 2 1\r\n5 1 2 1\r\n10 1 5 1\r\n6 1 9 1\r\n", "output": "3\r\n4\r\n4\r\n3\r\n3\r\n4\r\n3\r\n3\r\n4\r\n3\r\n"}, {"input": "2 5 1 0 1\r\n2\r\n\r\n1\r\n1 4 1 5\r\n", "output": "1\r\n"}, {"input": "2 10 1 1 1\r\n1\r\n10\r\n1\r\n1 5 1 8\r\n", "output": "3\r\n"}, {"input": "4 4 1 0 1\r\n1\r\n\r\n1\r\n1 2 1 4\r\n", "output": "2\r\n"}, {"input": "2 4 1 1 1\r\n1\r\n2\r\n1\r\n2 3 2 4\r\n", "output": "1\r\n"}, {"input": "1000 1000 1 1 10\r\n1\r\n2\r\n1\r\n1 900 1 1000\r\n", "output": "100\r\n"}, {"input": "2 4 1 1 1\r\n1\r\n4\r\n1\r\n1 2 1 3\r\n", "output": "1\r\n"}, {"input": "5 5 1 1 1\r\n3\r\n2\r\n1\r\n1 5 1 1\r\n", "output": "4\r\n"}]
| false |
stdio
| null | true |
813/F
|
813
|
F
|
PyPy 3-64
|
TESTS
| 2 | 62 | 1,228,800 |
148633666
|
from collections import deque
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def bfs(s):
q = deque()
q.append(s)
dist[s] = 0
ng = 0
while q:
i = q.popleft()
di = dist[i]
vi = visit[i]
for j, w0 in zip(G[i], W[i]):
if not w0:
continue
if not dist[j] ^ -1:
dist[j] = di ^ w0
visit[j] |= (vi | w0)
q.append(j)
else:
ng |= (dist[j] ^ di ^ z) & w0 & vi & visit[j]
return ng
n, q = map(int, input().split())
e = set()
G = [[] for _ in range(n + 1)]
W = [[] for _ in range(n + 1)]
d = [dict() for _ in range(n + 1)]
pow2 = [1]
for _ in range(63):
pow2.append(2 * pow2[-1])
w = [0] * 64
w[-1] = pow2[-1]
for i in range(62, -1, -1):
w[i] = w[i + 1] ^ pow2[i]
z = w[0]
z0 = pow2[-1]
ans = []
dist = [-1] * (n + 1)
visit = [0] * (n + 1)
for _ in range(1, q // 64 + 1):
for i in range(64):
x, y = map(int, input().split())
wi = w[i]
if not x in d[y]:
d[x][y] = len(G[x])
d[y][x] = len(G[y])
G[x].append(y)
G[y].append(x)
W[x].append(wi)
W[y].append(wi)
e.add(x)
e.add(y)
else:
W[x][d[x][y]] ^= wi
W[y][d[y][x]] ^= wi
ng = 0
for i in e:
if not dist[i] ^ -1:
ng |= bfs(i)
dist[i] = -1
visit[i] = 0
for i in pow2:
ans0 = "YES" if not i & ng else "NO"
ans.append(ans0)
for i in e:
Wi = W[i]
for j in range(len(Wi)):
Wi[j] = z if Wi[j] & z0 else 0
c = q % 64
for i in range(c):
x, y = map(int, input().split())
wi = w[i]
if not x in d[y]:
d[x][y] = len(G[x])
d[y][x] = len(G[y])
G[x].append(y)
G[y].append(x)
W[x].append(wi)
W[y].append(wi)
e.add(x)
e.add(y)
else:
W[x][d[x][y]] ^= wi
W[y][d[y][x]] ^= wi
ng = 0
for i in e:
if not dist[i] ^ -1:
ng |= bfs(i)
dist[i] = -1
visit[i] = 0
for i in pow2[:c]:
ans0 = "YES" if not i & ng else "NO"
ans.append(ans0)
sys.stdout.write("\n".join(ans))
| 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 |
813/F
|
813
|
F
|
PyPy 3
|
TESTS
| 1 | 61 | 0 |
154221537
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def get_root(s):
v = []
while not s == root[s]:
v.append(s)
s = root[s]
for i in v:
root[i] = s
return s
def unite(s, t):
rs, rt = get_root(s), get_root(t)
if not rs ^ rt:
return
if rank[s] == rank[t]:
rank[rs] += 1
if rank[s] >= rank[t]:
root[rt] = rs
else:
root[rs] = rt
return
def same(s, t):
return True if get_root(s) == get_root(t) else False
def get_root1(s):
v = []
while not s == root[s]:
v.append(s)
s = root[s]
for i in v:
ro0.append((i, root[i]))
root[i] = s
return s
def unite1(s, t):
rs, rt = get_root1(s), get_root1(t)
if not rs ^ rt:
return
if rank[s] == rank[t]:
ra0.append((rs, rank[rs]))
rank[rs] += 1
if rank[s] >= rank[t]:
ro0.append((rt, root[rt]))
root[rt] = rs
else:
ro0.append((rs, root[rs]))
root[rs] = rt
return
n, q = map(int, input().split())
s = set()
m = 800
u = q // m + min(q % m, 1)
l = q
n2 = 2 * n + 1
ans = []
e = []
s1 = []
for _ in range(u):
s0 = set()
z = []
for _ in range(min(l, m)):
x, y = map(int, input().split())
z.append((x, y))
if (x, y) in s:
s.remove((x, y))
s0.add((x, y))
root = [i for i in range(n2)]
rank = [1 for _ in range(n2)]
ok = 1
for x, y in s:
unite(x, y + n)
unite(y, x + n)
if same(x, x + n):
ok = 0
break
if not ok:
for x, y in s0:
s.add(x, y)
for x, y in z:
if (x, y) in s:
s.remove((x, y))
else:
s.add((x, y))
ans.append("\n".join(["NO"] * min(l, m)))
l -= m
continue
cnt = 0
ro0, ra0 = [], []
for x, y in s0:
if not same(x, y + n):
if same(x, x + n):
cnt -= 1
e.append((x, y, len(ro0), len(ra0)))
unite1(x, y + n)
unite1(y, x + n)
if same(x, x + n):
cnt += 1
s.add((x, y))
for x, y in z:
if not (x, y) in s:
if same(x, x + n):
cnt -= 1
e.append((x, y, len(ro0), len(ra0)))
if not same(x, y + n):
unite1(x, y + n)
unite1(y, x + n)
if same(x, x + n):
cnt += 1
s.add((x, y))
else:
while e:
x0, y0, ro1, ra1 = e.pop()
if not same(x0, x0 + n):
cnt += 1
while ro1 ^ len(ro0):
i, c = ro0.pop()
root[i] = c
while ra1 ^ len(ra0):
i, c = ra0.pop()
rank[i] = c
if not same(x0, x0 + n):
cnt -= 1
if x == x0 and y == y0:
break
s1.append((x0, y0))
while s1:
x0, y0 = s1.pop()
if same(x0, x0 + n):
cnt -= 1
e.append((x, y, len(ro0), len(ra0)))
if not same(x, y + n):
unite1(x, y + n)
unite1(y, x + n)
if same(x0, x0 + n):
cnt += 1
s.remove((x, y))
ans.append("YES" if not cnt else "NO")
sys.stdout.write("\n".join(ans))
| 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 |
925/B
|
925
|
B
|
Python 3
|
TESTS
| 0 | 78 | 307,200 |
84306663
|
n, x1, x2 = map(int, input().split())
cc = list(map(int, input().split()))
c = sorted(cc)
s = 0
f = True
middle = -1
for i in range(n):
s += c[i]
if f:
if s >= x1:
middle = i
s = 0
f = False
if f:
print('No')
elif s < x2:
print('No')
else:
print('Yes')
print(i + 1, n - i)
for i in range(n):
print(cc.index(c[i]) + 1, end=' ')
if i == middle:
print()
| 40 | 795 | 50,995,200 |
37714382
|
n, a, b = [int(x) for x in input().split()]
hs = [int(x) for x in input().split()]
hs = sorted(enumerate(hs), key=lambda x: x[1])
for i in range(1, n+1):
if hs[-i][1] * i >= a:
break
else:
print('No')
exit()
for j in range(i+1, n+1):
if hs[-j][1] * (j - i) >= b:
print('Yes')
print(i, j - i)
print(" ".join(map(str, [index+1 for index,_ in hs[-i:]])))
print(" ".join(map(str, [index+1 for index,_ in hs[-j:-i]])))
break
else:
for i in range(1, n+1):
if hs[-i][1] * i >= b:
break
else:
print('No')
exit()
for j in range(i+1, n+1):
if hs[-j][1] * (j - i) >= a:
print('Yes')
print(j - i, i)
print(" ".join(map(str, [index+1 for index,_ in hs[-j:-i]])))
print(" ".join(map(str, [index+1 for index,_ in hs[-i:]])))
break
else:
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 |
925/B
|
925
|
B
|
PyPy 3-64
|
TESTS
| 0 | 46 | 512,000 |
153450910
|
n, x, y = map(int,input().split())
c = list(map(int, input().split()))
left = -1
right = (1 << n) - 1
def check(hash):
res = 0
for i in range(n):
if (1 << i) & hash:
res += c[i]
return res
while right - left > 1:
mid = left + (right - left) // 2
if check(mid) >= x:
right = mid
else:
left = mid
p = 0
q = 0
pl = []
ql = []
for i in range(n):
if (1 << i) & right:
p += c[i]
pl.append(i + 1)
else:
q += c[i]
ql.append(i + 1)
# print(x,y,p,q)
if p >= x and q >= y:
print("Yes")
print(*pl)
print(*ql)
else:
print("No")
| 40 | 1,216 | 39,219,200 |
42132645
|
# python3
def readline(): return tuple(map(int, input().split()))
def ceil_div(num, den): return (num - 1) // den + 1
def main():
n, x1, x2 = readline()
c = readline()
xx = (x1, x2)
servers = sorted(enumerate(c, start=1), key=lambda p: p[1])
for (i, a) in enumerate(servers):
for (j, x) in enumerate(xx):
kj = ceil_div(x, a[1])
if i + kj < n and (n - i - kj) * servers[i + kj][1] >= sum(xx) - x:
print("Yes")
l1 = servers[i:i+kj]
l2 = servers[i+kj:]
if j: l1, l2 = l2, l1
print(len(l1), len(l2))
print(" ".join(str(d[0]) for d in l1))
print(" ".join(str(d[0]) for d in l2))
return
print("No")
main()
# Made By Mostafa_Khaled
|
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 |
850/A
|
850
|
A
|
Python 3
|
TESTS
| 2 | 93 | 307,200 |
53201855
|
from math import acos, pi
n = int(input())
tocke = [list(map(int, input().split())) for i in range(n)]
def vektor(t1, t2):
vector = []
for i in range(5):
vector.append(t2[i] - t1[i])
return vector
def skalarni_produkt(v1, v2):
vsota = 0
for i in range(5):
vsota += v1[i] * v2[i]
return vsota
def dolzina(v1):
return (skalarni_produkt(v1, v1)) ** (1 / 2)
konec = []
for i in range(n):
for j in range(n):
v1 = vektor(tocke[i], tocke[j])
for k in range(n):
v2 = vektor(tocke[i], tocke[k])
if not dolzina(v1) == 0 and not dolzina(v2) == 0:
t = skalarni_produkt(v1, v2) / (dolzina(v1) * dolzina(v2))
if t > 1 and t - 10 ** (-9) <= 1:
t = 1
if t < -1 and t + 10 ** (-9) >= -1:
t = -1
kot = acos(t)
if kot - pi / 2 >= 0 and not i in konec:
konec.append(i)
print(len(konec))
if not len(konec) == 0:
for i in konec:
print(i + 1)
| 19 | 62 | 0 |
30064873
|
def vec(c0, a, b):
an = 0
for i in range(5):
an += (a[i] - c0[i]) * (b[i] - c0[i])
return an
n = int(input())
if n > 80:
print(0)
exit(0)
a = [[0] * 5 for i in range(n)]
for i in range(n):
a[i] = [int(j) for j in input().split()]
ans = []
for i in range(n):
flag = True
for j in range(n):
for k in range(n):
if i != j and j != k and k != i and vec(a[i], a[j], a[k]) > 0:
flag = False
if flag:
ans.append(i)
print(len(ans))
for i in ans:
print(i + 1)
|
Codeforces Round 432 (Div. 1, based on IndiaHacks Final Round 2017)
|
CF
| 2,017 | 2 | 256 |
Five Dimensional Points
|
You are given set of n points in 5-dimensional space. The points are labeled from 1 to n. No two points coincide.
We will call point a bad if there are different points b and c, not equal to a, from the given set such that angle between vectors $${ \vec { a } } { \vec { b } }$$ and $${ \vec { a } } { \vec { c } }$$ is acute (i.e. strictly less than $$90^\circ$$). Otherwise, the point is called good.
The angle between vectors $$\overrightarrow{\tau}$$ and $${ \overrightarrow { y } }$$ in 5-dimensional space is defined as $$\arccos\left(\frac{\vec{x} \cdot \vec{y}}{|\vec{x}||\vec{y}|}\right)$$, where $$\vec{x} \cdot \vec{y} = x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4} + x_{5} y_{5}$$ is the scalar product and $$|\vec{x}| = \sqrt{\vec{x} \cdot \vec{x}}$$ is length of $$\overrightarrow{\tau}$$.
Given the list of points, print the indices of the good points in ascending order.
|
The first line of input contains a single integer n (1 ≤ n ≤ 103) — the number of points.
The next n lines of input contain five integers ai, bi, ci, di, ei (|ai|, |bi|, |ci|, |di|, |ei| ≤ 103) — the coordinates of the i-th point. All points are distinct.
|
First, print a single integer k — the number of good points.
Then, print k integers, each on their own line — the indices of the good points in ascending order.
| null |
In the first sample, the first point forms exactly a $$90^\circ$$ angle with all other pairs of points, so it is good.
In the second sample, along the cd plane, we can see the points look as follows:
We can see that all angles here are acute, so no points are good.
|
[{"input": "6\n0 0 0 0 0\n1 0 0 0 0\n0 1 0 0 0\n0 0 1 0 0\n0 0 0 1 0\n0 0 0 0 1", "output": "1\n1"}, {"input": "3\n0 0 1 2 0\n0 0 9 2 0\n0 0 5 9 0", "output": "0"}]
| 1,700 |
["brute force", "geometry", "math"]
| 19 |
[{"input": "6\r\n0 0 0 0 0\r\n1 0 0 0 0\r\n0 1 0 0 0\r\n0 0 1 0 0\r\n0 0 0 1 0\r\n0 0 0 0 1\r\n", "output": "1\r\n1\r\n"}, {"input": "3\r\n0 0 1 2 0\r\n0 0 9 2 0\r\n0 0 5 9 0\r\n", "output": "0\r\n"}, {"input": "1\r\n0 0 0 0 0\r\n", "output": "1\r\n1\r\n"}, {"input": "2\r\n0 1 2 3 4\r\n5 6 7 8 9\r\n", "output": "2\r\n1\r\n2\r\n"}, {"input": "10\r\n0 -110 68 -51 -155\r\n-85 -110 68 -51 -155\r\n85 -70 51 68 -230\r\n0 -40 51 68 75\r\n0 5 -51 -68 -190\r\n85 0 0 0 0\r\n85 -115 -68 51 35\r\n85 -75 -187 34 -40\r\n-85 -110 -136 102 -155\r\n85 -110 -17 119 -155\r\n", "output": "0\r\n"}, {"input": "6\r\n-305 -390 638 -623 343\r\n479 755 -343 144 89\r\n-268 843 -461 989 -301\r\n-986 -274 347 -847 -728\r\n278 718 -372 -674 270\r\n-477 562 -489 -858 611\r\n", "output": "0\r\n"}, {"input": "10\r\n-705 38 170 -768 689\r\n-705 86 248 -768 709\r\n-705 86 170 -742 709\r\n-705 86 144 -768 709\r\n-705 86 170 -820 709\r\n-705 106 170 -768 661\r\n-822 86 170 -768 709\r\n-705 98 170 -768 714\r\n-705 86 170 -768 709\r\n-601 86 170 -768 709\r\n", "output": "1\r\n9\r\n"}, {"input": "11\r\n358 -724 -232 53 -520\r\n486 -554 -328 53 -220\r\n358 -554 -232 -372 -520\r\n358 -554 -232 308 -520\r\n868 -554 448 53 -520\r\n478 -554 -322 53 -600\r\n358 296 -232 53 -520\r\n256 -554 -368 53 -520\r\n230 -554 -136 53 -820\r\n-182 -554 173 53 -160\r\n358 -554 -232 53 -520\r\n", "output": "1\r\n11\r\n"}, {"input": "8\r\n-559 581 509 257 343\r\n-544 451 569 277 343\r\n-451 451 434 401 343\r\n-559 451 509 257 83\r\n-664 451 89 117 343\r\n-559 451 509 257 993\r\n-715 451 509 374 343\r\n-811 451 684 -79 343\r\n", "output": "0\r\n"}, {"input": "11\r\n8 8 8 8 8\r\n2 2 2 2 2\r\n0 0 0 0 0\r\n6 6 6 6 6\r\n7 7 7 7 7\r\n10 10 10 10 10\r\n9 9 9 9 9\r\n3 3 3 3 3\r\n1 1 1 1 1\r\n5 5 5 5 5\r\n4 4 4 4 4\r\n", "output": "0\r\n"}, {"input": "7\r\n49 457 -650 325 -325\r\n0 0 325 325 0\r\n253 204 -325 0 -325\r\n204 -253 325 325 325\r\n408 -506 -325 -325 325\r\n49 457 -650 325 -650\r\n0 0 0 650 -325\r\n", "output": "0\r\n"}, {"input": "11\r\n1 0 0 0 0\r\n-1 0 0 0 0\r\n0 1 0 0 0\r\n0 -1 0 0 0\r\n0 0 1 0 0\r\n0 0 -1 0 0\r\n0 0 0 1 0\r\n0 0 0 -1 0\r\n0 0 0 0 1\r\n0 0 0 0 -1\r\n0 0 0 0 0\r\n", "output": "1\r\n11\r\n"}, {"input": "4\r\n0 0 0 0 0\r\n1 0 0 0 0\r\n0 1 0 0 0\r\n0 1 1 0 0\r\n", "output": "0\r\n"}]
| false |
stdio
| null | true |
518/C
|
518
|
C
|
Python 3
|
TESTS
| 0 | 46 | 6,963,200 |
126945010
|
n,m,k=map(int,input().split())
l1=[int(a) for a in input().split()]
l2=[int(a) for a in input().split()]
l3=[]
for i in range(n):
l3.append([l1[i],i])
l3.sort()
an=0
for i in range(m):
tg=l2[i]
an+=tg//k+tg%k
if l3[tg-1][1]!=0:
l3[tg-1][1]-=1
l3[l1[l3[tg-2][1]]][1]+=1
l1[tg-1],l1[tg-2]=l1[tg-2],l1[tg-1]
print(an)
| 43 | 124 | 20,684,800 |
185451156
|
from heapq import heapify, heappop, heappush, nlargest
from collections import Counter, defaultdict
from sys import stdin, stdout
from math import ceil, floor, sqrt
from functools import reduce,lru_cache
# n,m = map(int,stdin.readline().split())
# stdout.write(str(arr[x-y]-arr[x])+'\n')
# reps = int(stdin.readline())
# for _ in range(reps):
# s1,s2 = stdin.readline().strip().split()
# s = stdin.readline().strip()
# n = int(stdin.readline())
# n,m,k = map(int,stdin.readline().split())
# arr = list(map(int,stdin.readline().split()))
n,m,k = map(int,stdin.readline().split())
arr = list(map(int,stdin.readline().split()))
queries = list(map(int,stdin.readline().split()))
idx = [-1 for _ in range(n+1)]
for i,a in enumerate(arr):
idx[a] = i
ans = 0
for q in queries:
idxq = idx[q]
ans += idxq//k + 1
if idxq!=0:
prev = arr[idxq-1]
idx[prev], idx[q] = idx[q], idx[prev]
arr[idxq-1], arr[idxq] = arr[idxq], arr[idxq-1]
print(ans)
|
Codeforces Round 293 (Div. 2)
|
CF
| 2,015 | 1 | 256 |
Anya and Smartphone
|
Anya has bought a new smartphone that uses Berdroid operating system. The smartphone menu has exactly n applications, each application has its own icon. The icons are located on different screens, one screen contains k icons. The icons from the first to the k-th one are located on the first screen, from the (k + 1)-th to the 2k-th ones are on the second screen and so on (the last screen may be partially empty).
Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen t, Anya needs to make the following gestures: first she scrolls to the required screen number t, by making t - 1 gestures (if the icon is on the screen t), and then make another gesture — press the icon of the required application exactly once to launch it.
After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.
All applications are numbered from 1 to n. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.
Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?
Note that one application may be launched multiple times.
|
The first line of the input contains three numbers n, m, k (1 ≤ n, m, k ≤ 105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.
The next line contains n integers, permutation a1, a2, ..., an — the initial order of icons from left to right in the menu (from the first to the last one), ai — is the id of the application, whose icon goes i-th in the menu. Each integer from 1 to n occurs exactly once among ai.
The third line contains m integers b1, b2, ..., bm(1 ≤ bi ≤ n) — the ids of the launched applications in the planned order. One application may be launched multiple times.
|
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
| null |
In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.
After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.
After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.
After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.
In total, Anya makes 7 gestures.
|
[{"input": "8 3 3\n1 2 3 4 5 6 7 8\n7 8 1", "output": "7"}, {"input": "5 4 2\n3 1 5 2 4\n4 4 4 4", "output": "8"}]
| 1,600 |
["constructive algorithms", "data structures", "implementation"]
| 43 |
[{"input": "8 3 3\r\n1 2 3 4 5 6 7 8\r\n7 8 1\r\n", "output": "7\r\n"}, {"input": "5 4 2\r\n3 1 5 2 4\r\n4 4 4 4\r\n", "output": "8\r\n"}, {"input": "10 10 3\r\n1 2 3 4 5 6 7 8 9 10\r\n2 3 4 5 6 7 8 9 10 1\r\n", "output": "25\r\n"}, {"input": "10 12 3\r\n6 1 2 9 3 10 8 5 7 4\r\n3 9 9 4 8 2 3 8 10 8 3 4\r\n", "output": "29\r\n"}, {"input": "16 12 8\r\n6 11 12 5 1 10 16 3 13 15 9 4 8 2 14 7\r\n6 13 8 12 8 15 12 16 8 11 3 16\r\n", "output": "18\r\n"}, {"input": "4 3 5\r\n3 1 4 2\r\n4 2 4\r\n", "output": "3\r\n"}, {"input": "4 2 15\r\n2 4 3 1\r\n1 2\r\n", "output": "2\r\n"}, {"input": "71 96 75634\r\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71\r\n27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26\r\n", "output": "96\r\n"}, {"input": "71 81 2149\r\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 29 28 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71\r\n29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29\r\n", "output": "81\r\n"}]
| false |
stdio
| null | true |
685/B
|
685
|
B
|
PyPy 3-64
|
TESTS
| 2 | 872 | 70,656,000 |
145906613
|
import sys
import heapq
input = sys.stdin.buffer.readline
def process(n, P, Q):
g = [[] for i in range(n+1)]
for i in range(n-1):
i2 = i+2
j2 = P[i]
g[i2].append(j2)
g[j2].append(i2)
subtree_size = [0 for i in range(n+1)]
subtrees = [[] for i in range(n+1)]
centroids = [None for i in range(n+1)]
depths = [[1]]
while True:
next_s = []
for x in depths[-1]:
for y in g[x]:
if x==1 or y != P[x-2]:
next_s.append(y)
if len(next_s)==0:
break
depths.append(next_s)
while len(depths) > 0:
for x in depths[-1]:
subtree_size[x]+=1
if subtree_size[x]==1:
centroids[x] = x
else:
for x2 in g[x]:
if x==1 or x2 != P[x-2]:
c1 = centroids[x2]
size1 = subtree_size[c1]
if 2*size1 >= subtree_size[x]:
if centroids[x] is None:
centroids[x] = c1
c2 = P[c1-2]
size2 = subtree_size[c2]
if c2 != x and 2*size2 >= subtree_size[x]:
if centroids[x] is None:
centroids[x] = c2
if centroids[x] is None:
centroids[x] = x
if x > 1:
p = P[x-2]
subtree_size[p]+=subtree_size[x]
depths.pop()
for qi in Q:
sys.stdout.write(str(centroids[qi])+'\n')
n, q = [int(x) for x in input().split()]
P = [int(x) for x in input().split()]
Q = []
for i in range(q):
v = int(input())
Q.append(v)
process(n, P, Q)
| 49 | 2,401 | 47,001,600 |
18685578
|
def main():
import sys
E=sys.stdin
R=E.readline
n,q=map(int,R().split())
P=[0,0]+[int(i) for i in R().split()]
#n=q=300000
#P=[0]+list(range(n))
#exit()
C=[[]]
i=0
while i<n:
i+=1
C.append([])
i=2
j=len(P)
while i<j:
C[P[i]].append(i)
i+=1
i=Qi=0
W=[0]*(n+1)
S=[0]*(n+1)
Q=[0]*(n+1)
while i<n:
i+=1
W[i]=len(C[i])
if W[i]==0:
Q[Qi]=i
Qi+=1
A=[0]*(n+1)
while Qi:
Qi-=1
x=Q[Qi]
S[x]+=1
if P[x]:
y=P[x]
S[y]+=S[x]
W[y]-=1
if W[y]==0:
Q[Qi]=y
Qi+=1
a=S[x]>>1
A[x]=x
for z in C[x]:
if S[z]>a:
A[x]=A[z]
b=S[x]-S[A[x]]
while b>a:
A[x]=P[A[x]]
b=S[x]-S[A[x]]
break
while q:
q-=1
sys.stdout.write(str(A[int(R())])+'\n')
main()
|
Codeforces Round 359 (Div. 1)
|
CF
| 2,016 | 3 | 256 |
Kay and Snowflake
|
After the piece of a devilish mirror hit the Kay's eye, he is no longer interested in the beauty of the roses. Now he likes to watch snowflakes.
Once upon a time, he found a huge snowflake that has a form of the tree (connected acyclic graph) consisting of n nodes. The root of tree has index 1. Kay is very interested in the structure of this tree.
After doing some research he formed q queries he is interested in. The i-th query asks to find a centroid of the subtree of the node vi. Your goal is to answer all queries.
Subtree of a node is a part of tree consisting of this node and all it's descendants (direct or not). In other words, subtree of node v is formed by nodes u, such that node v is present on the path from u to root.
Centroid of a tree (or a subtree) is a node, such that if we erase it from the tree, the maximum size of the connected component will be at least two times smaller than the size of the initial tree (or a subtree).
|
The first line of the input contains two integers n and q (2 ≤ n ≤ 300 000, 1 ≤ q ≤ 300 000) — the size of the initial tree and the number of queries respectively.
The second line contains n - 1 integer p2, p3, ..., pn (1 ≤ pi ≤ n) — the indices of the parents of the nodes from 2 to n. Node 1 is a root of the tree. It's guaranteed that pi define a correct tree.
Each of the following q lines contain a single integer vi (1 ≤ vi ≤ n) — the index of the node, that define the subtree, for which we want to find a centroid.
|
For each query print the index of a centroid of the corresponding subtree. If there are many suitable nodes, print any of them. It's guaranteed, that each subtree has at least one centroid.
| null |
The first query asks for a centroid of the whole tree — this is node 3. If we delete node 3 the tree will split in four components, two of size 1 and two of size 2.
The subtree of the second node consists of this node only, so the answer is 2.
Node 3 is centroid of its own subtree.
The centroids of the subtree of the node 5 are nodes 5 and 6 — both answers are considered correct.
|
[{"input": "7 4\n1 1 3 3 5 3\n1\n2\n3\n5", "output": "3\n2\n3\n6"}]
| 1,900 |
["data structures", "dfs and similar", "dp", "trees"]
| 49 |
[{"input": "7 4\r\n1 1 3 3 5 3\r\n1\r\n2\r\n3\r\n5\r\n", "output": "3\r\n2\r\n3\r\n6\r\n"}, {"input": "2 2\r\n1\r\n1\r\n2\r\n", "output": "2\r\n2\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:
lines = f.read().splitlines()
ptr = 0
n, q = map(int, lines[ptr].split())
ptr +=1
parents = list(map(int, lines[ptr].split()))
ptr +=1
queries = []
for _ in range(q):
vi = int(lines[ptr])
queries.append(vi)
ptr +=1
# Build children list
children = [[] for _ in range(n+1)] # 1-based
for i in range(2, n+1):
p = parents[i-2]
children[p].append(i)
# Compute in_time, out_time, size via iterative DFS
in_time = [0]*(n+1)
out_time = [0]*(n+1)
size = [1]*(n+1)
time = 0
stack = [(1, False)] # (node, visited)
while stack:
node, visited = stack.pop()
if not visited:
time +=1
in_time[node] = time
stack.append( (node, True) )
for child in reversed(children[node]):
stack.append( (child, False) )
else:
for child in children[node]:
size[node] += size[child]
out_time[node] = time
# Read submission output
with open(submission_path) as f:
submission_lines = f.read().splitlines()
if len(submission_lines) != q:
print(0)
return
for i in range(q):
vi = queries[i]
line = submission_lines[i].strip()
if not line:
print(0)
return
try:
u = int(line)
except:
print(0)
return
if u <1 or u>n:
print(0)
return
if not (in_time[vi] <= in_time[u] <= out_time[vi]):
print(0)
return
s = size[vi]
parent_comp = s - size[u]
if parent_comp > s//2:
print(0)
return
for c in children[u]:
if size[c] > s//2:
print(0)
return
print(1)
if __name__ == "__main__":
main()
| true |
839/D
|
839
|
D
|
PyPy 3
|
TESTS
| 2 | 171 | 13,926,400 |
103996491
|
visited=[0]*(10**6+1)
lastfactor=[-1]*(10**6+1)
for j in range(2,10**6):
for k in range(2*j,10**6+1,j):
lastfactor[k]=j
n=int(input())
primes=[]
for i in range(2,10**6+1):
if lastfactor[i]==-1:
primes.append(i)
count=[0]*(10**6+1)
a=list(map(int,input().split()))
for i in range(n):
r=a[i]
curr=1
while curr*curr<=r:
if r%curr==0:
count[curr]+=1
count[r//curr]+=1
curr+=1
if (curr-1)*(curr-1)==r:
count[curr-1]-=1
ans=0
MOD=10**9+7
for i in range(2,10**6+1):
if count[i]==0:
continue
ans=(ans+i*count[i]*pow(2,count[i]-1,MOD))%MOD
for prime in primes:
if prime*i>10**6:
break
if count[prime*i]>0:
ans=(ans-(i*count[prime*i]*pow(2,count[prime*i]-1,MOD)))
print(ans)
| 47 | 655 | 38,707,200 |
155552887
|
import sys
mod=1000000007
m=int(input())#20w
a=list(map(int,sys.stdin.readline().split()))#100w
n=1000001#max a
answer=0
ans=[0]*n
cnt=[0]*n
for i in a:
cnt[i]+=1
for i in range(n-1,1,-1):
j=sum(cnt[i::i])
if j>0:
ans[i]=(j*(2**(j-1))-sum(ans[(2*i)::i]))%mod
answer=(answer+i*ans[i])%mod
print(answer)
|
Codeforces Round 428 (Div. 2)
|
CF
| 2,017 | 3 | 256 |
Winter is here
|
Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the rest of the world is fighting for the Iron Throne, he is going to get ready for the attack of the White Walkers.
He has created a method to know how strong his army is. Let the i-th soldier’s strength be ai. For some k he calls i1, i2, ..., ik a clan if i1 < i2 < i3 < ... < ik and gcd(ai1, ai2, ..., aik) > 1 . He calls the strength of that clan k·gcd(ai1, ai2, ..., aik). Then he defines the strength of his army by the sum of strengths of all possible clans.
Your task is to find the strength of his army. As the number may be very large, you have to print it modulo 1000000007 (109 + 7).
Greatest common divisor (gcd) of a sequence of integers is the maximum possible integer so that each element of the sequence is divisible by it.
|
The first line contains integer n (1 ≤ n ≤ 200000) — the size of the army.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000000) — denoting the strengths of his soldiers.
|
Print one integer — the strength of John Snow's army modulo 1000000007 (109 + 7).
| null |
In the first sample the clans are {1}, {2}, {1, 2} so the answer will be 1·3 + 1·3 + 2·3 = 12
|
[{"input": "3\n3 3 1", "output": "12"}, {"input": "4\n2 3 4 6", "output": "39"}]
| 2,200 |
["combinatorics", "dp", "math", "number theory"]
| 47 |
[{"input": "3\r\n3 3 1\r\n", "output": "12\r\n"}, {"input": "4\r\n2 3 4 6\r\n", "output": "39\r\n"}]
| false |
stdio
| null | true |
47/D
|
47
|
D
|
Python 3
|
TESTS
| 0 | 60 | 0 |
230675785
|
import sys
from itertools import product
n, m = map(int, sys.stdin.readline().split())
possible_codes = product('01', repeat=n)
valid_code_count = 0
for _ in range(m):
s, c = sys.stdin.readline().split()
c = int(c)
valid_code_count = 0
for code in possible_codes:
correct_digits = sum(s[i] == code[i] for i in range(n))
if correct_digits == c:
valid_code_count += 1
possible_codes = (code for code in possible_codes if sum(s[i] == code[i] for i in range(n)) == c)
print(valid_code_count)
| 26 | 466 | 3,686,400 |
181406585
|
from collections import defaultdict
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, m = map(int, input().split())
pow2 = [1]
for _ in range(20):
pow2.append(2 * pow2[-1])
n1, n2 = n // 2, (n + 1) // 2
x, y, z = [], [], []
for _ in range(m):
s, c = list(input().rstrip().decode().split())
x0, y0 = [], []
for i in range(n1):
x0.append(ord(s[i]) - 48)
for i in range(n1, n):
y0.append(ord(s[i]) - 48)
x.append(x0)
y.append(y0)
z.append(int(c))
cnt = defaultdict(lambda : 0)
u = [0] * n1
for i in range(pow2[n1]):
c = []
for j in range(n1):
u[j] = 1 if i & pow2[j] else 0
for j in range(m):
v = n1
for k, l in zip(u, x[j]):
v -= k ^ l
if v > z[j]:
break
c.append(z[j] - v)
if len(c) == m:
cnt[tuple(c)] += 1
ans = 0
u = [0] * n2
for i in range(pow2[n2]):
c = []
for j in range(n2):
u[j] = 1 if i & pow2[j] else 0
for j in range(m):
v = n2
for k, l in zip(u, y[j]):
v -= k ^ l
if v > z[j]:
break
c.append(v)
if len(c) == m:
c = tuple(c)
if c in cnt:
ans += cnt[c]
print(ans)
|
Codeforces Beta Round 44 (Div. 2)
|
CF
| 2,010 | 5 | 256 |
Safe
|
Vasya tries to break in a safe. He knows that a code consists of n numbers, and every number is a 0 or a 1. Vasya has made m attempts to enter the code. After each attempt the system told him in how many position stand the right numbers. It is not said in which positions the wrong numbers stand. Vasya has been so unlucky that he hasn’t entered the code where would be more than 5 correct numbers. Now Vasya is completely bewildered: he thinks there’s a mistake in the system and it is self-contradictory. Help Vasya — calculate how many possible code variants are left that do not contradict the previous system responses.
|
The first input line contains two integers n and m (6 ≤ n ≤ 35, 1 ≤ m ≤ 10) which represent the number of numbers in the code and the number of attempts made by Vasya. Then follow m lines, each containing space-separated si and ci which correspondingly indicate Vasya’s attempt (a line containing n numbers which are 0 or 1) and the system’s response (an integer from 0 to 5 inclusively).
|
Print the single number which indicates how many possible code variants that do not contradict the m system responses are left.
| null | null |
[{"input": "6 2\n000000 2\n010100 4", "output": "6"}, {"input": "6 3\n000000 2\n010100 4\n111100 0", "output": "0"}, {"input": "6 3\n000000 2\n010100 4\n111100 2", "output": "1"}]
| 2,200 |
["brute force"]
| 26 |
[{"input": "6 2\r\n000000 2\r\n010100 4\r\n", "output": "6\r\n"}, {"input": "6 3\r\n000000 2\r\n010100 4\r\n111100 0\r\n", "output": "0\r\n"}, {"input": "6 3\r\n000000 2\r\n010100 4\r\n111100 2\r\n", "output": "1\r\n"}, {"input": "6 1\r\n101011 2\r\n", "output": "15\r\n"}, {"input": "7 2\r\n1011111 2\r\n1001111 1\r\n", "output": "6\r\n"}, {"input": "6 4\r\n000110 2\r\n010001 2\r\n001111 2\r\n001100 2\r\n", "output": "1\r\n"}, {"input": "8 3\r\n00111100 5\r\n10100111 2\r\n10110101 2\r\n", "output": "6\r\n"}, {"input": "35 1\r\n00001111001110101000001101100010010 5\r\n", "output": "324632\r\n"}, {"input": "35 2\r\n00101101100111101110111010001101101 3\r\n00111111100101010110111010001101101 3\r\n", "output": "20\r\n"}, {"input": "35 1\r\n11000110100110101001100101001010110 2\r\n", "output": "595\r\n"}, {"input": "35 2\r\n00111111100000111101000110100111101 1\r\n00111111000000111101000010100111101 1\r\n", "output": "2\r\n"}, {"input": "35 6\r\n01100100110000001001100110001100011 5\r\n10000100110000011001110010001100011 5\r\n00101110100000010000100010001110011 4\r\n00110010101000011001100000001110011 5\r\n00100101110000011001101110001110011 4\r\n00110110110000011001101000000100011 5\r\n", "output": "1\r\n"}, {"input": "6 10\r\n110000 5\r\n010011 4\r\n110011 5\r\n110010 4\r\n000001 4\r\n010001 5\r\n110101 5\r\n110011 5\r\n110010 4\r\n011001 4\r\n", "output": "1\r\n"}]
| false |
stdio
| null | true |
620/F
|
620
|
F
|
PyPy 3-64
|
TESTS
| 0 | 46 | 0 |
197061937
|
n, q = map(int, input().split())
L = list(map(int, input().split()))
prefix_xor = [0]
for i in L:
prefix_xor.append(prefix_xor[-1] ^ i)
for _ in range(q):
l, r = map(int, input().split())
l -= 1; r -= 1
print(prefix_xor[r + 1] ^ prefix_xor[l])
| 22 | 8,064 | 121,446,400 |
224643493
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v, w):
return (u * z + v) * z + w
def update1(x, c):
i = x ^ l1
tree1[i], color1[i] = x, c
i >>= 1
for _ in range(w):
j, k = i << 1, i << 1 ^ 1
if color1[j] ^ c:
tree1[i] = tree1[k]
elif color1[k] ^ c:
tree1[i] = tree1[j]
else:
tree1[i] = min(tree1[j], tree1[k])
color1[i] = c
i >>= 1
return
def update2(x, c):
i = x ^ l1
tree2[i], color2[i] = x, c
i >>= 1
for _ in range(w):
j, k = i << 1, i << 1 ^ 1
if color2[j] ^ c:
tree2[i] = tree2[k]
elif color2[k] ^ c:
tree2[i] = tree2[j]
else:
tree2[i] = max(tree2[j], tree2[k])
color2[i] = c
i >>= 1
return
def xor_max1(x, r, c):
i = 1
if color1[i] ^ c or not tree1[i] < r:
return 0
for _ in range(w):
j, k = i << 1, i << 1 ^ 1
if color1[j] ^ c:
f0 = 1
elif color1[k] ^ c:
f0 = 0
elif tree1[k] < r and x ^ x0[tree1[j]] < x ^ x0[tree1[k]]:
f0 = 1
else:
f0 = 0
i = i << 1 ^ f0
return x ^ x0[i ^ l1]
def xor_max2(x, l, c):
i = 1
if color2[i] ^ c or not tree2[i] > l:
return 0
for _ in range(w):
j, k = i << 1, i << 1 ^ 1
if color2[j] ^ c:
f0 = 1
elif color2[k] ^ c:
f0 = 0
elif tree2[j] > l and x ^ x0[tree2[k]] < x ^ x0[tree2[j]]:
f0 = 0
else:
f0 = 1
i = i << 1 ^ f0
return x ^ x0[i ^ l1]
n, m = map(int, input().split())
a = list(map(int, input().split()))
x0 = [0] * (max(a) + 5)
for i in range(1, len(x0)):
x0[i] = x0[i - 1] ^ i
l0, r0 = [0] * m, [0] * m
m1 = 432
m2 = n // m1 + 3
q = []
s0 = [0] * (m2 + 1)
ans = [0] * m
z = max(m, n) + 5
for i in range(m):
l, r = map(int, input().split())
l, r = l - 1, r - 1
if l // m1 == r // m1:
ans0 = 0
for j in range(l, r + 1):
u = a[j]
for k in range(j, r + 1):
v = a[k]
ans1 = x0[u - 1] ^ x0[v] if u <= v else x0[u] ^ x0[v - 1]
ans0 = max(ans0, ans1)
ans[i] = ans0
continue
l0[i], r0[i] = l, r
x = l // m1
q.append(f(x, r, i))
s0[x + 1] += 1
ma = [0] * n
for i in range(n):
mai, u = 0, a[i]
for j in range(i, min(i // m1 * m1 + m1, n)):
v = a[j]
maj = x0[u - 1] ^ x0[v] if u <= v else x0[u] ^ x0[v - 1]
mai = max(mai, maj)
ma[i] = mai
for i in range(n - 2, -1, -1):
if (i + 1) % m1:
ma[i] = max(ma[i], ma[i + 1])
for i in range(1, m2 + 1):
s0[i] += s0[i - 1]
w = (max(a) + 1).bit_length()
l1 = pow(2, w)
l2 = 2 * l1
inf = pow(10, 9) + 1
tree1, color1 = [inf] * l2, [-1] * l2
tree2, color2 = [-inf] * l2, [-1] * l2
q.sort()
for c in range(m2):
if s0[c] == s0[c + 1]:
continue
r = (c + 1) * m1
c10, c11, c20, c21 = inf, inf, -inf, -inf
ans0 = 0
for i in range(s0[c], s0[c + 1]):
j = q[i] % z
ll, rr = l0[j], r0[j]
while r <= rr:
ar = a[r]
x1, x2 = x0[ar], x0[ar - 1]
if x1 > 1:
update2(ar, c)
elif x1 == 0:
c20 = max(c20, ar)
elif x1 == 1:
c21 = max(c21, ar)
if x2 > 1:
update1(ar - 1, c)
elif x2 == 0:
c10 = min(c10, ar - 1)
elif x2 == 1:
c11 = min(c11, ar - 1)
ans0 = max(ans0, xor_max1(x1, ar, c), xor_max2(x2, ar - 1, c))
if c10 < ar:
ans0 = max(ans0, x1)
if c11 < ar:
ans0 = max(ans0, x1 ^ 1)
if ar - 1 < c20:
ans0 = max(ans0, x2)
if ar - 1 < c21:
ans0 = max(ans0, x2 ^ 1)
r += 1
ans1 = max(ma[ll], ans0)
for l in range(ll, (c + 1) * m1):
al = a[l]
x1, x2 = x0[al], x0[al - 1]
ans1 = max(ans1, xor_max1(x1, al, c), xor_max2(x2, al - 1, c))
if c10 < al:
ans1 = max(ans1, x1)
if c11 < al:
ans1 = max(ans1, x1 ^ 1)
if al - 1 < c20:
ans1 = max(ans1, x2)
if al - 1 < c21:
ans1 = max(ans1, x2 ^ 1)
ans[j] = ans1
sys.stdout.write("\n".join(map(str, ans)))
|
Educational Codeforces Round 6
|
ICPC
| 2,016 | 10 | 512 |
Xors on Segments
|
You are given an array with n integers ai and m queries. Each query is described by two integers (lj, rj).
Let's define the function $$f(u,v) = u \oplus (u+1) \oplus \ldots \oplus v$$. The function is defined for only u ≤ v.
For each query print the maximal value of the function f(ax, ay) over all lj ≤ x, y ≤ rj, ax ≤ ay.
|
The first line contains two integers n, m (1 ≤ n ≤ 5·104, 1 ≤ m ≤ 5·103) — the size of the array and the number of the queries.
The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.
Each of the next m lines contains two integers lj, rj (1 ≤ lj ≤ rj ≤ n) – the parameters of the j-th query.
|
For each query print the value aj on a separate line — the maximal value of the function f(ax, ay) over all lj ≤ x, y ≤ rj, ax ≤ ay.
| null | null |
[{"input": "6 3\n1 2 3 4 5 6\n1 6\n2 5\n3 4", "output": "7\n7\n7"}, {"input": "1 1\n1\n1 1", "output": "1"}, {"input": "6 20\n10 21312 2314 214 1 322\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n2 2\n2 3\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6\n4 4\n4 5\n4 6\n5 5\n5 6\n6 6", "output": "10\n21313\n21313\n21313\n21313\n21313\n21312\n21313\n21313\n21313\n21313\n2314\n2315\n2315\n214\n215\n323\n1\n323\n322"}]
| 2,800 |
["data structures", "strings", "trees"]
| 22 |
[{"input": "6 3\r\n1 2 3 4 5 6\r\n1 6\r\n2 5\r\n3 4\r\n", "output": "7\r\n7\r\n7\r\n"}, {"input": "1 1\r\n1\r\n1 1\r\n", "output": "1\r\n"}, {"input": "6 20\r\n10 21312 2314 214 1 322\r\n1 1\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n1 6\r\n2 2\r\n2 3\r\n2 4\r\n2 5\r\n2 6\r\n3 4\r\n3 5\r\n3 6\r\n4 4\r\n4 5\r\n4 6\r\n5 5\r\n5 6\r\n6 6\r\n", "output": "10\r\n21313\r\n21313\r\n21313\r\n21313\r\n21313\r\n21312\r\n21313\r\n21313\r\n21313\r\n21313\r\n2314\r\n2315\r\n2315\r\n214\r\n215\r\n323\r\n1\r\n323\r\n322\r\n"}, {"input": "1 1\r\n1\r\n1 1\r\n", "output": "1\r\n"}, {"input": "5 10\r\n10 2 7 8 8\r\n2 5\r\n3 4\r\n5 5\r\n1 3\r\n3 4\r\n2 2\r\n2 5\r\n4 5\r\n5 5\r\n1 2\r\n", "output": "15\r\n15\r\n8\r\n12\r\n15\r\n2\r\n15\r\n8\r\n8\r\n10\r\n"}]
| false |
stdio
| null | true |
267/A
|
267
|
A
|
Python 3
|
TESTS
| 1 | 31 | 0 |
169341298
|
n = int(input())
for i in range(n):
(a, b) = map(int, input().split(' '))
if a == b:
print(1)
else:
operations = 0
bigger_number = max(a, b)
smaller_number = min(a, b)
while bigger_number != smaller_number:
if bigger_number % smaller_number == 0:
operations += smaller_number
bigger_number = (bigger_number - smaller_number) // smaller_number
else:
operations += bigger_number // smaller_number
bigger_number = bigger_number % smaller_number
if bigger_number < smaller_number:
smaller_number, bigger_number = bigger_number, smaller_number
print(operations + 1)
| 35 | 46 | 0 |
149972427
|
for _ in range(int(input())):
a,b = map(int,input().split())
i = 0
while True:
if a <= 0 or b <= 0:
break
if a > b:
i+=a//b
a-=a//b*b
else:
i+=b//a
b-=b//a*a
print(i)
|
Codeforces Testing Round 5
|
CF
| 2,013 | 1 | 256 |
Subtractions
|
You've got two numbers. As long as they are both larger than zero, they go through the same operation: subtract the lesser number from the larger one. If they equal substract one number from the another. For example, one operation transforms pair (4,17) to pair (4,13), it transforms (5,5) to (0,5).
You've got some number of pairs (ai, bi). How many operations will be performed for each of them?
|
The first line contains the number of pairs n (1 ≤ n ≤ 1000). Then follow n lines, each line contains a pair of positive integers ai, bi (1 ≤ ai, bi ≤ 109).
|
Print the sought number of operations for each pair on a single line.
| null | null |
[{"input": "2\n4 17\n7 987654321", "output": "8\n141093479"}]
| 900 |
["math", "number theory"]
| 35 |
[{"input": "2\r\n4 17\r\n7 987654321\r\n", "output": "8\r\n141093479\r\n"}, {"input": "10\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n7 987654321\r\n", "output": "141093479\r\n141093479\r\n141093479\r\n141093479\r\n141093479\r\n141093479\r\n141093479\r\n141093479\r\n141093479\r\n141093479\r\n"}, {"input": "1\r\n536870912 32\r\n", "output": "16777216\r\n"}, {"input": "20\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n1000000000 999999999\r\n", "output": "1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n1000000000\r\n"}, {"input": "3\r\n1000000000 1\r\n1000000000 1\r\n1 100000000\r\n", "output": "1000000000\r\n1000000000\r\n100000000\r\n"}]
| false |
stdio
| null | true |
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