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7
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82/B
82
B
Python 3
TESTS
32
1,714
2,867,200
56191977
n = int(input()) common,mas = set(),[] for i in range(n*(n-1)//2): lst = list(map(int,input().split())) lst.pop(0) for i,x in enumerate(lst): common.add(x) mas.append(set(lst)) d = {} for x in common:d[x]=False for x in common: if d[x]==False: res = set(common) for i,y in enumerate(mas): if x in y: res=res&y print(len(res),*res) for y in res:d[y]=True
34
312
2,662,400
5043606
from collections import defaultdict n = int(input()) if n == 2: t = input().split() print(1, t[1]) print(int(t[0]) - 1, ' '.join(t[2: ])) else: m = (n * (n - 1)) // 2 t = [0] * m p = defaultdict(list) for i in range(m): t[i] = list(map(int, input().split()))[1: ] for j in t[i]: if len(p[j]) < 2: p[j].append(i) t[i] = set(t[i]) p = set(tuple(q) for q in p.values()) t = [set(t[i]) & set(t[j]) for i, j in p] t = [str(len(q)) + ' ' + ' '.join(map(str, q)) for q in t] print('\n'.join(t))
Yandex.Algorithm 2011: Qualification 2
CF
2,011
2
256
Sets
Little Vasya likes very much to play with sets consisting of positive integers. To make the game more interesting, Vasya chose n non-empty sets in such a way, that no two of them have common elements. One day he wanted to show his friends just how interesting playing with numbers is. For that he wrote out all possible unions of two different sets on n·(n - 1) / 2 pieces of paper. Then he shuffled the pieces of paper. He had written out the numbers in the unions in an arbitrary order. For example, if n = 4, and the actual sets have the following form {1, 3}, {5}, {2, 4}, {7}, then the number of set pairs equals to six. The six pieces of paper can contain the following numbers: - 2, 7, 4. - 1, 7, 3; - 5, 4, 2; - 1, 3, 5; - 3, 1, 2, 4; - 5, 7. Then Vasya showed the pieces of paper to his friends, but kept the n sets secret from them. His friends managed to calculate which sets Vasya had thought of in the first place. And how about you, can you restore the sets by the given pieces of paper?
The first input file line contains a number n (2 ≤ n ≤ 200), n is the number of sets at Vasya's disposal. Then follow sets of numbers from the pieces of paper written on n·(n - 1) / 2 lines. Each set starts with the number ki (2 ≤ ki ≤ 200), which is the number of numbers written of the i-th piece of paper, and then follow ki numbers aij (1 ≤ aij ≤ 200). All the numbers on the lines are separated by exactly one space. It is guaranteed that the input data is constructed according to the above given rules from n non-intersecting sets.
Print on n lines Vasya's sets' description. The first number on the line shows how many numbers the current set has. Then the set should be recorded by listing its elements. Separate the numbers by spaces. Each number and each set should be printed exactly once. Print the sets and the numbers in the sets in any order. If there are several answers to that problem, print any of them. It is guaranteed that there is a solution.
null
null
[{"input": "4\n3 2 7 4\n3 1 7 3\n3 5 4 2\n3 1 3 5\n4 3 1 2 4\n2 5 7", "output": "1 7\n2 2 4\n2 1 3\n1 5"}, {"input": "4\n5 6 7 8 9 100\n4 7 8 9 1\n4 7 8 9 2\n3 1 6 100\n3 2 6 100\n2 1 2", "output": "3 7 8 9\n2 6 100\n1 1\n1 2"}, {"input": "3\n2 1 2\n2 1 3\n2 2 3", "output": "1 1\n1 2\n1 3"}]
1,700
["constructive algorithms", "hashing", "implementation"]
34
[{"input": "4\r\n3 2 7 4\r\n3 1 7 3\r\n3 5 4 2\r\n3 1 3 5\r\n4 3 1 2 4\r\n2 5 7\r\n", "output": "1 7 \r\n2 2 4 \r\n2 1 3 \r\n1 5 \r\n"}, {"input": "4\r\n5 6 7 8 9 100\r\n4 7 8 9 1\r\n4 7 8 9 2\r\n3 1 6 100\r\n3 2 6 100\r\n2 1 2\r\n", "output": "3 7 8 9 \r\n2 6 100 \r\n1 1 \r\n1 2 \r\n"}, {"input": "3\r\n2 1 2\r\n2 1 3\r\n2 2 3\r\n", "output": "1 1 \r\n1 2 \r\n1 3 \r\n"}, {"input": "3\r\n2 1 2\r\n10 1 90 80 70 60 50 40 30 20 10\r\n10 2 10 20 30 40 50 60 70 80 90\r\n", "output": "1 1 \r\n1 2 \r\n9 10 20 30 40 50 60 70 80 90 \r\n"}, {"input": "4\r\n4 56 44 53 43\r\n3 109 44 43\r\n3 109 56 53\r\n3 43 62 44\r\n3 62 56 53\r\n2 109 62\r\n", "output": "2 43 44 \r\n2 53 56 \r\n1 109 \r\n1 62 \r\n"}, {"input": "2\r\n2 1 2\r\n", "output": "1 2\r\n1 1"}, {"input": "2\r\n10 1 2 3 4 5 6 7 8 9 10\r\n", "output": "1 10\r\n9 1 2 3 4 5 6 7 8 9"}]
false
stdio
import sys def main(): input_path = sys.argv[1] submission_path = sys.argv[3] with open(input_path, 'r') as f: lines = [l.strip() for l in f if l.strip()] n = int(lines[0]) m = n * (n - 1) // 2 input_unions = set() for line in lines[1:1+m]: parts = list(map(int, line.split())) elements = parts[1:] input_unions.add(frozenset(elements)) with open(submission_path, 'r') as f: submission_lines = [l.strip() for l in f if l.strip()] if len(submission_lines) != n: print(0) return submission_sets = [] all_elements = set() for line in submission_lines: parts = list(map(int, line.split())) if not parts: print(0) return size = parts[0] if len(parts) - 1 != size: print(0) return elements = parts[1:] if len(elements) != len(set(elements)): print(0) return s = frozenset(elements) submission_sets.append(s) current = set(elements) if current & all_elements: print(0) return all_elements.update(current) for s in submission_sets: if not s: print(0) return generated = set() for i in range(n): for j in range(i+1, n): u = submission_sets[i].union(submission_sets[j]) generated.add(frozenset(u)) print(100 if generated == input_unions else 0) if __name__ == "__main__": main()
true
358/C
358
C
Python 3
TESTS
2
78
204,800
5042432
n = int( input() ) Q = 0 P = 0 Df = 0 Db = 0 l=[] for _ in range(n): x = int(input()) if x==0: if l!=[]: Q = max(l) for i in l: if i==Q: print("pushQueue") elif i>P and P<=Df: print("pushStack") P=i elif i>Df and Df<=P: print("pushFront") Df=i else: print("pushBack") cnt=0 s = "" if Q!=0: cnt+=1 s+=" popQueue" if P!=0: cnt+=1 s+=" popStack" if Df!=0: cnt+=1 s+=" popFront" print( str(cnt) + s ) Q=0 P=0 Df=0 Db=0 l=[] else: l.append(x)
26
826
12,800,000
149773930
n = int(input()) last_zero, largest, nd_largest, rd_largest = -1, -1, -1, -1 arr = [int(input()) for _ in range(n)] for i in range(n): if arr[i] == 0: filled_stk, filled_queue, filled_front = False, False, False for j in range(last_zero + 1, i): if arr[j] == largest and not filled_stk: filled_stk = True print('pushStack') elif arr[j] == nd_largest and not filled_queue: filled_queue = True print('pushQueue') elif arr[j] == rd_largest and not filled_front: filled_front = True print('pushFront') else: print('pushBack') print(int(filled_stk) + int(filled_queue) + int(filled_front), end='') if filled_stk: print(' popStack', end='') if filled_queue: print(' popQueue', end='') if filled_front: print(' popFront', end='') print('\n', end='') last_zero = i largest = nd_largest = rd_largest = -1 else: if arr[i] >= largest: largest, nd_largest, rd_largest = arr[i], largest, nd_largest elif arr[i] >= nd_largest: nd_largest, rd_largest = arr[i], nd_largest elif arr[i] > rd_largest: rd_largest = arr[i] for i in range(last_zero + 1, n): print('pushStack')
Codeforces Round 208 (Div. 2)
CF
2,013
2
256
Dima and Containers
Dima has a birthday soon! It's a big day! Saryozha's present to Dima is that Seryozha won't be in the room and won't disturb Dima and Inna as they celebrate the birthday. Inna's present to Dima is a stack, a queue and a deck. Inna wants her present to show Dima how great a programmer he is. For that, she is going to give Dima commands one by one. There are two types of commands: 1. Add a given number into one of containers. For the queue and the stack, you can add elements only to the end. For the deck, you can add elements to the beginning and to the end. 2. Extract a number from each of at most three distinct containers. Tell all extracted numbers to Inna and then empty all containers. In the queue container you can extract numbers only from the beginning. In the stack container you can extract numbers only from the end. In the deck number you can extract numbers from the beginning and from the end. You cannot extract numbers from empty containers. Every time Dima makes a command of the second type, Inna kisses Dima some (possibly zero) number of times. Dima knows Inna perfectly well, he is sure that this number equals the sum of numbers he extracts from containers during this operation. As we've said before, Dima knows Inna perfectly well and he knows which commands Inna will give to Dima and the order of the commands. Help Dima find the strategy that lets him give as more kisses as possible for his birthday!
The first line contains integer n (1 ≤ n ≤ 105) — the number of Inna's commands. Then n lines follow, describing Inna's commands. Each line consists an integer: 1. Integer a (1 ≤ a ≤ 105) means that Inna gives Dima a command to add number a into one of containers. 2. Integer 0 shows that Inna asks Dima to make at most three extractions from different containers.
Each command of the input must correspond to one line of the output — Dima's action. For the command of the first type (adding) print one word that corresponds to Dima's choice: - pushStack — add to the end of the stack; - pushQueue — add to the end of the queue; - pushFront — add to the beginning of the deck; - pushBack — add to the end of the deck. For a command of the second type first print an integer k (0 ≤ k ≤ 3), that shows the number of extract operations, then print k words separated by space. The words can be: - popStack — extract from the end of the stack; - popQueue — extract from the beginning of the line; - popFront — extract from the beginning from the deck; - popBack — extract from the end of the deck. The printed operations mustn't extract numbers from empty containers. Also, they must extract numbers from distinct containers. The printed sequence of actions must lead to the maximum number of kisses. If there are multiple sequences of actions leading to the maximum number of kisses, you are allowed to print any of them.
null
null
[{"input": "10\n0\n1\n0\n1\n2\n0\n1\n2\n3\n0", "output": "0\npushStack\n1 popStack\npushStack\npushQueue\n2 popStack popQueue\npushStack\npushQueue\npushFront\n3 popStack popQueue popFront"}, {"input": "4\n1\n2\n3\n0", "output": "pushStack\npushQueue\npushFront\n3 popStack popQueue popFront"}]
2,000
["constructive algorithms", "greedy", "implementation"]
26
[{"input": "10\r\n0\r\n1\r\n0\r\n1\r\n2\r\n0\r\n1\r\n2\r\n3\r\n0\r\n", "output": "0\r\npushStack\r\n1 popStack\r\npushStack\r\npushQueue\r\n2 popStack popQueue\r\npushStack\r\npushQueue\r\npushFront\r\n3 popStack popQueue popFront\r\n"}, {"input": "4\r\n1\r\n2\r\n3\r\n0\r\n", "output": "pushStack\r\npushQueue\r\npushFront\r\n3 popStack popQueue popFront\r\n"}, {"input": "2\r\n0\r\n1\r\n", "output": "0\r\npushQueue\r\n"}, {"input": "5\r\n1\r\n1\r\n1\r\n2\r\n1\r\n", "output": "pushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\n"}, {"input": "5\r\n3\r\n2\r\n3\r\n1\r\n0\r\n", "output": "pushStack\r\npushQueue\r\npushFront\r\npushBack\r\n3 popStack popQueue popFront\r\n"}, {"input": "49\r\n8735\r\n95244\r\n50563\r\n33648\r\n10711\r\n30217\r\n49166\r\n28240\r\n0\r\n97232\r\n12428\r\n16180\r\n58610\r\n61112\r\n74423\r\n56323\r\n43327\r\n0\r\n12549\r\n48493\r\n43086\r\n69266\r\n27033\r\n37338\r\n43900\r\n5570\r\n25293\r\n44517\r\n7183\r\n41969\r\n31944\r\n32247\r\n96959\r\n44890\r\n98237\r\n52601\r\n29081\r\n93641\r\n14980\r\n29539\r\n84672\r\n57310\r\n91014\r\n31721\r\n6944\r\n67672\r\n22040\r\n86269\r\n86709\r\n", "output": "pushBack\npushStack\npushQueue\npushBack\npushBack\npushBack\npushFront\npushBack\n3 popStack popQueue popFront\npushStack\npushBack\npushBack\npushBack\npushFront\npushQueue\npushBack\npushBack\n3 popStack popQueue popFront\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushStack\npushBack\npushBack\npushFront\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushQueue\npushBack\npushBack\npushBack\npushBack\n"}, {"input": "55\r\n73792\r\n39309\r\n73808\r\n47389\r\n34803\r\n87947\r\n32460\r\n14649\r\n70151\r\n35816\r\n8272\r\n78886\r\n71345\r\n61907\r\n16977\r\n85362\r\n0\r\n43792\r\n8118\r\n83254\r\n89459\r\n32230\r\n87068\r\n82617\r\n94847\r\n83528\r\n37629\r\n31438\r\n97413\r\n62260\r\n13651\r\n47564\r\n43543\r\n61292\r\n51025\r\n64106\r\n0\r\n19282\r\n35422\r\n19657\r\n95170\r\n10266\r\n43771\r\n3190\r\n93962\r\n11747\r\n43021\r\n91531\r\n88370\r\n1760\r\n10950\r\n77059\r\n61741\r\n52965\r\n10445\r\n", "output": "pushBack\npushBack\npushBack\npushBack\npushBack\npushStack\npushBack\npushBack\npushBack\npushBack\npushBack\npushFront\npushBack\npushBack\npushBack\npushQueue\n3 popStack popQueue popFront\npushBack\npushBack\npushBack\npushFront\npushBack\npushBack\npushBack\npushQueue\npushBack\npushBack\npushBack\npushStack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\npushBack\n3 popStack popQueue popFront\npushBack\npushBack\npushBack\npushBack\npushFront\npushBack\npushQueue\npushBack\npushBack\npushBack\npushBack\npushBack\npushStack\npushBack\npushBack\npushBack\npushBack\npushBack\n"}, {"input": "10\r\n1\r\n2\r\n3\r\n5\r\n4\r\n9\r\n8\r\n6\r\n7\r\n0\r\n", "output": "pushBack\r\npushBack\r\npushBack\r\npushBack\r\npushBack\r\npushStack\r\npushQueue\r\npushBack\r\npushFront\r\n3 popStack popQueue popFront\r\n"}, {"input": "10\r\n1\r\n3\r\n4\r\n2\r\n6\r\n8\r\n5\r\n7\r\n10\r\n9\r\n", "output": "pushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\npushQueue\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}]
false
stdio
import sys from collections import deque def main(input_path, output_path, submission_path): with open(input_path) as f: input_lines = f.read().splitlines() n = int(input_lines[0]) input_commands = input_lines[1:n+1] with open(submission_path) as f: submission_lines = [line.strip() for line in f] if len(submission_lines) != len(input_commands): print(0) return stack = [] queue = [] deck = deque() sub_idx = 0 for cmd in input_commands: if cmd == '0': if sub_idx >= len(submission_lines): print(0) return line = submission_lines[sub_idx] sub_idx += 1 parts = line.split() if not parts: print(0) return try: k = int(parts[0]) except: print(0) return if k < 0 or k > 3 or len(parts) != k + 1: print(0) return ops = parts[1:] # Compute max possible sum using copies stack_copy = list(stack) queue_copy = list(queue) deck_copy = deque(deck) contributions = [] if stack_copy: contributions.append(stack_copy[-1]) if queue_copy: contributions.append(queue_copy[0]) if deck_copy: deck_val = max(deck_copy[0], deck_copy[-1]) if deck_copy else 0 contributions.append(deck_val) contributions.sort(reverse=True) max_sum = sum(contributions[:3]) # Validate submission's ops temp_stack = list(stack) temp_queue = list(queue) temp_deck = deque(deck) sum_sub = 0 used = set() valid = True for op in ops: container = None if op == 'popStack': if not temp_stack: valid = False break sum_sub += temp_stack.pop() container = 'stack' elif op == 'popQueue': if not temp_queue: valid = False break sum_sub += temp_queue.pop(0) container = 'queue' elif op == 'popFront': if not temp_deck: valid = False break sum_sub += temp_deck.popleft() container = 'deck' elif op == 'popBack': if not temp_deck: valid = False break sum_sub += temp_deck.pop() container = 'deck' else: valid = False break if container in used: valid = False break used.add(container) if not valid or sum_sub != max_sum: print(0) return # Clear containers stack.clear() queue.clear() deck.clear() else: a = int(cmd) if sub_idx >= len(submission_lines): print(0) return line = submission_lines[sub_idx] sub_idx += 1 if line == 'pushStack': stack.append(a) elif line == 'pushQueue': queue.append(a) elif line == 'pushFront': deck.appendleft(a) elif line == 'pushBack': deck.append(a) else: print(0) return print(1) if __name__ == '__main__': main(sys.argv[1], sys.argv[2], sys.argv[3])
true
247/B
250
B
PyPy 3-64
TESTS
1
154
0
165702430
for i in range(int(input())): a = input().split(':') for j in range(len(a)): if len(a[j]) == 0: while len(a) < 8: a.insert(j, '') for j in range(8): a[j] = '0' * (4 - len(a[j])) + a[j] print(*a, sep=':')
40
92
0
187122060
n = int(input()) for i in range(n): s = input().split(':') if (s[0] == ''): s[0] = '0'*4 if s[-1] == '': s[-1] = '0'*4 idx = -1 for i in range(len(s)): if s[i] == '': idx = i break for i in range(len(s)): if (len(s[i]) < 4): s[i] = (4-len(s[i]))*'0'+s[i] if idx != -1: s[idx:idx+1] = (8-len(s)+1)*['0'*4] print(':'.join(s))
CROC-MBTU 2012, Final Round
CF
2,012
2
256
Restoring IPv6
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full. Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address. Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0. You can see examples of zero block shortenings below: - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::"; - "a56f:0000:0000:0124:0001:0000:1234:0ff0"  →  "a56f::0124:0001:0000:1234:0ff0"; - "a56f:0000:0000:0000:0001:0000:1234:0ff0"  →  "a56f:0000::0000:0001:0000:1234:0ff0"; - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::0000"; - "0000:0000:0000:0000:0000:0000:0000:0000"  →  "::". It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon. The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short. You've got several short records of IPv6 addresses. Restore their full record.
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100). Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:". It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
null
null
[{"input": "6\na56f:d3:0:0124:01:f19a:1000:00\na56f:00d3:0000:0124:0001::\na56f::0124:0001:0000:1234:0ff0\na56f:0000::0000:0001:0000:1234:0ff0\n::\n0ea::4d:f4:6:0", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\na56f:00d3:0000:0124:0001:0000:0000:0000\na56f:0000:0000:0124:0001:0000:1234:0ff0\na56f:0000:0000:0000:0001:0000:1234:0ff0\n0000:0000:0000:0000:0000:0000:0000:0000\n00ea:0000:0000:0000:004d:00f4:0006:0000"}]
1,500
[]
40
[{"input": "6\r\na56f:d3:0:0124:01:f19a:1000:00\r\na56f:00d3:0000:0124:0001::\r\na56f::0124:0001:0000:1234:0ff0\r\na56f:0000::0000:0001:0000:1234:0ff0\r\n::\r\n0ea::4d:f4:6:0\r\n", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\r\na56f:00d3:0000:0124:0001:0000:0000:0000\r\na56f:0000:0000:0124:0001:0000:1234:0ff0\r\na56f:0000:0000:0000:0001:0000:1234:0ff0\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n00ea:0000:0000:0000:004d:00f4:0006:0000\r\n"}, {"input": "10\r\n1::7\r\n0:0::1\r\n::1ed\r\n::30:44\r\n::eaf:ff:000b\r\n56fe::\r\ndf0:3df::\r\nd03:ab:0::\r\n85::0485:0\r\n::\r\n", "output": "0001:0000:0000:0000:0000:0000:0000:0007\n0000:0000:0000:0000:0000:0000:0000:0001\n0000:0000:0000:0000:0000:0000:0000:01ed\n0000:0000:0000:0000:0000:0000:0030:0044\n0000:0000:0000:0000:0000:0eaf:00ff:000b\n56fe:0000:0000:0000:0000:0000:0000:0000\n0df0:03df:0000:0000:0000:0000:0000:0000\n0d03:00ab:0000:0000:0000:0000:0000:0000\n0085:0000:0000:0000:0000:0000:0485:0000\n0000:0000:0000:0000:0000:0000:0000:0000\n"}, {"input": "6\r\n0:00:000:0000::\r\n1:01:001:0001::\r\nf:0f:00f:000f::\r\n1:10:100:1000::\r\nf:f0:f00:f000::\r\nf:ff:fff:ffff::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0001:0001:0001:0001:0000:0000:0000:0000\r\n000f:000f:000f:000f:0000:0000:0000:0000\r\n0001:0010:0100:1000:0000:0000:0000:0000\r\n000f:00f0:0f00:f000:0000:0000:0000:0000\r\n000f:00ff:0fff:ffff:0000:0000:0000:0000\r\n"}, {"input": "3\r\n::\r\n::\r\n::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n"}, {"input": "4\r\n1:2:3:4:5:6:7:8\r\n0:0:0:0:0:0:0:0\r\nf:0f:00f:000f:ff:0ff:00ff:fff\r\n0fff:0ff0:0f0f:f0f:0f0:f0f0:f00f:ff0f\r\n", "output": "0001:0002:0003:0004:0005:0006:0007:0008\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n000f:000f:000f:000f:00ff:00ff:00ff:0fff\r\n0fff:0ff0:0f0f:0f0f:00f0:f0f0:f00f:ff0f\r\n"}]
false
stdio
null
true
246/E
246
E
Python 3
TESTS
1
60
307,200
229854455
import heapq import itertools import sys # from typing import Counter # import itertools # import math # import os # import random from collections import defaultdict from functools import lru_cache from bisect import bisect_left, bisect_right # from heapq import heapify, heappop, heappush # from io import BytesIO, IOBase # from string import * # region fastio input = lambda: sys.stdin.readline().rstrip() sint = lambda: int(input()) mint = lambda: map(int, input().split()) ints = lambda: list(map(int, input().split())) # MOD = 998244353 MOD = 10 ** 9 + 7 # DIR = ((-1, 0), (0, 1), (1, 0), (0, -1)) sys.setrecursionlimit(3 * 10 ** 5) from types import GeneratorType def bootstrap(f, stack=[]): def wrappedfunc(*args, **kwargs): if stack: return f(*args, **kwargs) else: to = f(*args, **kwargs) while True: if type(to) is GeneratorType: stack.append(to) to = next(to) else: stack.pop() if not stack: break to = stack[-1].send(to) return to return wrappedfunc def solve() -> None: n = sint() adj = [[] for _ in range(n)] a = [None] * n for i in range(n): s, f = input().split() f = int(f) a[i] = s if f: adj[f - 1].append(i) queries = [[] for _ in range(n)] m = sint() for i in range(m): x, k = mint() queries[x - 1].append((k, i)) ans = [0] * m @bootstrap def dfs(u, fa, d): nonlocal ans dus = defaultdict(set) dus[d].add(a[u]) for v in adj[u]: if v != fa: dvs = yield dfs(v, u, d + 1) if len(dus) < len(dvs): dus, dvs = dvs, dus for k, st in dvs.items(): dus[k] |= st dvs.clear() for k, index in queries[u]: ans[index] = len(dus[d + k]) yield dus # return dus dfs(0, -1, 0) for x in ans: print(x) if __name__ == '__main__': solve()
72
1,652
89,702,400
230808843
import os,sys,random,threading from random import randint from copy import deepcopy from io import BytesIO, IOBase from types import GeneratorType from functools import lru_cache, reduce from bisect import bisect_left, bisect_right from collections import Counter, defaultdict, deque from itertools import accumulate, combinations, permutations from heapq import heapify, heappop, heappush from typing import Generic, Iterable, Iterator, TypeVar, Union, List from string import ascii_lowercase, ascii_uppercase from math import ceil, floor, sqrt, pi, factorial, gcd, log, log10, log2, inf from decimal import Decimal, getcontext BUFSIZE = 4096 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin = IOWrapper(sys.stdin) sys.stdout = IOWrapper(sys.stdout) mod = int(1e9 + 7) #998244353 inf = int(1e20) input = lambda: sys.stdin.readline().rstrip("\r\n") MI = lambda :map(int,input().split()) li = lambda :list(MI()) ii = lambda :int(input()) py = lambda :print("YES") pn = lambda :print("NO") DIRS = [(0, 1), (1, 0), (0, -1), (-1, 0)] # 右下左上 DIRS8 = [(0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1), (-1, 0),(-1, 1)] # →↘↓↙←↖↑↗ class BIT1: """单点修改,区间和查询""" __slots__ = "size", "bit", "tree" def __init__(self, n: int): self.size = n self.bit = n.bit_length() self.tree = [0]*(n+1) def add(self, index: int, delta: int) -> None: # index 必须大于0 while index <= self.size: self.tree[index]+=delta index += index & -index def _query(self, index: int) -> int: res = 0 while index > 0: res += self.tree[index] index -= index & -index return res def query(self, left: int, right: int) -> int: return self._query(right) - self._query(left - 1) n=ii() g=[[] for _ in range(n+5)] name=[None]*(n+5) for i in range(1,n+1): s,fa=input().split() name[i]=s fa=int(fa) g[fa].append(i) dep_node=[[] for _ in range(n+5)] stack = [(0, 0)] tmp = [] st = [None]*(n+5) #节点找dfn ed = [None]*(n+5) st[0] = 0 dep = [0]*(n+5) #深度 rank=[0]*(n+5) #dfn找节点 while stack: u, s = stack.pop() if s: ed[u] = len(tmp) - 1 else: st[u] = len(tmp) dep_node[dep[u]].append(st[u]) rank[st[u]]=u tmp.append(u) stack.append((u, 1)) for v in g[u]: if st[v] is None: stack.append((v, 0)) dep[v]=dep[u]+1 query=[[] for _ in range(n+5)] q=ii() res=[0]*q for i in range(q): v,k=li() if dep[v]+k<n+5: query[dep[v]+k].append([i,v]) for d in range(n+5): if len(dep_node[d])==0: continue arr=[] for i,v in query[d]: j=bisect_left(dep_node[d],st[v]) if j>=len(dep_node[d]) or dep_node[d][j]>ed[v]: continue j2=bisect_left(dep_node[d],ed[v]) if j2>=len(dep_node[d]) or dep_node[d][j2]>ed[v]: j2-=1 arr.append([j,j2,i]) bit=BIT1(len(dep_node[d])+5) arr.sort(key=lambda x:-x[1]) idx={} for i in range(len(dep_node[d])): if len(arr)==0: break node=rank[dep_node[d][i]] s=name[node] if s in idx: bit.add(idx[s]+1,-1) idx[s]=i bit.add(i+1,1) while arr and arr[-1][1]<=i: L,R,pos=arr.pop() res[pos]=bit.query(L+1,R+1) for i in res: print(i)
Codeforces Round 151 (Div. 2)
CF
2,012
3
256
Blood Cousins Return
Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique. We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b. We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b. In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0). We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a. Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.
The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n). It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.
Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.
null
null
[{"input": "6\npasha 0\ngerald 1\ngerald 1\nvalera 2\nigor 3\nolesya 1\n5\n1 1\n1 2\n1 3\n3 1\n6 1", "output": "2\n2\n0\n1\n0"}, {"input": "6\nvalera 0\nvalera 1\nvalera 1\ngerald 0\nvalera 4\nkolya 4\n7\n1 1\n1 2\n2 1\n2 2\n4 1\n5 1\n6 1", "output": "1\n0\n0\n0\n2\n0\n0"}]
2,400
["binary search", "data structures", "dfs and similar", "dp", "sortings"]
72
[{"input": "6\r\npasha 0\r\ngerald 1\r\ngerald 1\r\nvalera 2\r\nigor 3\r\nolesya 1\r\n5\r\n1 1\r\n1 2\r\n1 3\r\n3 1\r\n6 1\r\n", "output": "2\r\n2\r\n0\r\n1\r\n0\r\n"}, {"input": "6\r\nvalera 0\r\nvalera 1\r\nvalera 1\r\ngerald 0\r\nvalera 4\r\nkolya 4\r\n7\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n4 1\r\n5 1\r\n6 1\r\n", "output": "1\r\n0\r\n0\r\n0\r\n2\r\n0\r\n0\r\n"}, {"input": "1\r\nc 0\r\n20\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "2\r\ndd 0\r\nh 0\r\n20\r\n2 1\r\n2 1\r\n1 1\r\n1 1\r\n1 1\r\n2 1\r\n2 1\r\n1 1\r\n1 1\r\n2 1\r\n2 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n2 1\r\n2 1\r\n1 1\r\n1 1\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "3\r\na 0\r\nhj 0\r\ng 2\r\n20\r\n1 1\r\n2 1\r\n2 1\r\n1 1\r\n2 1\r\n2 1\r\n1 1\r\n1 1\r\n1 1\r\n3 1\r\n2 1\r\n1 1\r\n2 1\r\n3 1\r\n3 1\r\n2 1\r\n3 1\r\n2 1\r\n1 1\r\n2 1\r\n", "output": "0\r\n1\r\n1\r\n0\r\n1\r\n1\r\n0\r\n0\r\n0\r\n0\r\n1\r\n0\r\n1\r\n0\r\n0\r\n1\r\n0\r\n1\r\n0\r\n1\r\n"}, {"input": "3\r\ne 0\r\nb 0\r\nkg 0\r\n20\r\n1 1\r\n3 1\r\n3 1\r\n2 1\r\n1 1\r\n3 1\r\n2 1\r\n1 1\r\n2 1\r\n2 1\r\n2 1\r\n3 1\r\n3 1\r\n3 1\r\n1 1\r\n1 1\r\n2 1\r\n1 1\r\n2 1\r\n3 1\r\n", "output": "0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "4\r\nbf 0\r\nc 0\r\njc 2\r\ni 3\r\n20\r\n3 1\r\n3 1\r\n4 1\r\n2 1\r\n1 1\r\n4 1\r\n4 1\r\n2 1\r\n2 1\r\n3 1\r\n4 1\r\n4 1\r\n1 1\r\n2 1\r\n4 1\r\n1 1\r\n1 1\r\n4 1\r\n1 1\r\n4 1\r\n", "output": "1\r\n1\r\n0\r\n1\r\n0\r\n0\r\n0\r\n1\r\n1\r\n1\r\n0\r\n0\r\n0\r\n1\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n"}, {"input": "4\r\nfh 0\r\neg 0\r\nc 1\r\nhb 3\r\n20\r\n4 1\r\n3 1\r\n1 1\r\n4 1\r\n2 1\r\n4 1\r\n3 1\r\n4 1\r\n3 1\r\n3 1\r\n2 1\r\n1 1\r\n4 1\r\n2 1\r\n2 1\r\n3 1\r\n3 1\r\n4 1\r\n1 1\r\n3 1\r\n", "output": "0\r\n1\r\n1\r\n0\r\n0\r\n0\r\n1\r\n0\r\n1\r\n1\r\n0\r\n1\r\n0\r\n0\r\n0\r\n1\r\n1\r\n0\r\n1\r\n1\r\n"}]
false
stdio
null
true
247/B
250
B
PyPy 3
TESTS
1
186
0
203520865
for _ in range((int)(input())) : s = input().split(':') for _ in range(len(s)) : if s[_] == "" : while len(s) < 8 : s.insert(_ , "") for _ in range(8) : s[_] = s[_].zfill(4) print(":".join(s))
40
92
614,400
208397906
import ipaddress def solve(): src = input() addr = ipaddress.ip_address(src) print(addr.exploded) def main(): n = int(input()) for _ in range(n): solve() main()
CROC-MBTU 2012, Final Round
CF
2,012
2
256
Restoring IPv6
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full. Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address. Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0. You can see examples of zero block shortenings below: - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::"; - "a56f:0000:0000:0124:0001:0000:1234:0ff0"  →  "a56f::0124:0001:0000:1234:0ff0"; - "a56f:0000:0000:0000:0001:0000:1234:0ff0"  →  "a56f:0000::0000:0001:0000:1234:0ff0"; - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::0000"; - "0000:0000:0000:0000:0000:0000:0000:0000"  →  "::". It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon. The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short. You've got several short records of IPv6 addresses. Restore their full record.
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100). Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:". It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
null
null
[{"input": "6\na56f:d3:0:0124:01:f19a:1000:00\na56f:00d3:0000:0124:0001::\na56f::0124:0001:0000:1234:0ff0\na56f:0000::0000:0001:0000:1234:0ff0\n::\n0ea::4d:f4:6:0", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\na56f:00d3:0000:0124:0001:0000:0000:0000\na56f:0000:0000:0124:0001:0000:1234:0ff0\na56f:0000:0000:0000:0001:0000:1234:0ff0\n0000:0000:0000:0000:0000:0000:0000:0000\n00ea:0000:0000:0000:004d:00f4:0006:0000"}]
1,500
[]
40
[{"input": "6\r\na56f:d3:0:0124:01:f19a:1000:00\r\na56f:00d3:0000:0124:0001::\r\na56f::0124:0001:0000:1234:0ff0\r\na56f:0000::0000:0001:0000:1234:0ff0\r\n::\r\n0ea::4d:f4:6:0\r\n", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\r\na56f:00d3:0000:0124:0001:0000:0000:0000\r\na56f:0000:0000:0124:0001:0000:1234:0ff0\r\na56f:0000:0000:0000:0001:0000:1234:0ff0\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n00ea:0000:0000:0000:004d:00f4:0006:0000\r\n"}, {"input": "10\r\n1::7\r\n0:0::1\r\n::1ed\r\n::30:44\r\n::eaf:ff:000b\r\n56fe::\r\ndf0:3df::\r\nd03:ab:0::\r\n85::0485:0\r\n::\r\n", "output": "0001:0000:0000:0000:0000:0000:0000:0007\n0000:0000:0000:0000:0000:0000:0000:0001\n0000:0000:0000:0000:0000:0000:0000:01ed\n0000:0000:0000:0000:0000:0000:0030:0044\n0000:0000:0000:0000:0000:0eaf:00ff:000b\n56fe:0000:0000:0000:0000:0000:0000:0000\n0df0:03df:0000:0000:0000:0000:0000:0000\n0d03:00ab:0000:0000:0000:0000:0000:0000\n0085:0000:0000:0000:0000:0000:0485:0000\n0000:0000:0000:0000:0000:0000:0000:0000\n"}, {"input": "6\r\n0:00:000:0000::\r\n1:01:001:0001::\r\nf:0f:00f:000f::\r\n1:10:100:1000::\r\nf:f0:f00:f000::\r\nf:ff:fff:ffff::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0001:0001:0001:0001:0000:0000:0000:0000\r\n000f:000f:000f:000f:0000:0000:0000:0000\r\n0001:0010:0100:1000:0000:0000:0000:0000\r\n000f:00f0:0f00:f000:0000:0000:0000:0000\r\n000f:00ff:0fff:ffff:0000:0000:0000:0000\r\n"}, {"input": "3\r\n::\r\n::\r\n::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n"}, {"input": "4\r\n1:2:3:4:5:6:7:8\r\n0:0:0:0:0:0:0:0\r\nf:0f:00f:000f:ff:0ff:00ff:fff\r\n0fff:0ff0:0f0f:f0f:0f0:f0f0:f00f:ff0f\r\n", "output": "0001:0002:0003:0004:0005:0006:0007:0008\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n000f:000f:000f:000f:00ff:00ff:00ff:0fff\r\n0fff:0ff0:0f0f:0f0f:00f0:f0f0:f00f:ff0f\r\n"}]
false
stdio
null
true
82/B
82
B
Python 3
TESTS
32
278
307,200
14960585
c, v = [[0] * 201 for i in range(201)], [] n = int(input()) for i in range(n * (n - 1) // 2): a = list(map(int, input().split()))[1:] for x in a: for y in a: c[x][y] += 1 for i, ci in enumerate(c): if ci[i]: a = [j for j, cij in enumerate(ci) if cij == n - 1] v.append(a) for x in a: c[x][x] = 0 for vi in v: print(len(vi), *vi)
34
560
9,523,200
197184607
def solve(): n = int(input()) all_elements, set_list = set(), [] for i in range(n * (n - 1) // 2): inp_list = list(map(int, input().split())) inp_list.pop(0) set_list.append(set(inp_list)) all_elements |= set(inp_list) if n == 2: item = list(set_list[0]) print(len(item) - 1, *item[1:]) print(1, item[0]) return stat = {x: False for x in all_elements} ans = [] for el in all_elements: if not stat[el]: res = set(all_elements) for i, y in enumerate(set_list): if el in y: res = res & y ans.append([len(res)] + list(res)) for i in res: stat[i] = True for el in ans: for i in el: print(i, end=" ") print() solve()
Yandex.Algorithm 2011: Qualification 2
CF
2,011
2
256
Sets
Little Vasya likes very much to play with sets consisting of positive integers. To make the game more interesting, Vasya chose n non-empty sets in such a way, that no two of them have common elements. One day he wanted to show his friends just how interesting playing with numbers is. For that he wrote out all possible unions of two different sets on n·(n - 1) / 2 pieces of paper. Then he shuffled the pieces of paper. He had written out the numbers in the unions in an arbitrary order. For example, if n = 4, and the actual sets have the following form {1, 3}, {5}, {2, 4}, {7}, then the number of set pairs equals to six. The six pieces of paper can contain the following numbers: - 2, 7, 4. - 1, 7, 3; - 5, 4, 2; - 1, 3, 5; - 3, 1, 2, 4; - 5, 7. Then Vasya showed the pieces of paper to his friends, but kept the n sets secret from them. His friends managed to calculate which sets Vasya had thought of in the first place. And how about you, can you restore the sets by the given pieces of paper?
The first input file line contains a number n (2 ≤ n ≤ 200), n is the number of sets at Vasya's disposal. Then follow sets of numbers from the pieces of paper written on n·(n - 1) / 2 lines. Each set starts with the number ki (2 ≤ ki ≤ 200), which is the number of numbers written of the i-th piece of paper, and then follow ki numbers aij (1 ≤ aij ≤ 200). All the numbers on the lines are separated by exactly one space. It is guaranteed that the input data is constructed according to the above given rules from n non-intersecting sets.
Print on n lines Vasya's sets' description. The first number on the line shows how many numbers the current set has. Then the set should be recorded by listing its elements. Separate the numbers by spaces. Each number and each set should be printed exactly once. Print the sets and the numbers in the sets in any order. If there are several answers to that problem, print any of them. It is guaranteed that there is a solution.
null
null
[{"input": "4\n3 2 7 4\n3 1 7 3\n3 5 4 2\n3 1 3 5\n4 3 1 2 4\n2 5 7", "output": "1 7\n2 2 4\n2 1 3\n1 5"}, {"input": "4\n5 6 7 8 9 100\n4 7 8 9 1\n4 7 8 9 2\n3 1 6 100\n3 2 6 100\n2 1 2", "output": "3 7 8 9\n2 6 100\n1 1\n1 2"}, {"input": "3\n2 1 2\n2 1 3\n2 2 3", "output": "1 1\n1 2\n1 3"}]
1,700
["constructive algorithms", "hashing", "implementation"]
34
[{"input": "4\r\n3 2 7 4\r\n3 1 7 3\r\n3 5 4 2\r\n3 1 3 5\r\n4 3 1 2 4\r\n2 5 7\r\n", "output": "1 7 \r\n2 2 4 \r\n2 1 3 \r\n1 5 \r\n"}, {"input": "4\r\n5 6 7 8 9 100\r\n4 7 8 9 1\r\n4 7 8 9 2\r\n3 1 6 100\r\n3 2 6 100\r\n2 1 2\r\n", "output": "3 7 8 9 \r\n2 6 100 \r\n1 1 \r\n1 2 \r\n"}, {"input": "3\r\n2 1 2\r\n2 1 3\r\n2 2 3\r\n", "output": "1 1 \r\n1 2 \r\n1 3 \r\n"}, {"input": "3\r\n2 1 2\r\n10 1 90 80 70 60 50 40 30 20 10\r\n10 2 10 20 30 40 50 60 70 80 90\r\n", "output": "1 1 \r\n1 2 \r\n9 10 20 30 40 50 60 70 80 90 \r\n"}, {"input": "4\r\n4 56 44 53 43\r\n3 109 44 43\r\n3 109 56 53\r\n3 43 62 44\r\n3 62 56 53\r\n2 109 62\r\n", "output": "2 43 44 \r\n2 53 56 \r\n1 109 \r\n1 62 \r\n"}, {"input": "2\r\n2 1 2\r\n", "output": "1 2\r\n1 1"}, {"input": "2\r\n10 1 2 3 4 5 6 7 8 9 10\r\n", "output": "1 10\r\n9 1 2 3 4 5 6 7 8 9"}]
false
stdio
import sys def main(): input_path = sys.argv[1] submission_path = sys.argv[3] with open(input_path, 'r') as f: lines = [l.strip() for l in f if l.strip()] n = int(lines[0]) m = n * (n - 1) // 2 input_unions = set() for line in lines[1:1+m]: parts = list(map(int, line.split())) elements = parts[1:] input_unions.add(frozenset(elements)) with open(submission_path, 'r') as f: submission_lines = [l.strip() for l in f if l.strip()] if len(submission_lines) != n: print(0) return submission_sets = [] all_elements = set() for line in submission_lines: parts = list(map(int, line.split())) if not parts: print(0) return size = parts[0] if len(parts) - 1 != size: print(0) return elements = parts[1:] if len(elements) != len(set(elements)): print(0) return s = frozenset(elements) submission_sets.append(s) current = set(elements) if current & all_elements: print(0) return all_elements.update(current) for s in submission_sets: if not s: print(0) return generated = set() for i in range(n): for j in range(i+1, n): u = submission_sets[i].union(submission_sets[j]) generated.add(frozenset(u)) print(100 if generated == input_unions else 0) if __name__ == "__main__": main()
true
82/B
82
B
PyPy 3-64
TESTS
32
436
9,625,600
159106321
I = lambda:list(map(int, input().split())) unions = [] n = int(input()) loc = {} s1, s2 = -1, -1 for i in range(n*(n-1)//2): lst = I()[1:] for num in lst: if num in loc: s1, s2 = loc[num], i else: loc[num] = i unions.append(lst) sets = [] sets.append([v for v in unions[s1] if v in unions[s2]]) for union in unions: sub = [v for v in union if v not in sets[0]] if sub != union: sets.append(sub) for s in sets: print(len(s), ' '.join([str(num) for num in s]), sep = ' ')
34
186
2,457,600
4454031
n=int(input()) L=[] for i in range((n*(n-1))//2): A=input().split() L.append(A[1:]) x=L[0][0] Set=list(L[0]) Set.remove(x) for i in range(1,(n*(n-1))//2): if(x in L[i]): for item in L[i]: if(item in Set): Set.remove(item) break x=Set[0] Sets=[] Sets.append(list(Set)) for i in range((n*(n-1))//2): if(x in L[i]): Sets.append(list(L[i])) for item in Set: Sets[-1].remove(item) for item in Sets: print(len(item),end="") for z in item: print(" "+str(z),end="") print()
Yandex.Algorithm 2011: Qualification 2
CF
2,011
2
256
Sets
Little Vasya likes very much to play with sets consisting of positive integers. To make the game more interesting, Vasya chose n non-empty sets in such a way, that no two of them have common elements. One day he wanted to show his friends just how interesting playing with numbers is. For that he wrote out all possible unions of two different sets on n·(n - 1) / 2 pieces of paper. Then he shuffled the pieces of paper. He had written out the numbers in the unions in an arbitrary order. For example, if n = 4, and the actual sets have the following form {1, 3}, {5}, {2, 4}, {7}, then the number of set pairs equals to six. The six pieces of paper can contain the following numbers: - 2, 7, 4. - 1, 7, 3; - 5, 4, 2; - 1, 3, 5; - 3, 1, 2, 4; - 5, 7. Then Vasya showed the pieces of paper to his friends, but kept the n sets secret from them. His friends managed to calculate which sets Vasya had thought of in the first place. And how about you, can you restore the sets by the given pieces of paper?
The first input file line contains a number n (2 ≤ n ≤ 200), n is the number of sets at Vasya's disposal. Then follow sets of numbers from the pieces of paper written on n·(n - 1) / 2 lines. Each set starts with the number ki (2 ≤ ki ≤ 200), which is the number of numbers written of the i-th piece of paper, and then follow ki numbers aij (1 ≤ aij ≤ 200). All the numbers on the lines are separated by exactly one space. It is guaranteed that the input data is constructed according to the above given rules from n non-intersecting sets.
Print on n lines Vasya's sets' description. The first number on the line shows how many numbers the current set has. Then the set should be recorded by listing its elements. Separate the numbers by spaces. Each number and each set should be printed exactly once. Print the sets and the numbers in the sets in any order. If there are several answers to that problem, print any of them. It is guaranteed that there is a solution.
null
null
[{"input": "4\n3 2 7 4\n3 1 7 3\n3 5 4 2\n3 1 3 5\n4 3 1 2 4\n2 5 7", "output": "1 7\n2 2 4\n2 1 3\n1 5"}, {"input": "4\n5 6 7 8 9 100\n4 7 8 9 1\n4 7 8 9 2\n3 1 6 100\n3 2 6 100\n2 1 2", "output": "3 7 8 9\n2 6 100\n1 1\n1 2"}, {"input": "3\n2 1 2\n2 1 3\n2 2 3", "output": "1 1\n1 2\n1 3"}]
1,700
["constructive algorithms", "hashing", "implementation"]
34
[{"input": "4\r\n3 2 7 4\r\n3 1 7 3\r\n3 5 4 2\r\n3 1 3 5\r\n4 3 1 2 4\r\n2 5 7\r\n", "output": "1 7 \r\n2 2 4 \r\n2 1 3 \r\n1 5 \r\n"}, {"input": "4\r\n5 6 7 8 9 100\r\n4 7 8 9 1\r\n4 7 8 9 2\r\n3 1 6 100\r\n3 2 6 100\r\n2 1 2\r\n", "output": "3 7 8 9 \r\n2 6 100 \r\n1 1 \r\n1 2 \r\n"}, {"input": "3\r\n2 1 2\r\n2 1 3\r\n2 2 3\r\n", "output": "1 1 \r\n1 2 \r\n1 3 \r\n"}, {"input": "3\r\n2 1 2\r\n10 1 90 80 70 60 50 40 30 20 10\r\n10 2 10 20 30 40 50 60 70 80 90\r\n", "output": "1 1 \r\n1 2 \r\n9 10 20 30 40 50 60 70 80 90 \r\n"}, {"input": "4\r\n4 56 44 53 43\r\n3 109 44 43\r\n3 109 56 53\r\n3 43 62 44\r\n3 62 56 53\r\n2 109 62\r\n", "output": "2 43 44 \r\n2 53 56 \r\n1 109 \r\n1 62 \r\n"}, {"input": "2\r\n2 1 2\r\n", "output": "1 2\r\n1 1"}, {"input": "2\r\n10 1 2 3 4 5 6 7 8 9 10\r\n", "output": "1 10\r\n9 1 2 3 4 5 6 7 8 9"}]
false
stdio
import sys def main(): input_path = sys.argv[1] submission_path = sys.argv[3] with open(input_path, 'r') as f: lines = [l.strip() for l in f if l.strip()] n = int(lines[0]) m = n * (n - 1) // 2 input_unions = set() for line in lines[1:1+m]: parts = list(map(int, line.split())) elements = parts[1:] input_unions.add(frozenset(elements)) with open(submission_path, 'r') as f: submission_lines = [l.strip() for l in f if l.strip()] if len(submission_lines) != n: print(0) return submission_sets = [] all_elements = set() for line in submission_lines: parts = list(map(int, line.split())) if not parts: print(0) return size = parts[0] if len(parts) - 1 != size: print(0) return elements = parts[1:] if len(elements) != len(set(elements)): print(0) return s = frozenset(elements) submission_sets.append(s) current = set(elements) if current & all_elements: print(0) return all_elements.update(current) for s in submission_sets: if not s: print(0) return generated = set() for i in range(n): for j in range(i+1, n): u = submission_sets[i].union(submission_sets[j]) generated.add(frozenset(u)) print(100 if generated == input_unions else 0) if __name__ == "__main__": main()
true
82/B
82
B
PyPy 3
TESTS
32
530
27,852,800
129293680
def process(n, A): d = {} n2 = n*(n-1)//2 for i in range(n2): row = A[i] m = len(row) for j in range(1, m): c = row[j] if c not in d: d[c] = set([]) d[c].add(i) if n==2: row = A[0] S1 = [] S2 = [] for x in row: if x==row[0]: S1.append(x) else: S2.append(x) S1 = [len(S1)]+S1 S2 = [len(S2)]+S2 return [S1, S2] for c in d: break L = list(d[c]) i1 = L[0] i2 = L[1] s1 = set(A[i1][1:]).intersection(set(A[i2][1:])) S1 = [] for x in A[i1][1:]: if x in s1: S1.append(x) answer = [[len(S1)]+S1] for i in L: S2 = [] for x in A[i][1:]: if x not in s1: S2.append(x) S2 = [len(S2)]+S2 answer.append(S2) return answer n = int(input()) A = [] n2 = n*(n-1)//2 for i in range(n2): row = [int(x) for x in input().split()] A.append(row) answer = process(n, A) for x in answer: print(' '.join(map(str, x)))
34
186
10,649,600
126790096
#Med_Neji #x,y=map( lambda x:int(x)//abs(int(x)) ,input().split()) #x,y=map(int,input().split()) #n=int(input()) #l=list(map(int,input().split())) #s=input() #x=0 n=int(input()) l = [(set(input().split()[1:])) for _ in range(n*(n-1)//2)] if n==2: l=list(l[0]) print(1,l[0]) print(len(l)-1,*l[1:]) q=set() for i in range(n*(n-1)//2-1): for j in range(i+1,n*(n-1)//2): p=l[i]&l[j] if p!=set(): q.add(" ".join(sorted(list(p)))) q.add(" ".join(sorted(list(l[i]^p)))) q.add(" ".join(sorted(list(l[j]^p)))) if len(q)==n: break else: continue break for i in q: if i!="": print(len(i.split()),i)
Yandex.Algorithm 2011: Qualification 2
CF
2,011
2
256
Sets
Little Vasya likes very much to play with sets consisting of positive integers. To make the game more interesting, Vasya chose n non-empty sets in such a way, that no two of them have common elements. One day he wanted to show his friends just how interesting playing with numbers is. For that he wrote out all possible unions of two different sets on n·(n - 1) / 2 pieces of paper. Then he shuffled the pieces of paper. He had written out the numbers in the unions in an arbitrary order. For example, if n = 4, and the actual sets have the following form {1, 3}, {5}, {2, 4}, {7}, then the number of set pairs equals to six. The six pieces of paper can contain the following numbers: - 2, 7, 4. - 1, 7, 3; - 5, 4, 2; - 1, 3, 5; - 3, 1, 2, 4; - 5, 7. Then Vasya showed the pieces of paper to his friends, but kept the n sets secret from them. His friends managed to calculate which sets Vasya had thought of in the first place. And how about you, can you restore the sets by the given pieces of paper?
The first input file line contains a number n (2 ≤ n ≤ 200), n is the number of sets at Vasya's disposal. Then follow sets of numbers from the pieces of paper written on n·(n - 1) / 2 lines. Each set starts with the number ki (2 ≤ ki ≤ 200), which is the number of numbers written of the i-th piece of paper, and then follow ki numbers aij (1 ≤ aij ≤ 200). All the numbers on the lines are separated by exactly one space. It is guaranteed that the input data is constructed according to the above given rules from n non-intersecting sets.
Print on n lines Vasya's sets' description. The first number on the line shows how many numbers the current set has. Then the set should be recorded by listing its elements. Separate the numbers by spaces. Each number and each set should be printed exactly once. Print the sets and the numbers in the sets in any order. If there are several answers to that problem, print any of them. It is guaranteed that there is a solution.
null
null
[{"input": "4\n3 2 7 4\n3 1 7 3\n3 5 4 2\n3 1 3 5\n4 3 1 2 4\n2 5 7", "output": "1 7\n2 2 4\n2 1 3\n1 5"}, {"input": "4\n5 6 7 8 9 100\n4 7 8 9 1\n4 7 8 9 2\n3 1 6 100\n3 2 6 100\n2 1 2", "output": "3 7 8 9\n2 6 100\n1 1\n1 2"}, {"input": "3\n2 1 2\n2 1 3\n2 2 3", "output": "1 1\n1 2\n1 3"}]
1,700
["constructive algorithms", "hashing", "implementation"]
34
[{"input": "4\r\n3 2 7 4\r\n3 1 7 3\r\n3 5 4 2\r\n3 1 3 5\r\n4 3 1 2 4\r\n2 5 7\r\n", "output": "1 7 \r\n2 2 4 \r\n2 1 3 \r\n1 5 \r\n"}, {"input": "4\r\n5 6 7 8 9 100\r\n4 7 8 9 1\r\n4 7 8 9 2\r\n3 1 6 100\r\n3 2 6 100\r\n2 1 2\r\n", "output": "3 7 8 9 \r\n2 6 100 \r\n1 1 \r\n1 2 \r\n"}, {"input": "3\r\n2 1 2\r\n2 1 3\r\n2 2 3\r\n", "output": "1 1 \r\n1 2 \r\n1 3 \r\n"}, {"input": "3\r\n2 1 2\r\n10 1 90 80 70 60 50 40 30 20 10\r\n10 2 10 20 30 40 50 60 70 80 90\r\n", "output": "1 1 \r\n1 2 \r\n9 10 20 30 40 50 60 70 80 90 \r\n"}, {"input": "4\r\n4 56 44 53 43\r\n3 109 44 43\r\n3 109 56 53\r\n3 43 62 44\r\n3 62 56 53\r\n2 109 62\r\n", "output": "2 43 44 \r\n2 53 56 \r\n1 109 \r\n1 62 \r\n"}, {"input": "2\r\n2 1 2\r\n", "output": "1 2\r\n1 1"}, {"input": "2\r\n10 1 2 3 4 5 6 7 8 9 10\r\n", "output": "1 10\r\n9 1 2 3 4 5 6 7 8 9"}]
false
stdio
import sys def main(): input_path = sys.argv[1] submission_path = sys.argv[3] with open(input_path, 'r') as f: lines = [l.strip() for l in f if l.strip()] n = int(lines[0]) m = n * (n - 1) // 2 input_unions = set() for line in lines[1:1+m]: parts = list(map(int, line.split())) elements = parts[1:] input_unions.add(frozenset(elements)) with open(submission_path, 'r') as f: submission_lines = [l.strip() for l in f if l.strip()] if len(submission_lines) != n: print(0) return submission_sets = [] all_elements = set() for line in submission_lines: parts = list(map(int, line.split())) if not parts: print(0) return size = parts[0] if len(parts) - 1 != size: print(0) return elements = parts[1:] if len(elements) != len(set(elements)): print(0) return s = frozenset(elements) submission_sets.append(s) current = set(elements) if current & all_elements: print(0) return all_elements.update(current) for s in submission_sets: if not s: print(0) return generated = set() for i in range(n): for j in range(i+1, n): u = submission_sets[i].union(submission_sets[j]) generated.add(frozenset(u)) print(100 if generated == input_unions else 0) if __name__ == "__main__": main()
true
247/B
250
B
Python 3
TESTS
1
124
409,600
14751560
import re n = int(input()) for i in range(n): s = input().strip() group_count = s.count(':') + 1 if group_count < 8: s = re.sub('::', ':' + (9 - group_count) * '0000:', s) groups = s.split(':') for i, group in enumerate(groups): if len(group) < 4: groups[i] = (4 - len(group)) * '0' + group print(':'.join(groups))
40
92
614,400
208398186
import ipaddress for _ in range(int(input())): print(ipaddress.ip_address(input()).exploded)
CROC-MBTU 2012, Final Round
CF
2,012
2
256
Restoring IPv6
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full. Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address. Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0. You can see examples of zero block shortenings below: - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::"; - "a56f:0000:0000:0124:0001:0000:1234:0ff0"  →  "a56f::0124:0001:0000:1234:0ff0"; - "a56f:0000:0000:0000:0001:0000:1234:0ff0"  →  "a56f:0000::0000:0001:0000:1234:0ff0"; - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::0000"; - "0000:0000:0000:0000:0000:0000:0000:0000"  →  "::". It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon. The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short. You've got several short records of IPv6 addresses. Restore their full record.
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100). Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:". It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
null
null
[{"input": "6\na56f:d3:0:0124:01:f19a:1000:00\na56f:00d3:0000:0124:0001::\na56f::0124:0001:0000:1234:0ff0\na56f:0000::0000:0001:0000:1234:0ff0\n::\n0ea::4d:f4:6:0", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\na56f:00d3:0000:0124:0001:0000:0000:0000\na56f:0000:0000:0124:0001:0000:1234:0ff0\na56f:0000:0000:0000:0001:0000:1234:0ff0\n0000:0000:0000:0000:0000:0000:0000:0000\n00ea:0000:0000:0000:004d:00f4:0006:0000"}]
1,500
[]
40
[{"input": "6\r\na56f:d3:0:0124:01:f19a:1000:00\r\na56f:00d3:0000:0124:0001::\r\na56f::0124:0001:0000:1234:0ff0\r\na56f:0000::0000:0001:0000:1234:0ff0\r\n::\r\n0ea::4d:f4:6:0\r\n", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\r\na56f:00d3:0000:0124:0001:0000:0000:0000\r\na56f:0000:0000:0124:0001:0000:1234:0ff0\r\na56f:0000:0000:0000:0001:0000:1234:0ff0\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n00ea:0000:0000:0000:004d:00f4:0006:0000\r\n"}, {"input": "10\r\n1::7\r\n0:0::1\r\n::1ed\r\n::30:44\r\n::eaf:ff:000b\r\n56fe::\r\ndf0:3df::\r\nd03:ab:0::\r\n85::0485:0\r\n::\r\n", "output": "0001:0000:0000:0000:0000:0000:0000:0007\n0000:0000:0000:0000:0000:0000:0000:0001\n0000:0000:0000:0000:0000:0000:0000:01ed\n0000:0000:0000:0000:0000:0000:0030:0044\n0000:0000:0000:0000:0000:0eaf:00ff:000b\n56fe:0000:0000:0000:0000:0000:0000:0000\n0df0:03df:0000:0000:0000:0000:0000:0000\n0d03:00ab:0000:0000:0000:0000:0000:0000\n0085:0000:0000:0000:0000:0000:0485:0000\n0000:0000:0000:0000:0000:0000:0000:0000\n"}, {"input": "6\r\n0:00:000:0000::\r\n1:01:001:0001::\r\nf:0f:00f:000f::\r\n1:10:100:1000::\r\nf:f0:f00:f000::\r\nf:ff:fff:ffff::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0001:0001:0001:0001:0000:0000:0000:0000\r\n000f:000f:000f:000f:0000:0000:0000:0000\r\n0001:0010:0100:1000:0000:0000:0000:0000\r\n000f:00f0:0f00:f000:0000:0000:0000:0000\r\n000f:00ff:0fff:ffff:0000:0000:0000:0000\r\n"}, {"input": "3\r\n::\r\n::\r\n::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n"}, {"input": "4\r\n1:2:3:4:5:6:7:8\r\n0:0:0:0:0:0:0:0\r\nf:0f:00f:000f:ff:0ff:00ff:fff\r\n0fff:0ff0:0f0f:f0f:0f0:f0f0:f00f:ff0f\r\n", "output": "0001:0002:0003:0004:0005:0006:0007:0008\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n000f:000f:000f:000f:00ff:00ff:00ff:0fff\r\n0fff:0ff0:0f0f:0f0f:00f0:f0f0:f00f:ff0f\r\n"}]
false
stdio
null
true
247/B
250
B
Python 3
TESTS
1
124
4,608,000
20167972
n = int(input()) ips = [input() for j in range(n)] for ip in ips: colonNum = ip.count(":") if colonNum < 7: ip = ip.replace("::", ":" * (9 - colonNum)) comps = ip.split(":") for i in range(len(comps)): comps[i] = "0" * (4 - len(comps[i])) + comps[i] print(":".join(comps))
40
124
0
11596106
import sys def solve(): n = int(input()) for test in range(n): line = input().split(':') i = 0 while i < len(line): if i > 0 and line[i] == '' and line[i-1] == '': line.pop(i) i-=1 i+=1 for i, string in enumerate(line): if string == '': line[i] = modify(string, want = (32 - 4 * (len(line) - 1))) else: line[i] = modify(string) tempstr = ''.join(line) chunked = [tempstr[i: i + 4] for i in range(0, 32, 4)] print(':'.join(chunked)) def modify(string, want = 4): while len(string) < want: string='0' + string return string # print(line) # res = "" # otherblocks = 0 # sawprev = False # for c in line: # if c != ':' and sawprev == False: # otherblocks+=1 # sawprev = True # if c == ':': # sawprev = False # print(otherblocks) if sys.hexversion == 50594544 : sys.stdin = open("test.txt") solve()
CROC-MBTU 2012, Final Round
CF
2,012
2
256
Restoring IPv6
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full. Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address. Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0. You can see examples of zero block shortenings below: - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::"; - "a56f:0000:0000:0124:0001:0000:1234:0ff0"  →  "a56f::0124:0001:0000:1234:0ff0"; - "a56f:0000:0000:0000:0001:0000:1234:0ff0"  →  "a56f:0000::0000:0001:0000:1234:0ff0"; - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::0000"; - "0000:0000:0000:0000:0000:0000:0000:0000"  →  "::". It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon. The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short. You've got several short records of IPv6 addresses. Restore their full record.
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100). Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:". It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
null
null
[{"input": "6\na56f:d3:0:0124:01:f19a:1000:00\na56f:00d3:0000:0124:0001::\na56f::0124:0001:0000:1234:0ff0\na56f:0000::0000:0001:0000:1234:0ff0\n::\n0ea::4d:f4:6:0", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\na56f:00d3:0000:0124:0001:0000:0000:0000\na56f:0000:0000:0124:0001:0000:1234:0ff0\na56f:0000:0000:0000:0001:0000:1234:0ff0\n0000:0000:0000:0000:0000:0000:0000:0000\n00ea:0000:0000:0000:004d:00f4:0006:0000"}]
1,500
[]
40
[{"input": "6\r\na56f:d3:0:0124:01:f19a:1000:00\r\na56f:00d3:0000:0124:0001::\r\na56f::0124:0001:0000:1234:0ff0\r\na56f:0000::0000:0001:0000:1234:0ff0\r\n::\r\n0ea::4d:f4:6:0\r\n", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\r\na56f:00d3:0000:0124:0001:0000:0000:0000\r\na56f:0000:0000:0124:0001:0000:1234:0ff0\r\na56f:0000:0000:0000:0001:0000:1234:0ff0\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n00ea:0000:0000:0000:004d:00f4:0006:0000\r\n"}, {"input": "10\r\n1::7\r\n0:0::1\r\n::1ed\r\n::30:44\r\n::eaf:ff:000b\r\n56fe::\r\ndf0:3df::\r\nd03:ab:0::\r\n85::0485:0\r\n::\r\n", "output": "0001:0000:0000:0000:0000:0000:0000:0007\n0000:0000:0000:0000:0000:0000:0000:0001\n0000:0000:0000:0000:0000:0000:0000:01ed\n0000:0000:0000:0000:0000:0000:0030:0044\n0000:0000:0000:0000:0000:0eaf:00ff:000b\n56fe:0000:0000:0000:0000:0000:0000:0000\n0df0:03df:0000:0000:0000:0000:0000:0000\n0d03:00ab:0000:0000:0000:0000:0000:0000\n0085:0000:0000:0000:0000:0000:0485:0000\n0000:0000:0000:0000:0000:0000:0000:0000\n"}, {"input": "6\r\n0:00:000:0000::\r\n1:01:001:0001::\r\nf:0f:00f:000f::\r\n1:10:100:1000::\r\nf:f0:f00:f000::\r\nf:ff:fff:ffff::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0001:0001:0001:0001:0000:0000:0000:0000\r\n000f:000f:000f:000f:0000:0000:0000:0000\r\n0001:0010:0100:1000:0000:0000:0000:0000\r\n000f:00f0:0f00:f000:0000:0000:0000:0000\r\n000f:00ff:0fff:ffff:0000:0000:0000:0000\r\n"}, {"input": "3\r\n::\r\n::\r\n::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n"}, {"input": "4\r\n1:2:3:4:5:6:7:8\r\n0:0:0:0:0:0:0:0\r\nf:0f:00f:000f:ff:0ff:00ff:fff\r\n0fff:0ff0:0f0f:f0f:0f0:f0f0:f00f:ff0f\r\n", "output": "0001:0002:0003:0004:0005:0006:0007:0008\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n000f:000f:000f:000f:00ff:00ff:00ff:0fff\r\n0fff:0ff0:0f0f:0f0f:00f0:f0f0:f00f:ff0f\r\n"}]
false
stdio
null
true
247/B
250
B
Python 3
TESTS
1
154
0
51964779
n=int(input()) for i in range(n): s=input().split(":") t=True for i in range(len(s)): if 0<len(s[i])<4: s[i]=(4-len(s[i]))*'0'+s[i] elif t and len(s[i])==0: s[i]="0000:"*(8-len(s))+"0000" t=False elif len(s[i])==0: s[i]="0000" for i in range(len(s)-1): print(s[i],end=":") print(s[len(s)-1])
40
124
0
117566553
for _ in range(int(input())): D = input().split(':') if not D[0]: D = D[1:] if not D[-1]: D.pop() n = len(D) for i in range(n): D[i] = D[i].zfill(4) if D[i] else '0000:' * (8 - n) + '0000' print(':'.join(D))
CROC-MBTU 2012, Final Round
CF
2,012
2
256
Restoring IPv6
An IPv6-address is a 128-bit number. For convenience, this number is recorded in blocks of 16 bits in hexadecimal record, the blocks are separated by colons — 8 blocks in total, each block has four hexadecimal digits. Here is an example of the correct record of a IPv6 address: "0124:5678:90ab:cdef:0124:5678:90ab:cdef". We'll call such format of recording an IPv6-address full. Besides the full record of an IPv6 address there is a short record format. The record of an IPv6 address can be shortened by removing one or more leading zeroes at the beginning of each block. However, each block should contain at least one digit in the short format. For example, the leading zeroes can be removed like that: "a56f:00d3:0000:0124:0001:f19a:1000:0000" → "a56f:d3:0:0124:01:f19a:1000:00". There are more ways to shorten zeroes in this IPv6 address. Some IPv6 addresses contain long sequences of zeroes. Continuous sequences of 16-bit zero blocks can be shortened to "::". A sequence can consist of one or several consecutive blocks, with all 16 bits equal to 0. You can see examples of zero block shortenings below: - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::"; - "a56f:0000:0000:0124:0001:0000:1234:0ff0"  →  "a56f::0124:0001:0000:1234:0ff0"; - "a56f:0000:0000:0000:0001:0000:1234:0ff0"  →  "a56f:0000::0000:0001:0000:1234:0ff0"; - "a56f:00d3:0000:0124:0001:0000:0000:0000"  →  "a56f:00d3:0000:0124:0001::0000"; - "0000:0000:0000:0000:0000:0000:0000:0000"  →  "::". It is not allowed to shorten zero blocks in the address more than once. This means that the short record can't contain the sequence of characters "::" more than once. Otherwise, it will sometimes be impossible to determine the number of zero blocks, each represented by a double colon. The format of the record of the IPv6 address after removing the leading zeroes and shortening the zero blocks is called short. You've got several short records of IPv6 addresses. Restore their full record.
The first line contains a single integer n — the number of records to restore (1 ≤ n ≤ 100). Each of the following n lines contains a string — the short IPv6 addresses. Each string only consists of string characters "0123456789abcdef:". It is guaranteed that each short address is obtained by the way that is described in the statement from some full IPv6 address.
For each short IPv6 address from the input print its full record on a separate line. Print the full records for the short IPv6 addresses in the order, in which the short records follow in the input.
null
null
[{"input": "6\na56f:d3:0:0124:01:f19a:1000:00\na56f:00d3:0000:0124:0001::\na56f::0124:0001:0000:1234:0ff0\na56f:0000::0000:0001:0000:1234:0ff0\n::\n0ea::4d:f4:6:0", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\na56f:00d3:0000:0124:0001:0000:0000:0000\na56f:0000:0000:0124:0001:0000:1234:0ff0\na56f:0000:0000:0000:0001:0000:1234:0ff0\n0000:0000:0000:0000:0000:0000:0000:0000\n00ea:0000:0000:0000:004d:00f4:0006:0000"}]
1,500
[]
40
[{"input": "6\r\na56f:d3:0:0124:01:f19a:1000:00\r\na56f:00d3:0000:0124:0001::\r\na56f::0124:0001:0000:1234:0ff0\r\na56f:0000::0000:0001:0000:1234:0ff0\r\n::\r\n0ea::4d:f4:6:0\r\n", "output": "a56f:00d3:0000:0124:0001:f19a:1000:0000\r\na56f:00d3:0000:0124:0001:0000:0000:0000\r\na56f:0000:0000:0124:0001:0000:1234:0ff0\r\na56f:0000:0000:0000:0001:0000:1234:0ff0\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n00ea:0000:0000:0000:004d:00f4:0006:0000\r\n"}, {"input": "10\r\n1::7\r\n0:0::1\r\n::1ed\r\n::30:44\r\n::eaf:ff:000b\r\n56fe::\r\ndf0:3df::\r\nd03:ab:0::\r\n85::0485:0\r\n::\r\n", "output": "0001:0000:0000:0000:0000:0000:0000:0007\n0000:0000:0000:0000:0000:0000:0000:0001\n0000:0000:0000:0000:0000:0000:0000:01ed\n0000:0000:0000:0000:0000:0000:0030:0044\n0000:0000:0000:0000:0000:0eaf:00ff:000b\n56fe:0000:0000:0000:0000:0000:0000:0000\n0df0:03df:0000:0000:0000:0000:0000:0000\n0d03:00ab:0000:0000:0000:0000:0000:0000\n0085:0000:0000:0000:0000:0000:0485:0000\n0000:0000:0000:0000:0000:0000:0000:0000\n"}, {"input": "6\r\n0:00:000:0000::\r\n1:01:001:0001::\r\nf:0f:00f:000f::\r\n1:10:100:1000::\r\nf:f0:f00:f000::\r\nf:ff:fff:ffff::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0001:0001:0001:0001:0000:0000:0000:0000\r\n000f:000f:000f:000f:0000:0000:0000:0000\r\n0001:0010:0100:1000:0000:0000:0000:0000\r\n000f:00f0:0f00:f000:0000:0000:0000:0000\r\n000f:00ff:0fff:ffff:0000:0000:0000:0000\r\n"}, {"input": "3\r\n::\r\n::\r\n::\r\n", "output": "0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n"}, {"input": "4\r\n1:2:3:4:5:6:7:8\r\n0:0:0:0:0:0:0:0\r\nf:0f:00f:000f:ff:0ff:00ff:fff\r\n0fff:0ff0:0f0f:f0f:0f0:f0f0:f00f:ff0f\r\n", "output": "0001:0002:0003:0004:0005:0006:0007:0008\r\n0000:0000:0000:0000:0000:0000:0000:0000\r\n000f:000f:000f:000f:00ff:00ff:00ff:0fff\r\n0fff:0ff0:0f0f:0f0f:00f0:f0f0:f00f:ff0f\r\n"}]
false
stdio
null
true
665/B
665
B
Python 3
TESTS
3
46
4,608,000
17403780
#!/usr/bin/python3 n, m, k = map(int, input().split()) positionOf = [x-1 for x in map(int, input().split())] sol = 0 for i in range(n): v = [x-1 for x in map(int, input().split())] for x in v: sol += positionOf[x] + 1 for y in range(k): if positionOf[y] < positionOf[x]: positionOf[y] += 1 positionOf[x] = 0 print(sol)
10
46
0
186638540
n, m, k = map(int, input().split()) a = list(map(int,input().split())) ans = m*n for i in range(n): b = list(map(int,input().split())) for j in range(m): pos = a.index(b[j]) ans += pos del a[pos] a.insert(0,b[j]) print(ans)
Educational Codeforces Round 12
ICPC
2,016
1
256
Shopping
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online. The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order. Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer. When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating. Your task is to calculate the total time it takes for Ayush to process all the orders. You can assume that the market has endless stock.
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market. The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k. Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Print the only integer t — the total time needed for Ayush to process all the orders.
null
Customer 1 wants the items 1 and 5. pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5]. pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2]. Time taken for the first customer is 3 + 5 = 8. Customer 2 wants the items 3 and 1. pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2]. pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2]. Time taken for the second customer is 3 + 3 = 6. Total time is 8 + 6 = 14. Formally pos(x) is the index of x in the current row.
[{"input": "2 2 5\n3 4 1 2 5\n1 5\n3 1", "output": "14"}]
1,400
["brute force"]
10
[{"input": "2 2 5\r\n3 4 1 2 5\r\n1 5\r\n3 1\r\n", "output": "14\r\n"}, {"input": "4 4 4\r\n1 2 3 4\r\n3 4 2 1\r\n4 3 2 1\r\n4 1 2 3\r\n4 1 2 3\r\n", "output": "59\r\n"}, {"input": "1 1 1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "10 1 100\r\n1 55 67 75 40 86 24 84 82 26 81 23 70 79 51 54 21 78 31 98 68 93 66 88 99 65 20 52 35 85 16 12 94 100 59 56 18 33 47 46 71 8 38 57 2 92 3 95 6 4 87 22 48 80 15 29 11 45 72 76 44 60 91 90 39 74 41 36 13 27 53 83 32 5 30 63 89 64 49 17 9 97 69 14 50 77 37 96 10 42 28 34 61 19 73 7 62 43 58 25\r\n33\r\n69\r\n51\r\n7\r\n68\r\n70\r\n1\r\n35\r\n24\r\n7\r\n", "output": "335\r\n"}, {"input": "100 1 1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "100\r\n"}, {"input": "3 2 3\r\n3 1 2\r\n1 2\r\n2 1\r\n2 3\r\n", "output": "13\r\n"}, {"input": "10 10 10\r\n3 4 1 2 8 9 5 10 6 7\r\n9 10 7 8 6 1 2 3 4 5\r\n2 5 3 6 1 4 9 7 8 10\r\n2 9 1 8 4 7 5 10 6 3\r\n10 9 7 1 3 6 2 8 5 4\r\n2 5 1 3 7 10 4 9 8 6\r\n6 1 8 7 9 2 3 5 4 10\r\n1 3 2 8 6 9 4 10 5 7\r\n5 2 4 8 6 1 10 9 3 7\r\n5 1 7 10 4 6 2 8 9 3\r\n2 1 3 9 7 10 6 4 8 5\r\n", "output": "771\r\n"}]
false
stdio
null
true
632/F
632
F
PyPy 3-64
TESTS
2
46
0
216944293
import sys def input(): return sys.stdin.readline().strip() n = int(input()) l = [0]*n for x in range(n): l[x] = tuple(input().split()) l2 = list(zip(*l)) for x in range(n): if l[x] != l2[x]: print("NOT MAGIC") break else: print("MAGIC")
34
3,306
474,521,600
142642726
from collections import defaultdict, 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) dist = [-1] * n dist[s] = 0 while q: i = q.popleft() di = dist[i] for j, c in G[i]: if dist[j] == -1: dist[j] = max(di, c) q.append(j) return dist n = int(input()) a = [list(map(int, input().split())) for _ in range(n)] d = defaultdict(lambda : []) ok = 1 for i in range(n): ai = a[i] for j in range(i + 1, n): if ai[j] ^ a[j][i]: ok = 0 break d[ai[j]].append(i * n + j) if ai[i]: ok = 0 if not ok: break if ok: root = [i for i in range(n)] rank = [1 for _ in range(n)] size = [1 for _ in range(n)] G = [[] for _ in range(n)] c = 1 for i in sorted(d.keys()): for x in d[i]: u, v = x // n, x % n if not same(u, v): unite(u, v) auv = a[u][v] G[u].append((v, auv)) G[v].append((u, auv)) c += 1 if not c ^ n: break if not c ^ n: break for i in range(n): dist = bfs(i) ai = a[i] for x, y in zip(dist, ai): if x < y: ok = 0 break if not ok: break ans = "MAGIC" if ok else "NOT MAGIC" print(ans)
Educational Codeforces Round 9
ICPC
2,016
5
512
Magic Matrix
You're given a matrix A of size n × n. Let's call the matrix with nonnegative elements magic if it is symmetric (so aij = aji), aii = 0 and aij ≤ max(aik, ajk) for all triples i, j, k. Note that i, j, k do not need to be distinct. Determine if the matrix is magic. As the input/output can reach very huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
The first line contains integer n (1 ≤ n ≤ 2500) — the size of the matrix A. Each of the next n lines contains n integers aij (0 ≤ aij < 109) — the elements of the matrix A. Note that the given matrix not necessarily is symmetric and can be arbitrary.
Print ''MAGIC" (without quotes) if the given matrix A is magic. Otherwise print ''NOT MAGIC".
null
null
[{"input": "3\n0 1 2\n1 0 2\n2 2 0", "output": "MAGIC"}, {"input": "2\n0 1\n2 3", "output": "NOT MAGIC"}, {"input": "4\n0 1 2 3\n1 0 3 4\n2 3 0 5\n3 4 5 0", "output": "NOT MAGIC"}]
2,400
["brute force", "divide and conquer", "graphs", "matrices", "trees"]
34
[{"input": "3\r\n0 1 2\r\n1 0 2\r\n2 2 0\r\n", "output": "MAGIC\r\n"}, {"input": "2\r\n0 1\r\n2 3\r\n", "output": "NOT MAGIC\r\n"}, {"input": "4\r\n0 1 2 3\r\n1 0 3 4\r\n2 3 0 5\r\n3 4 5 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "5\r\n0 2 5 9 5\r\n2 0 5 9 5\r\n5 5 0 9 4\r\n9 9 9 0 9\r\n5 5 4 9 0\r\n", "output": "MAGIC\r\n"}, {"input": "10\r\n0 16 5 14 14 17 14 14 9 14\r\n16 0 16 16 16 17 16 16 16 16\r\n5 16 0 14 14 17 14 14 9 14\r\n14 16 14 0 7 17 3 8 14 12\r\n14 16 14 7 0 17 7 8 14 12\r\n17 17 17 17 17 0 17 17 17 17\r\n14 16 14 3 7 17 0 8 14 12\r\n14 16 14 8 8 17 8 0 14 12\r\n9 16 9 14 14 17 14 14 0 14\r\n14 16 14 12 12 17 12 12 14 0\r\n", "output": "MAGIC\r\n"}, {"input": "5\r\n0 2 9 10 10\r\n2 0 5 2 3\r\n9 5 0 1 1\r\n10 2 1 0 7\r\n10 3 1 7 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "10\r\n0 18 0 12 20 3 14 12 13 2\r\n18 0 6 12 7 20 1 9 13 10\r\n0 6 0 15 17 9 16 15 1 0\r\n12 12 15 0 0 8 19 20 11 11\r\n20 7 17 0 0 3 5 14 8 3\r\n3 20 9 8 3 0 7 16 20 17\r\n14 1 16 19 5 7 0 14 18 14\r\n12 9 15 20 14 16 14 0 6 19\r\n13 13 1 11 8 20 18 6 0 13\r\n2 10 0 11 3 17 14 19 13 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "2\r\n1 1\r\n1 1\r\n", "output": "NOT MAGIC\r\n"}, {"input": "3\r\n0 999999998 999999998\r\n999999998 0 999999999\r\n999999998 999999999 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "5\r\n0 3 7 1 1\r\n3 0 7 1 1\r\n7 7 0 7 7\r\n1 1 7 0 1\r\n1 1 7 1 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "5\r\n0 2 9 1 1\r\n2 0 9 1 1\r\n9 9 0 9 9\r\n1 1 9 0 1\r\n1 1 9 1 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "3\r\n0 1 2\r\n0 0 2\r\n2 2 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "3\r\n1 2 3\r\n2 1 3\r\n3 3 1\r\n", "output": "NOT MAGIC\r\n"}, {"input": "4\r\n0 9 9 9\r\n9 0 1 2\r\n9 1 0 3\r\n9 2 3 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "2\r\n2 2\r\n2 2\r\n", "output": "NOT MAGIC\r\n"}, {"input": "4\r\n0 1 2 9\r\n1 0 3 9\r\n2 3 0 9\r\n9 9 9 0\r\n", "output": "NOT MAGIC\r\n"}, {"input": "4\r\n0 0 0 4\r\n0 0 3 4\r\n0 3 0 4\r\n4 4 4 0\r\n", "output": "NOT MAGIC\r\n"}]
false
stdio
null
true
367/A
367
A
PyPy 3-64
TESTS
2
46
0
174047011
import sys input = sys.stdin.readline s = input()[:-1] d = [[0, 0, 0]] for i in s: e = d[-1].copy() e[ord(i)-120] += 1 d.append(e) for _ in range(int(input())): a, b = map(int, input().split()) q = [] for i in range(3): q.append(d[b][i]-d[a-1][i]) q.sort() if q[2] - q[0] > 1: print('NO') else: print('YES')
38
624
6,451,200
5285846
import sys s=sys.stdin.readline().split()[0] m=int(sys.stdin.readline()) Numx=[] Numy=[] Numz=[] x=0 y=0 z=0 for i in range(len(s)): if(s[i]=='x'): x+=1 if(s[i]=='y'): y+=1 if(s[i]=='z'): z+=1 Numx.append(x) Numy.append(y) Numz.append(z) Ans="" for M in range(m): s,e=map(int,sys.stdin.readline().split()) if(e-s+1<=2): Ans+="YES\n" continue s-=1 e-=1 x=Numx[e] y=Numy[e] z=Numz[e] if(s!=0): x-=Numx[s-1] y-=Numy[s-1] z-=Numz[s-1] if(x==y==z): Ans+="YES\n" continue L=[x,y,z] L.sort() if(L[0]==L[1] and L[2]==L[1]+1): Ans+="YES\n" continue if(L[1]==L[2] and L[0]==L[1]-1): Ans+="YES\n" else: Ans+="NO\n" sys.stdout.write(Ans)
Codeforces Round 215 (Div. 1)
CF
2,013
1
256
Sereja and Algorithm
Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q1q2... qk. The algorithm consists of two steps: 1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2. 2. Rearrange the letters of the found subsequence randomly and go to step 1. Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string. Sereja wants to test his algorithm. For that, he has string s = s1s2... sn, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring slisli + 1... sri (1 ≤ li ≤ ri ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (li, ri) determine if the algorithm works correctly on this test or not.
The first line contains non-empty string s, its length (n) doesn't exceed 105. It is guaranteed that string s only contains characters: 'x', 'y', 'z'. The second line contains integer m (1 ≤ m ≤ 105) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
null
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
[{"input": "zyxxxxxxyyz\n5\n5 5\n1 3\n1 11\n1 4\n3 6", "output": "YES\nYES\nNO\nYES\nNO"}]
1,500
["data structures", "implementation"]
38
[{"input": "zyxxxxxxyyz\r\n5\r\n5 5\r\n1 3\r\n1 11\r\n1 4\r\n3 6\r\n", "output": "YES\r\nYES\r\nNO\r\nYES\r\nNO\r\n"}, {"input": "yxzyzxzzxyyzzxxxzyyzzyzxxzxyzyyzxyzxyxxyzxyxzyzxyzxyyxzzzyzxyyxyzxxy\r\n10\r\n17 67\r\n6 35\r\n12 45\r\n56 56\r\n14 30\r\n25 54\r\n1 1\r\n46 54\r\n3 33\r\n19 40\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\nYES\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "xxxxyyxyyzzyxyxzxyzyxzyyyzyzzxxxxzyyzzzzyxxxxzzyzzyzx\r\n5\r\n4 4\r\n3 3\r\n1 24\r\n3 28\r\n18 39\r\n", "output": "YES\r\nYES\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "yzxyzxyzxzxzyzxyzyzzzyxzyz\r\n9\r\n4 6\r\n2 7\r\n3 5\r\n14 24\r\n3 13\r\n2 24\r\n2 5\r\n2 14\r\n3 15\r\n", "output": "YES\r\nYES\r\nYES\r\nNO\r\nYES\r\nNO\r\nYES\r\nNO\r\nNO\r\n"}, {"input": "zxyzxyzyyzxzzxyzxyzx\r\n15\r\n7 10\r\n17 17\r\n6 7\r\n8 14\r\n4 7\r\n11 18\r\n12 13\r\n1 1\r\n3 8\r\n1 1\r\n9 17\r\n4 4\r\n5 11\r\n3 15\r\n1 1\r\n", "output": "NO\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nYES\r\nNO\r\nYES\r\nYES\r\nNO\r\nYES\r\n"}, {"input": "x\r\n1\r\n1 1\r\n", "output": "YES\r\n"}]
false
stdio
null
true
840/B
840
B
PyPy 3
PRETESTS
3
93
0
29576435
import sys sys.setrecursionlimit(500000) n, m = map(int, sys.stdin.readline().split()) d = list(map(int, sys.stdin.readline().split())) gph = [[] for _ in range(n)] for _ in range(m): u, v = map(int, sys.stdin.readline().split()) u -= 1 v -= 1 gph[u].append((v, _)) gph[v].append((u, _)) t = -1 if d.count(1) % 2 == 1: if 0 not in d: print(-1) exit(0) t = d.index(0) ans = [False] * m vis = [False] * n def go (u): vis[u] = True ret = (d[u] == 1) or (u == t) for v, i in gph[u]: if not vis[v] and go(v): ret = not ret ans[i] = True return ret go(0) print(ans.count(True)) print("\n".join([str(i+1) for i in range(m) if ans[i]]))
73
2,979
87,756,800
167362143
import bisect import copy import decimal import fractions import heapq import itertools import math import random import sys import time from collections import Counter,deque,defaultdict from functools import lru_cache,reduce from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max def _heappush_max(heap,item): heap.append(item) heapq._siftdown_max(heap, 0, len(heap)-1) def _heappushpop_max(heap, item): if heap and item < heap[0]: item, heap[0] = heap[0], item heapq._siftup_max(heap, 0) return item from math import gcd as GCD read=sys.stdin.read readline=sys.stdin.readline readlines=sys.stdin.readlines class Graph: def __init__(self,V,edges=False,graph=False,directed=False,weighted=False,inf=float("inf")): self.V=V self.directed=directed self.weighted=weighted self.inf=inf if graph: self.graph=graph self.edges=[] for i in range(self.V): if self.weighted: for j,d in self.graph[i]: if self.directed or not self.directed and i<=j: self.edges.append((i,j,d)) else: for j in self.graph[i]: if self.directed or not self.directed and i<=j: self.edges.append((i,j)) else: self.edges=edges self.graph=[[] for i in range(self.V)] if weighted: for i,j,d in self.edges: self.graph[i].append((j,d)) if not self.directed: self.graph[j].append((i,d)) else: for i,j in self.edges: self.graph[i].append(j) if not self.directed: self.graph[j].append(i) def SIV_DFS(self,s,bipartite_graph=False,cycle_detection=False,directed_acyclic=False,euler_tour=False,linked_components=False,lowlink=False,parents=False,postorder=False,preorder=False,subtree_size=False,topological_sort=False,unweighted_dist=False,weighted_dist=False): seen=[False]*self.V finished=[False]*self.V if directed_acyclic or cycle_detection or topological_sort: dag=True if euler_tour: et=[] if linked_components: lc=[] if lowlink: order=[None]*self.V ll=[None]*self.V idx=0 if parents or cycle_detection or lowlink or subtree_size: ps=[None]*self.V if postorder or topological_sort: post=[] if preorder: pre=[] if subtree_size: ss=[1]*self.V if unweighted_dist or bipartite_graph: uwd=[self.inf]*self.V uwd[s]=0 if weighted_dist: wd=[self.inf]*self.V wd[s]=0 stack=[(s,0)] if self.weighted else [s] while stack: if self.weighted: x,d=stack.pop() else: x=stack.pop() if not seen[x]: seen[x]=True stack.append((x,d) if self.weighted else x) if euler_tour: et.append(x) if linked_components: lc.append(x) if lowlink: order[x]=idx ll[x]=idx idx+=1 if preorder: pre.append(x) for y in self.graph[x]: if self.weighted: y,d=y if not seen[y]: stack.append((y,d) if self.weighted else y) if parents or cycle_detection or lowlink or subtree_size: ps[y]=x if unweighted_dist or bipartite_graph: uwd[y]=uwd[x]+1 if weighted_dist: wd[y]=wd[x]+d elif not finished[y]: if (directed_acyclic or cycle_detection or topological_sort) and dag: dag=False if cycle_detection: cd=(y,x) elif not finished[x]: finished[x]=True if euler_tour: et.append(~x) if lowlink: bl=True for y in self.graph[x]: if self.weighted: y,d=y if ps[x]==y and bl: bl=False continue ll[x]=min(ll[x],order[y]) if x!=s: ll[ps[x]]=min(ll[ps[x]],ll[x]) if postorder or topological_sort: post.append(x) if subtree_size: for y in self.graph[x]: if self.weighted: y,d=y if y==ps[x]: continue ss[x]+=ss[y] if bipartite_graph: bg=[[],[]] for tpl in self.edges: x,y=tpl[:2] if self.weighted else tpl if uwd[x]==self.inf or uwd[y]==self.inf: continue if not uwd[x]%2^uwd[y]%2: bg=False break else: for x in range(self.V): if uwd[x]==self.inf: continue bg[uwd[x]%2].append(x) retu=() if bipartite_graph: retu+=(bg,) if cycle_detection: if dag: cd=[] else: y,x=cd cd=self.Route_Restoration(y,x,ps) retu+=(cd,) if directed_acyclic: retu+=(dag,) if euler_tour: retu+=(et,) if linked_components: retu+=(lc,) if lowlink: retu=(ll,) if parents: retu+=(ps,) if postorder: retu+=(post,) if preorder: retu+=(pre,) if subtree_size: retu+=(ss,) if topological_sort: if dag: tp_sort=post[::-1] else: tp_sort=[] retu+=(tp_sort,) if unweighted_dist: retu+=(uwd,) if weighted_dist: retu+=(wd,) if len(retu)==1: retu=retu[0] return retu N,M=map(int,readline().split()) D=list(map(int,readline().split())) edges=[] idx={} for i in range(M): u,v=map(int,readline().split()) u-=1;v-=1 edges.append((u,v)) idx[(u,v)]=i idx[(v,u)]=i G=Graph(N,edges=edges) parents,tour=G.SIV_DFS(0,parents=True,postorder=True) if D.count(1)%2: for i in range(N): if D[i]==-1: D[i]=1 break for i in range(N): if D[i]==-1: D[i]=0 if D.count(1)%2: print(-1) exit() ans_lst=[] for x in tour[:-1]: if D[x]: ans_lst.append(idx[(x,parents[x])]+1) D[x]^=1 D[parents[x]]^=1 print(len(ans_lst)) if ans_lst: print(*ans_lst,sep="\n")
Codeforces Round 429 (Div. 1)
CF
2,017
3
256
Leha and another game about graph
Leha plays a computer game, where is on each level is given a connected graph with n vertices and m edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer di, which can be equal to 0, 1 or - 1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, di = - 1 or it's degree modulo 2 is equal to di. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them.
The first line contains two integers n, m (1 ≤ n ≤ 3·105, n - 1 ≤ m ≤ 3·105) — number of vertices and edges. The second line contains n integers d1, d2, ..., dn ( - 1 ≤ di ≤ 1) — numbers on the vertices. Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n) — edges. It's guaranteed, that graph in the input is connected.
Print - 1 in a single line, if solution doesn't exist. Otherwise in the first line k — number of edges in a subset. In the next k lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1.
null
In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1.
[{"input": "1 0\n1", "output": "-1"}, {"input": "4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4", "output": "0"}, {"input": "2 1\n1 1\n1 2", "output": "1\n1"}, {"input": "3 3\n0 -1 1\n1 2\n2 3\n1 3", "output": "1\n2"}]
2,100
["constructive algorithms", "data structures", "dfs and similar", "dp", "graphs"]
73
[{"input": "1 0\r\n1\r\n", "output": "-1\r\n"}, {"input": "4 5\r\n0 0 0 -1\r\n1 2\r\n2 3\r\n3 4\r\n1 4\r\n2 4\r\n", "output": "0\r\n"}, {"input": "2 1\r\n1 1\r\n1 2\r\n", "output": "1\r\n1\r\n"}, {"input": "3 3\r\n0 -1 1\r\n1 2\r\n2 3\r\n1 3\r\n", "output": "1\r\n2\r\n"}, {"input": "10 10\r\n-1 -1 -1 -1 -1 -1 -1 -1 -1 -1\r\n6 7\r\n8 3\r\n6 4\r\n4 2\r\n9 2\r\n5 10\r\n9 8\r\n10 7\r\n5 1\r\n6 2\r\n", "output": "0\r\n"}, {"input": "3 2\r\n1 0 1\r\n1 2\r\n2 3\r\n", "output": "2\r\n2\r\n1\r\n"}]
false
stdio
import sys def main(input_path, output_path, submission_output_path): # Read input with open(input_path, 'r') as f: n, m = map(int, f.readline().split()) d = list(map(int, f.readline().split())) edges = [tuple(map(int, f.readline().split())) for _ in range(m)] # Read submission output with open(submission_output_path, 'r') as f: lines = f.read().strip().split('\n') if not lines: print(0) return first_line = lines[0].strip() if first_line == '-1': sum_d = sum(x for x in d if x != -1) has_neg1 = any(x == -1 for x in d) exists = (sum_d % 2 == 0) or (has_neg1 and (sum_d % 2 == 1)) print(0 if exists else 1) return else: try: k = int(first_line) if len(lines) < k + 1: print(0) return selected = list(map(int, lines[1:1 + k])) except: print(0) return # Check edge indices if any(e < 1 or e > m for e in selected): print(0) return if len(set(selected)) != len(selected): print(0) return # Compute degrees degree = [0] * (n + 1) # 1-based for e in selected: u, v = edges[e - 1] degree[u] += 1 degree[v] += 1 # Check each vertex valid = True for i in range(n): di = d[i] if di == -1: continue if degree[i + 1] % 2 != di: valid = False break print(1 if valid else 0) if __name__ == '__main__': input_path = sys.argv[1] output_path = sys.argv[2] submission_output_path = sys.argv[3] main(input_path, output_path, submission_output_path)
true
665/B
665
B
Python 3
TESTS
3
62
4,608,000
18257545
import math [n, m, k] = map(int,input().split(" ")) pos = list(map(int,input().split(" "))) queue = [0] * k for i in range(k): # print(i, pos[i]) queue[pos[i]-1] = i+1 # print(queue) time = 0 for i in range(n): items = list(map(int,input().split(" "))) for j in range(m): item = items[j] # print(item, queue.index(item)) time += queue.index(item) + 1 queue.remove(item) queue.insert(0,item) print(time)
10
61
0
215054002
n , m , k = map ( int , input () . split () ) p = list ( map ( int , input () . split () ) ) ans = 0 for _ in range ( n ) : arr = list ( map ( int , input () . split () ) ) for i in arr : pos = p . index ( i ) ans += pos + 1 p = [p [pos]] + p[:pos] + p[pos + 1:] print (ans)
Educational Codeforces Round 12
ICPC
2,016
1
256
Shopping
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online. The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order. Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer. When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating. Your task is to calculate the total time it takes for Ayush to process all the orders. You can assume that the market has endless stock.
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market. The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k. Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Print the only integer t — the total time needed for Ayush to process all the orders.
null
Customer 1 wants the items 1 and 5. pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5]. pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2]. Time taken for the first customer is 3 + 5 = 8. Customer 2 wants the items 3 and 1. pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2]. pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2]. Time taken for the second customer is 3 + 3 = 6. Total time is 8 + 6 = 14. Formally pos(x) is the index of x in the current row.
[{"input": "2 2 5\n3 4 1 2 5\n1 5\n3 1", "output": "14"}]
1,400
["brute force"]
10
[{"input": "2 2 5\r\n3 4 1 2 5\r\n1 5\r\n3 1\r\n", "output": "14\r\n"}, {"input": "4 4 4\r\n1 2 3 4\r\n3 4 2 1\r\n4 3 2 1\r\n4 1 2 3\r\n4 1 2 3\r\n", "output": "59\r\n"}, {"input": "1 1 1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "10 1 100\r\n1 55 67 75 40 86 24 84 82 26 81 23 70 79 51 54 21 78 31 98 68 93 66 88 99 65 20 52 35 85 16 12 94 100 59 56 18 33 47 46 71 8 38 57 2 92 3 95 6 4 87 22 48 80 15 29 11 45 72 76 44 60 91 90 39 74 41 36 13 27 53 83 32 5 30 63 89 64 49 17 9 97 69 14 50 77 37 96 10 42 28 34 61 19 73 7 62 43 58 25\r\n33\r\n69\r\n51\r\n7\r\n68\r\n70\r\n1\r\n35\r\n24\r\n7\r\n", "output": "335\r\n"}, {"input": "100 1 1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "100\r\n"}, {"input": "3 2 3\r\n3 1 2\r\n1 2\r\n2 1\r\n2 3\r\n", "output": "13\r\n"}, {"input": "10 10 10\r\n3 4 1 2 8 9 5 10 6 7\r\n9 10 7 8 6 1 2 3 4 5\r\n2 5 3 6 1 4 9 7 8 10\r\n2 9 1 8 4 7 5 10 6 3\r\n10 9 7 1 3 6 2 8 5 4\r\n2 5 1 3 7 10 4 9 8 6\r\n6 1 8 7 9 2 3 5 4 10\r\n1 3 2 8 6 9 4 10 5 7\r\n5 2 4 8 6 1 10 9 3 7\r\n5 1 7 10 4 6 2 8 9 3\r\n2 1 3 9 7 10 6 4 8 5\r\n", "output": "771\r\n"}]
false
stdio
null
true
82/B
82
B
PyPy 3-64
TESTS
32
186
7,577,600
196929606
from copy import copy from sys import stdin n = int(stdin.readline()) previous = [] neigh_by_element = dict() for i in range(n * (n - 1) // 2): m = 0 arr = set() for z, p in enumerate(map(int, stdin.readline().strip().split())): if z == 0: continue arr.add(p) for elem in arr: if elem not in neigh_by_element: neigh_by_element[elem] = copy(arr) else: neigh_by_element[elem].intersection_update(arr) ans = set() for v in neigh_by_element.values(): ans.add(frozenset(v)) for a in ans: print(len(a), *a)
34
248
7,577,600
196930821
from copy import copy from sys import stdin n = int(stdin.readline()) if n == 2: a, *arr = map(int, stdin.readline().strip().split()) first = arr[:len(arr) // 2] second = arr[len(arr) // 2:] print(len(first), *first) print(len(second), *second) exit() neigh_by_element = dict() for i in range(n * (n - 1) // 2): m = 0 received_set = set() for z, p in enumerate(map(int, stdin.readline().strip().split())): if z == 0: continue received_set.add(p) for elem in received_set: if elem not in neigh_by_element: neigh_by_element[elem] = copy(received_set) else: neigh_by_element[elem].intersection_update(received_set) ans = set() for v in neigh_by_element.values(): ans.add(frozenset(v)) for a in ans: print(len(a), *a)
Yandex.Algorithm 2011: Qualification 2
CF
2,011
2
256
Sets
Little Vasya likes very much to play with sets consisting of positive integers. To make the game more interesting, Vasya chose n non-empty sets in such a way, that no two of them have common elements. One day he wanted to show his friends just how interesting playing with numbers is. For that he wrote out all possible unions of two different sets on n·(n - 1) / 2 pieces of paper. Then he shuffled the pieces of paper. He had written out the numbers in the unions in an arbitrary order. For example, if n = 4, and the actual sets have the following form {1, 3}, {5}, {2, 4}, {7}, then the number of set pairs equals to six. The six pieces of paper can contain the following numbers: - 2, 7, 4. - 1, 7, 3; - 5, 4, 2; - 1, 3, 5; - 3, 1, 2, 4; - 5, 7. Then Vasya showed the pieces of paper to his friends, but kept the n sets secret from them. His friends managed to calculate which sets Vasya had thought of in the first place. And how about you, can you restore the sets by the given pieces of paper?
The first input file line contains a number n (2 ≤ n ≤ 200), n is the number of sets at Vasya's disposal. Then follow sets of numbers from the pieces of paper written on n·(n - 1) / 2 lines. Each set starts with the number ki (2 ≤ ki ≤ 200), which is the number of numbers written of the i-th piece of paper, and then follow ki numbers aij (1 ≤ aij ≤ 200). All the numbers on the lines are separated by exactly one space. It is guaranteed that the input data is constructed according to the above given rules from n non-intersecting sets.
Print on n lines Vasya's sets' description. The first number on the line shows how many numbers the current set has. Then the set should be recorded by listing its elements. Separate the numbers by spaces. Each number and each set should be printed exactly once. Print the sets and the numbers in the sets in any order. If there are several answers to that problem, print any of them. It is guaranteed that there is a solution.
null
null
[{"input": "4\n3 2 7 4\n3 1 7 3\n3 5 4 2\n3 1 3 5\n4 3 1 2 4\n2 5 7", "output": "1 7\n2 2 4\n2 1 3\n1 5"}, {"input": "4\n5 6 7 8 9 100\n4 7 8 9 1\n4 7 8 9 2\n3 1 6 100\n3 2 6 100\n2 1 2", "output": "3 7 8 9\n2 6 100\n1 1\n1 2"}, {"input": "3\n2 1 2\n2 1 3\n2 2 3", "output": "1 1\n1 2\n1 3"}]
1,700
["constructive algorithms", "hashing", "implementation"]
34
[{"input": "4\r\n3 2 7 4\r\n3 1 7 3\r\n3 5 4 2\r\n3 1 3 5\r\n4 3 1 2 4\r\n2 5 7\r\n", "output": "1 7 \r\n2 2 4 \r\n2 1 3 \r\n1 5 \r\n"}, {"input": "4\r\n5 6 7 8 9 100\r\n4 7 8 9 1\r\n4 7 8 9 2\r\n3 1 6 100\r\n3 2 6 100\r\n2 1 2\r\n", "output": "3 7 8 9 \r\n2 6 100 \r\n1 1 \r\n1 2 \r\n"}, {"input": "3\r\n2 1 2\r\n2 1 3\r\n2 2 3\r\n", "output": "1 1 \r\n1 2 \r\n1 3 \r\n"}, {"input": "3\r\n2 1 2\r\n10 1 90 80 70 60 50 40 30 20 10\r\n10 2 10 20 30 40 50 60 70 80 90\r\n", "output": "1 1 \r\n1 2 \r\n9 10 20 30 40 50 60 70 80 90 \r\n"}, {"input": "4\r\n4 56 44 53 43\r\n3 109 44 43\r\n3 109 56 53\r\n3 43 62 44\r\n3 62 56 53\r\n2 109 62\r\n", "output": "2 43 44 \r\n2 53 56 \r\n1 109 \r\n1 62 \r\n"}, {"input": "2\r\n2 1 2\r\n", "output": "1 2\r\n1 1"}, {"input": "2\r\n10 1 2 3 4 5 6 7 8 9 10\r\n", "output": "1 10\r\n9 1 2 3 4 5 6 7 8 9"}]
false
stdio
import sys def main(): input_path = sys.argv[1] submission_path = sys.argv[3] with open(input_path, 'r') as f: lines = [l.strip() for l in f if l.strip()] n = int(lines[0]) m = n * (n - 1) // 2 input_unions = set() for line in lines[1:1+m]: parts = list(map(int, line.split())) elements = parts[1:] input_unions.add(frozenset(elements)) with open(submission_path, 'r') as f: submission_lines = [l.strip() for l in f if l.strip()] if len(submission_lines) != n: print(0) return submission_sets = [] all_elements = set() for line in submission_lines: parts = list(map(int, line.split())) if not parts: print(0) return size = parts[0] if len(parts) - 1 != size: print(0) return elements = parts[1:] if len(elements) != len(set(elements)): print(0) return s = frozenset(elements) submission_sets.append(s) current = set(elements) if current & all_elements: print(0) return all_elements.update(current) for s in submission_sets: if not s: print(0) return generated = set() for i in range(n): for j in range(i+1, n): u = submission_sets[i].union(submission_sets[j]) generated.add(frozenset(u)) print(100 if generated == input_unions else 0) if __name__ == "__main__": main()
true
367/A
367
A
Python 3
TESTS
2
61
307,200
103952039
s=input() xc=yc=zc=0 xlis,ylis,zlis=[0],[0],[0] for i in s: if i=='x': xc+=1 elif i=='y': yc+=1 else: zc+=1 xlis.append(xc) ylis.append(yc) zlis.append(zc) for __ in range(int(input())): l,r=map(int,input().split()) xc=xlis[r]-xlis[l-1] yc=ylis[r]-ylis[l-1] zc=zlis[r]-zlis[l-1] lis=[xc,yc,zc] lis.sort() if lis[2]-lis[0]>1: print('NO') else: print('YES')
38
639
6,451,200
5250794
import sys s=sys.stdin.readline().split()[0] m=int(sys.stdin.readline()) Numx=[] Numy=[] Numz=[] x=0 y=0 z=0 for i in range(len(s)): if(s[i]=='x'): x+=1 if(s[i]=='y'): y+=1 if(s[i]=='z'): z+=1 Numx.append(x) Numy.append(y) Numz.append(z) Ans="" for M in range(m): s,e=map(int,sys.stdin.readline().split()) if(e-s+1<=2): Ans+="YES\n" continue s-=1 e-=1 x=Numx[e] y=Numy[e] z=Numz[e] if(s!=0): x-=Numx[s-1] y-=Numy[s-1] z-=Numz[s-1] if(x==y==z): Ans+="YES\n" continue L=[x,y,z] L.sort() if(L[0]==L[1] and L[2]==L[1]+1): Ans+="YES\n" continue if(L[1]==L[2] and L[0]==L[1]-1): Ans+="YES\n" else: Ans+="NO\n" sys.stdout.write(Ans)
Codeforces Round 215 (Div. 1)
CF
2,013
1
256
Sereja and Algorithm
Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q1q2... qk. The algorithm consists of two steps: 1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2. 2. Rearrange the letters of the found subsequence randomly and go to step 1. Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string. Sereja wants to test his algorithm. For that, he has string s = s1s2... sn, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring slisli + 1... sri (1 ≤ li ≤ ri ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (li, ri) determine if the algorithm works correctly on this test or not.
The first line contains non-empty string s, its length (n) doesn't exceed 105. It is guaranteed that string s only contains characters: 'x', 'y', 'z'. The second line contains integer m (1 ≤ m ≤ 105) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
null
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
[{"input": "zyxxxxxxyyz\n5\n5 5\n1 3\n1 11\n1 4\n3 6", "output": "YES\nYES\nNO\nYES\nNO"}]
1,500
["data structures", "implementation"]
38
[{"input": "zyxxxxxxyyz\r\n5\r\n5 5\r\n1 3\r\n1 11\r\n1 4\r\n3 6\r\n", "output": "YES\r\nYES\r\nNO\r\nYES\r\nNO\r\n"}, {"input": "yxzyzxzzxyyzzxxxzyyzzyzxxzxyzyyzxyzxyxxyzxyxzyzxyzxyyxzzzyzxyyxyzxxy\r\n10\r\n17 67\r\n6 35\r\n12 45\r\n56 56\r\n14 30\r\n25 54\r\n1 1\r\n46 54\r\n3 33\r\n19 40\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\nYES\r\nNO\r\nYES\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "xxxxyyxyyzzyxyxzxyzyxzyyyzyzzxxxxzyyzzzzyxxxxzzyzzyzx\r\n5\r\n4 4\r\n3 3\r\n1 24\r\n3 28\r\n18 39\r\n", "output": "YES\r\nYES\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "yzxyzxyzxzxzyzxyzyzzzyxzyz\r\n9\r\n4 6\r\n2 7\r\n3 5\r\n14 24\r\n3 13\r\n2 24\r\n2 5\r\n2 14\r\n3 15\r\n", "output": "YES\r\nYES\r\nYES\r\nNO\r\nYES\r\nNO\r\nYES\r\nNO\r\nNO\r\n"}, {"input": "zxyzxyzyyzxzzxyzxyzx\r\n15\r\n7 10\r\n17 17\r\n6 7\r\n8 14\r\n4 7\r\n11 18\r\n12 13\r\n1 1\r\n3 8\r\n1 1\r\n9 17\r\n4 4\r\n5 11\r\n3 15\r\n1 1\r\n", "output": "NO\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\nYES\r\nNO\r\nYES\r\nYES\r\nNO\r\nYES\r\n"}, {"input": "x\r\n1\r\n1 1\r\n", "output": "YES\r\n"}]
false
stdio
null
true
639/A
639
A
Python 3
PRETESTS
7
1,278
15,462,400
16998331
str1 = input().split() n = int(str1[0]) k = int(str1[1]) q = int(str1[2]) friends = input().split() online = set() for i in range(q): str1 = input().split() if str1[0] == '2': if int(str1[1]) in online: print("YES") else: print("NO") else: online.add(int(str1[1])) if len(online) > k: minelem = int(str1[1]) for on in online: if friends[on - 1] < friends[minelem - 1]: minelem = on online.remove(minelem)
44
265
20,480,000
188466426
from sys import stdin; inp = stdin.readline from math import dist, ceil, floor, sqrt, log from collections import defaultdict, Counter def IA(sep=' '): return list(map(int, inp().split(sep))) def FA(): return list(map(float, inp().split())) def SA(): return inp().split() def I(): return int(inp()) def F(): return float(inp()) def S(): return input() def O(l:list): return ' '.join(map(str, l)) from itertools import compress def main(): n, k, q = IA() a = IA() best = set() for _ in range(q): t, index = IA() f = a[index-1] if t == 1: if len(best) < k: best.add(f) else: mi = min(best) if f > mi: best.remove(mi) best.add(f) else: if f in best: print('YES') else: print('NO') if __name__ == '__main__': main()
VK Cup 2016 - Round 1
CF
2,016
2
256
Bear and Displayed Friends
Limak is a little polar bear. He loves connecting with other bears via social networks. He has n friends and his relation with the i-th of them is described by a unique integer ti. The bigger this value is, the better the friendship is. No two friends have the same value ti. Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time. The system displays friends who are online. On the screen there is space to display at most k friends. If there are more than k friends online then the system displays only k best of them — those with biggest ti. Your task is to handle queries of two types: - "1 id" — Friend id becomes online. It's guaranteed that he wasn't online before. - "2 id" — Check whether friend id is displayed by the system. Print "YES" or "NO" in a separate line. Are you able to help Limak and answer all queries of the second type?
The first line contains three integers n, k and q (1 ≤ n, q ≤ 150 000, 1 ≤ k ≤ min(6, n)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 109) where ti describes how good is Limak's relation with the i-th friend. The i-th of the following q lines contains two integers typei and idi (1 ≤ typei ≤ 2, 1 ≤ idi ≤ n) — the i-th query. If typei = 1 then a friend idi becomes online. If typei = 2 then you should check whether a friend idi is displayed. It's guaranteed that no two queries of the first type will have the same idi becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (typei = 2) so the output won't be empty.
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
null
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries: 1. "1 3" — Friend 3 becomes online. 2. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO". 3. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES". 4. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3. 5. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given k = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (t1 < t2, t3) so friend 1 won't be displayed 6. "2 1" — Print "NO". 7. "2 2" — Print "YES". 8. "2 3" — Print "YES".
[{"input": "4 2 8\n300 950 500 200\n1 3\n2 4\n2 3\n1 1\n1 2\n2 1\n2 2\n2 3", "output": "NO\nYES\nNO\nYES\nYES"}, {"input": "6 3 9\n50 20 51 17 99 24\n1 3\n1 4\n1 5\n1 2\n2 4\n2 2\n1 1\n2 4\n2 3", "output": "NO\nYES\nNO\nYES"}]
1,200
["implementation"]
44
[{"input": "4 2 8\r\n300 950 500 200\r\n1 3\r\n2 4\r\n2 3\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n2 3\r\n", "output": "NO\r\nYES\r\nNO\r\nYES\r\nYES\r\n"}, {"input": "6 3 9\r\n50 20 51 17 99 24\r\n1 3\r\n1 4\r\n1 5\r\n1 2\r\n2 4\r\n2 2\r\n1 1\r\n2 4\r\n2 3\r\n", "output": "NO\r\nYES\r\nNO\r\nYES\r\n"}, {"input": "6 3 10\r\n62417580 78150524 410053501 582708235 630200761 760672946\r\n2 2\r\n1 5\r\n1 2\r\n1 4\r\n2 4\r\n2 1\r\n2 1\r\n1 6\r\n2 5\r\n2 6\r\n", "output": "NO\r\nYES\r\nNO\r\nNO\r\nYES\r\nYES\r\n"}, {"input": "20 2 15\r\n12698951 55128070 116962690 156763505 188535242 194018601 269939893 428710623 442819431 483000923 516768937 552903993 633087286 656092270 671535141 714291344 717660646 846508634 879748146 937368929\r\n2 7\r\n1 2\r\n2 4\r\n1 19\r\n1 12\r\n1 5\r\n2 18\r\n2 11\r\n1 16\r\n2 1\r\n2 3\r\n2 19\r\n1 17\r\n2 9\r\n2 6\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\nNO\r\nNO\r\nYES\r\nNO\r\nNO\r\n"}, {"input": "1 1 1\r\n1000000000\r\n2 1\r\n", "output": "NO\r\n"}]
false
stdio
null
true
665/B
665
B
PyPy 3
TESTS
3
280
6,656,000
59228344
import io, sys, atexit, os import math as ma from sys import exit from decimal import Decimal as dec from itertools import permutations from itertools import combinations def li (): return list (map (int, input ().split ())) def num (): return map (int, input ().split ()) def nu (): return int (input ()) def find_gcd ( x, y ): while (y): x, y = y, x % y return x def lcm ( x, y ): gg = find_gcd (x, y) return (x * y // gg) def printDivisors ( n ): # Note that this loop runs till square root i = 1 cc=0 while i <= ma.sqrt (n): if (n % i == 0): # If divisors are equal, print only one if (n / i == i): cc+=1 else: # Otherwise print both cc+=2 i = i + 1 return cc mm = 1000000007 yp = 0 def solve (): t = 1 for tt in range (t): n,m,k=num() a=li() ff=[0]*k for i in range(k): ff[a[i]-1]=i ss=0 for i in range(n): b=li() for j in range(m): for ll in range(k): if(b[j]-1==ff[ll]): ss+=ll+1 ff=[b[j]-1]+ff[0:ll]+ff[ll+1:] break print(ss) if __name__ == "__main__": solve ()
10
61
4,710,400
17420179
def main(): n, m, k = map(int, input().split()) aa, t = list(map(int, input().split())), n * m * k aa.reverse() u, v, w = aa.index, aa.remove, aa.append for _ in range(n): for a in map(int, input().split()): t -= u(a) v(a) w(a) print(t) if __name__ == '__main__': main()
Educational Codeforces Round 12
ICPC
2,016
1
256
Shopping
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online. The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order. Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer. When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating. Your task is to calculate the total time it takes for Ayush to process all the orders. You can assume that the market has endless stock.
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market. The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k. Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Print the only integer t — the total time needed for Ayush to process all the orders.
null
Customer 1 wants the items 1 and 5. pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5]. pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2]. Time taken for the first customer is 3 + 5 = 8. Customer 2 wants the items 3 and 1. pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2]. pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2]. Time taken for the second customer is 3 + 3 = 6. Total time is 8 + 6 = 14. Formally pos(x) is the index of x in the current row.
[{"input": "2 2 5\n3 4 1 2 5\n1 5\n3 1", "output": "14"}]
1,400
["brute force"]
10
[{"input": "2 2 5\r\n3 4 1 2 5\r\n1 5\r\n3 1\r\n", "output": "14\r\n"}, {"input": "4 4 4\r\n1 2 3 4\r\n3 4 2 1\r\n4 3 2 1\r\n4 1 2 3\r\n4 1 2 3\r\n", "output": "59\r\n"}, {"input": "1 1 1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "10 1 100\r\n1 55 67 75 40 86 24 84 82 26 81 23 70 79 51 54 21 78 31 98 68 93 66 88 99 65 20 52 35 85 16 12 94 100 59 56 18 33 47 46 71 8 38 57 2 92 3 95 6 4 87 22 48 80 15 29 11 45 72 76 44 60 91 90 39 74 41 36 13 27 53 83 32 5 30 63 89 64 49 17 9 97 69 14 50 77 37 96 10 42 28 34 61 19 73 7 62 43 58 25\r\n33\r\n69\r\n51\r\n7\r\n68\r\n70\r\n1\r\n35\r\n24\r\n7\r\n", "output": "335\r\n"}, {"input": "100 1 1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "100\r\n"}, {"input": "3 2 3\r\n3 1 2\r\n1 2\r\n2 1\r\n2 3\r\n", "output": "13\r\n"}, {"input": "10 10 10\r\n3 4 1 2 8 9 5 10 6 7\r\n9 10 7 8 6 1 2 3 4 5\r\n2 5 3 6 1 4 9 7 8 10\r\n2 9 1 8 4 7 5 10 6 3\r\n10 9 7 1 3 6 2 8 5 4\r\n2 5 1 3 7 10 4 9 8 6\r\n6 1 8 7 9 2 3 5 4 10\r\n1 3 2 8 6 9 4 10 5 7\r\n5 2 4 8 6 1 10 9 3 7\r\n5 1 7 10 4 6 2 8 9 3\r\n2 1 3 9 7 10 6 4 8 5\r\n", "output": "771\r\n"}]
false
stdio
null
true
24/A
24
A
PyPy 3
TESTS
4
280
0
56988831
n = int(input()) e = [ [] for i in range(n)] e2 = {} C = 0 for i in range(n): a,b,c = map(int,input().split()) a-=1 b-=1 e[a].append((b,c)) e2[a] = (b,c) e[b].append((a,c)) C += c r = e[0][0] l = e[0][1] right = 0 left = 0 rr = [r[0]] for i in range(n-1): for node in e[r[0]]: if node[0] in rr: continue if r[0] in e2.keys(): left += node[1] right += node[1] rr.append(node[0]) r = node right += e[r[0]][0][1] if r[0] in e2.keys(): left += r[1] print(min( abs(right - left) , left))
21
92
0
187852605
n = int(input()) lst = [] for _ in range(n): a, b, c = map(int, input().split()) lst.append([a,b,c]) pos1, pos2, f1, f2 = lst[0][1], lst[0][0], lst[0][0], lst[0][1] sc1, sc2 = 0, lst[0][2] del lst[0] lst2 = lst*1 while pos1 != f1: for x in range(len(lst)): if lst[x][0] == pos1: pos1 = lst[x][1] del lst[x] break if lst[x][1] == pos1: pos1 = lst[x][0] sc1 += lst[x][2] del lst[x] break while pos2 != f2: for x in range(len(lst2)): if lst2[x][0] == pos2: pos2 = lst2[x][1] del lst2[x] break if lst2[x][1] == pos2: pos2 = lst2[x][0] sc2 += lst2[x][2] del lst2[x] break print(min(sc1, sc2))
Codeforces Beta Round 24
ICPC
2,010
2
256
Ring road
Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Berland decided to keep up with new trends. Formerly all n cities of Berland were connected by n two-way roads in the ring, i. e. each city was connected directly to exactly two other cities, and from each city it was possible to get to any other city. Government of Berland introduced one-way traffic on all n roads, but it soon became clear that it's impossible to get from some of the cities to some others. Now for each road is known in which direction the traffic is directed at it, and the cost of redirecting the traffic. What is the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other?
The first line contains integer n (3 ≤ n ≤ 100) — amount of cities (and roads) in Berland. Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 100) — road is directed from city ai to city bi, redirecting the traffic costs ci.
Output single integer — the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other.
null
null
[{"input": "3\n1 3 1\n1 2 1\n3 2 1", "output": "1"}, {"input": "3\n1 3 1\n1 2 5\n3 2 1", "output": "2"}, {"input": "6\n1 5 4\n5 3 8\n2 4 15\n1 6 16\n2 3 23\n4 6 42", "output": "39"}, {"input": "4\n1 2 9\n2 3 8\n3 4 7\n4 1 5", "output": "0"}]
1,400
["graphs"]
21
[{"input": "3\r\n1 3 1\r\n1 2 1\r\n3 2 1\r\n", "output": "1\r\n"}, {"input": "3\r\n1 3 1\r\n1 2 5\r\n3 2 1\r\n", "output": "2\r\n"}, {"input": "6\r\n1 5 4\r\n5 3 8\r\n2 4 15\r\n1 6 16\r\n2 3 23\r\n4 6 42\r\n", "output": "39\r\n"}, {"input": "4\r\n1 2 9\r\n2 3 8\r\n3 4 7\r\n4 1 5\r\n", "output": "0\r\n"}, {"input": "5\r\n5 3 89\r\n2 3 43\r\n4 2 50\r\n1 4 69\r\n1 5 54\r\n", "output": "143\r\n"}, {"input": "10\r\n1 8 16\r\n6 1 80\r\n6 5 27\r\n5 7 86\r\n7 9 72\r\n4 9 20\r\n4 3 54\r\n3 2 57\r\n10 2 61\r\n8 10 90\r\n", "output": "267\r\n"}, {"input": "17\r\n8 12 43\r\n13 12 70\r\n7 13 68\r\n11 7 19\r\n5 11 24\r\n5 1 100\r\n4 1 10\r\n3 4 68\r\n2 3 46\r\n15 2 58\r\n15 6 38\r\n6 9 91\r\n9 10 72\r\n14 10 32\r\n14 17 97\r\n17 16 67\r\n8 16 40\r\n", "output": "435\r\n"}, {"input": "22\r\n18 22 46\r\n18 21 87\r\n5 21 17\r\n5 10 82\r\n10 12 81\r\n17 12 98\r\n16 17 17\r\n16 13 93\r\n4 13 64\r\n4 11 65\r\n15 11 18\r\n6 15 35\r\n6 7 61\r\n7 19 12\r\n19 1 65\r\n8 1 32\r\n8 2 46\r\n9 2 19\r\n9 3 58\r\n3 14 65\r\n20 14 67\r\n20 22 2\r\n", "output": "413\r\n"}, {"input": "3\r\n3 1 1\r\n2 1 1\r\n2 3 1\r\n", "output": "1\r\n"}]
false
stdio
null
true
24/A
24
A
PyPy 3-64
TESTS
4
124
0
225896932
n = int(input()) sl,el,c1,c2 = [],[],0,0 for i in range(n): s,e,c = map(int,input().split()) if s not in sl and e not in el: sl.append(s) el.append(e) c1 += c else: c2 += c print(min(c1,c2))
21
92
0
203666447
def main(): n = int(input()) road_cost = {} road_dict = {i: [] for i in range(1, n + 1)} total_cost = 0 for _ in range(n): ai, bi, ci = map(int, input().split()) road_dict[ai].append(bi) road_dict[bi].append(ai) road_cost[(ai, bi)] = ci total_cost += ci cost = 0 current_city = 0 while current_city != 1: if current_city == 0: current_city = 1 next_city = road_dict[1][0] else: road_dict[current_city].remove(prev_city) next_city = road_dict[current_city][0] if (current_city, next_city) not in road_cost: cost += road_cost[(next_city, current_city)] prev_city = current_city current_city = next_city print(min(cost, total_cost - cost)) if __name__ == "__main__": main()
Codeforces Beta Round 24
ICPC
2,010
2
256
Ring road
Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Berland decided to keep up with new trends. Formerly all n cities of Berland were connected by n two-way roads in the ring, i. e. each city was connected directly to exactly two other cities, and from each city it was possible to get to any other city. Government of Berland introduced one-way traffic on all n roads, but it soon became clear that it's impossible to get from some of the cities to some others. Now for each road is known in which direction the traffic is directed at it, and the cost of redirecting the traffic. What is the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other?
The first line contains integer n (3 ≤ n ≤ 100) — amount of cities (and roads) in Berland. Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 100) — road is directed from city ai to city bi, redirecting the traffic costs ci.
Output single integer — the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other.
null
null
[{"input": "3\n1 3 1\n1 2 1\n3 2 1", "output": "1"}, {"input": "3\n1 3 1\n1 2 5\n3 2 1", "output": "2"}, {"input": "6\n1 5 4\n5 3 8\n2 4 15\n1 6 16\n2 3 23\n4 6 42", "output": "39"}, {"input": "4\n1 2 9\n2 3 8\n3 4 7\n4 1 5", "output": "0"}]
1,400
["graphs"]
21
[{"input": "3\r\n1 3 1\r\n1 2 1\r\n3 2 1\r\n", "output": "1\r\n"}, {"input": "3\r\n1 3 1\r\n1 2 5\r\n3 2 1\r\n", "output": "2\r\n"}, {"input": "6\r\n1 5 4\r\n5 3 8\r\n2 4 15\r\n1 6 16\r\n2 3 23\r\n4 6 42\r\n", "output": "39\r\n"}, {"input": "4\r\n1 2 9\r\n2 3 8\r\n3 4 7\r\n4 1 5\r\n", "output": "0\r\n"}, {"input": "5\r\n5 3 89\r\n2 3 43\r\n4 2 50\r\n1 4 69\r\n1 5 54\r\n", "output": "143\r\n"}, {"input": "10\r\n1 8 16\r\n6 1 80\r\n6 5 27\r\n5 7 86\r\n7 9 72\r\n4 9 20\r\n4 3 54\r\n3 2 57\r\n10 2 61\r\n8 10 90\r\n", "output": "267\r\n"}, {"input": "17\r\n8 12 43\r\n13 12 70\r\n7 13 68\r\n11 7 19\r\n5 11 24\r\n5 1 100\r\n4 1 10\r\n3 4 68\r\n2 3 46\r\n15 2 58\r\n15 6 38\r\n6 9 91\r\n9 10 72\r\n14 10 32\r\n14 17 97\r\n17 16 67\r\n8 16 40\r\n", "output": "435\r\n"}, {"input": "22\r\n18 22 46\r\n18 21 87\r\n5 21 17\r\n5 10 82\r\n10 12 81\r\n17 12 98\r\n16 17 17\r\n16 13 93\r\n4 13 64\r\n4 11 65\r\n15 11 18\r\n6 15 35\r\n6 7 61\r\n7 19 12\r\n19 1 65\r\n8 1 32\r\n8 2 46\r\n9 2 19\r\n9 3 58\r\n3 14 65\r\n20 14 67\r\n20 22 2\r\n", "output": "413\r\n"}, {"input": "3\r\n3 1 1\r\n2 1 1\r\n2 3 1\r\n", "output": "1\r\n"}]
false
stdio
null
true
837/G
837
G
PyPy 3-64
TESTS
9
1,403
42,905,600
198697694
import bisect import heapq import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline n = int(input()) m = 333 s = [0] * (2 * n) x0, c1, c2, c3 = [], [], [], [] for i in range(n): x1, x2, y1, a, b, y2 = map(int, input().split()) s[2 * i], s[2 * i + 1] = x1, x2 x0.append((x1, x2)) c1.append((0, y1)) c2.append((a, b)) c3.append((0, y2)) s = list(set(sorted(s))) d = dict() l = len(s) for i in range(l): d[s[i]] = i + 1 a, b = [], [] for i in range(n // m): a0, b0 = [0] * m, [0] * m h = [] for j in range(i * m, (i + 1) * m): x1, x2 = x0[j] heapq.heappush(h, d[x1] * n + j) heapq.heappush(h, d[x2] * n + j) _, y1 = c1[j] b0[j % m] = y1 a1, b1 = [-1] * (l + 1), [-1] * (l + 1) a1[0], b1[0] = 0, 0 visit = [0] * m sa, sb = 0, sum(b0) while h: u, v = divmod(heapq.heappop(h), n) w = v % m sa -= a0[w] sb -= b0[w] if not visit[w]: visit[w] = 1 a2, b2 = c2[v] else: a2, b2 = c3[v] sa += a2 sb += b2 a0[w], b0[w] = a2, b2 a1[u], b1[u] = sa, sb for j in range(1, l + 1): if a1[j] == -1: a1[j], b1[j] = a1[j - 1], b1[j - 1] a.append(a1) b.append(b1) q = int(input()) ans = [] last, mod = 0, pow(10, 9) for _ in range(q): l, r, x = map(int, input().split()) x = (x + last) % mod ans0 = 0 l -= 1 while l % m and l < r: x1, x2 = x0[l] if x <= x1: a0, b0 = c1[l] elif x <= x2: a0, b0 = c2[l] else: a0, b0 = c3[l] ans0 += a0 * x + b0 l += 1 i, j = l // m, bisect.bisect_left(s, x) while l + m < r: ans0 += a[i][j] * x + b[i][j] i += 1 l += m while l < r: x1, x2 = x0[l] if x <= x1: a0, b0 = c1[l] elif x <= x2: a0, b0 = c2[l] else: a0, b0 = c3[l] ans0 += a0 * x + b0 l += 1 ans.append(ans0) last = ans0 sys.stdout.write("\n".join(map(str, ans)))
25
2,917
232,038,400
227942774
import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline def get_segment(s, t): s, t = s + l1, t + l1 ans = [] while s <= t: if s & 1: ans.append(s) s += 1 s >>= 1 if not t & 1: ans.append(t) t -= 1 t >>= 1 return ans n = int(input()) l1 = pow(2, (n + 1).bit_length()) l2 = 2 * l1 s0 = [0] * (l2 + 1) s0[1], s0[2] = 0, 3 * l1 for i in range(2, l2): u = (s0[(i >> 1) + 1] - s0[i >> 1]) >> 1 s0[i + 1] = s0[i] + u z = s0[l2] inf = pow(10, 9) + 1 a, b, s = [0] * z, [0] * z, [inf] * z for i in range(1, n + 1): x1, x2, y1, a0, b0, y2 = map(int, input().split()) j = s0[i + l1] a[j + 1] = a0 b[j], b[j + 1], b[j + 2] = y1, b0, y2 s[j], s[j + 1] = x1, x2 for i in range(l1 - 1, 0, -1): u, v, w = s0[2 * i], s0[2 * i + 1], s0[i] ru, rv = s0[2 * i + 1], s0[2 * i + 2] while u < ru or v < rv: a[w], b[w] = a[u] + a[v], b[u] + b[v] k = min(s[u], s[v]) if s[u] == k: u += 1 if s[v] == k: v += 1 s[w] = k if k == inf: break w += 1 pow2 = [1] for _ in range(20): pow2.append(2 * pow2[-1]) m = int(input()) ans = [0] * m last, mod = 0, pow(10, 9) for i in range(m): l, r, x = map(int, input().split()) x = (x + last) % mod seg = get_segment(l, r) ans0 = 0 for j in seg: k, r0 = s0[j], s0[j + 1] c = (k - r0).bit_length() for u in range(c, -1, -1): if k + pow2[u] < r0 and s[k + pow2[u]] < x: k += pow2[u] if s[k] < x and k < r0: k += 1 ans0 += a[k] * x + b[k] ans[i] = ans0 last = ans0 sys.stdout.write("\n".join(map(str, ans)))
Educational Codeforces Round 26
ICPC
2,017
5
1,024
Functions On The Segments
You have an array f of n functions.The function fi(x) (1 ≤ i ≤ n) is characterized by parameters: x1, x2, y1, a, b, y2 and take values: - y1, if x ≤ x1. - a·x + b, if x1 < x ≤ x2. - y2, if x > x2. There are m queries. Each query is determined by numbers l, r and x. For a query with number i (1 ≤ i ≤ m), you need to calculate the sum of all fj(xi) where l ≤ j ≤ r. The value of xi is calculated as follows: xi = (x + last) mod 109, where last is the answer to the query with number i - 1. The value of last equals 0 if i = 1.
First line contains one integer number n (1 ≤ n ≤ 75000). Each of the next n lines contains six integer numbers: x1, x2, y1, a, b, y2 (0 ≤ x1 < x2 ≤ 2·105, 0 ≤ y1, y2 ≤ 109, 0 ≤ a, b ≤ 104). Next line contains one integer number m (1 ≤ m ≤ 500000). Each of the next m lines contains three integer numbers: l, r and x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109).
null
null
null
[{"input": "1\n1 2 1 4 5 10\n1\n1 1 2", "output": "13"}, {"input": "3\n2 5 1 1 1 4\n3 6 8 2 5 7\n1 3 5 1 4 10\n3\n1 3 3\n2 3 2\n1 2 5", "output": "19\n17\n11"}]
2,500
["data structures"]
25
[{"input": "1\r\n1 2 1 4 5 10\r\n1\r\n1 1 2\r\n", "output": "13\r\n"}, {"input": "3\r\n2 5 1 1 1 4\r\n3 6 8 2 5 7\r\n1 3 5 1 4 10\r\n3\r\n1 3 3\r\n2 3 2\r\n1 2 5\r\n", "output": "19\r\n17\r\n11\r\n"}, {"input": "7\r\n4 8 562244866 6 8 140807945\r\n5 7 415374420 7 6 596093578\r\n3 4 766370993 3 7 973128805\r\n4 6 841321398 3 2 893107667\r\n4 9 968214628 5 8 329310838\r\n0 10 64315145 9 4 716996081\r\n1 6 568407659 3 0 22184171\r\n15\r\n4 7 647898558\r\n6 7 972924557\r\n7 7 222835064\r\n4 7 605696049\r\n3 3 549028549\r\n7 7 716270684\r\n2 4 716558406\r\n3 3 681131761\r\n2 2 177328926\r\n3 6 298280462\r\n6 7 256719966\r\n3 3 554401527\r\n5 6 136725488\r\n5 6 457040333\r\n7 7 176796906\r\n", "output": "1961598757\r\n739180252\r\n22184171\r\n1961598757\r\n973128805\r\n22184171\r\n2462330050\r\n973128805\r\n596093578\r\n2912543391\r\n739180252\r\n973128805\r\n1046306919\r\n1046306919\r\n22184171\r\n"}, {"input": "8\r\n1 10 753310968 2 4 688105437\r\n1 6 193355966 4 7 598962222\r\n0 8 600395827 5 8 840898713\r\n0 3 421099093 10 3 239232128\r\n7 8 984393026 4 0 820591274\r\n2 3 99820619 1 4 750632847\r\n8 10 582155129 0 6 387245981\r\n5 6 667916797 0 4 3445518\r\n15\r\n5 5 361972828\r\n7 8 705501679\r\n8 8 890961558\r\n6 6 865121916\r\n7 7 595820781\r\n6 7 782831246\r\n6 6 295277930\r\n2 5 251627063\r\n2 6 439397863\r\n7 7 104400662\r\n6 8 219241770\r\n3 4 429934639\r\n5 6 858415301\r\n2 4 36071843\r\n4 7 976548276\r\n", "output": "820591274\r\n390691499\r\n3445518\r\n750632847\r\n387245981\r\n1137878828\r\n750632847\r\n2499684337\r\n3250317184\r\n387245981\r\n1141324346\r\n1080130841\r\n1571224121\r\n1679093063\r\n2197702230\r\n"}, {"input": "9\r\n3 10 280813558 6 3 193933732\r\n3 9 971337514 8 4 896798158\r\n3 7 803016882 1 6 413701329\r\n2 10 74505717 6 8 658985518\r\n2 10 571422 8 3 680467929\r\n9 10 766729875 4 9 415673394\r\n5 6 669531526 3 0 752307791\r\n4 7 98000452 6 5 681772410\r\n8 10 722448957 9 3 55646997\r\n15\r\n3 5 232894565\r\n6 6 13973148\r\n9 9 401454228\r\n4 5 253126950\r\n6 7 708194225\r\n1 2 868537175\r\n2 3 32141176\r\n7 9 949367089\r\n1 9 596725344\r\n7 7 624167310\r\n1 1 835000987\r\n9 9 526102998\r\n4 7 50218935\r\n1 7 695691438\r\n5 6 878938576\r\n", "output": "1753154776\r\n415673394\r\n55646997\r\n1339453447\r\n1167981185\r\n1090731890\r\n1310499487\r\n1489727198\r\n4749287258\r\n752307791\r\n193933732\r\n55646997\r\n2507434632\r\n4011867851\r\n1096141323\r\n"}]
false
stdio
null
true
843/A
843
A
Python 3
TESTS
2
30
0
195043692
n = int(input()) a = list(map(int, input().split())) sorted_a = sorted(a) ind = 0 num = a[0] ans = [] k = 0 if a[0] == sorted_a[0] else 1 for i in range(1, n): num = max(a[ind:i + 1]) k += 1 if a[i] != sorted_a[i] else 0 if sorted_a[i] == num and ind != i: ans.append([k, ind + 1, i + 1]) for j in range(ind + 1, i): if a[j] != sorted_a[j]: ans[ind].append(j + 1) ind = i + 1 k = 0 elif a[i] == sorted_a[i]: ans.append((1, i + 1)) print(len(ans)) for i in ans: print(*i)
71
389
44,441,600
129813603
import io import os import sys class UnionFind: def __init__(self, n): self.cnt = n self.par = list(range(n)) self.ext = [1] * (n) def find(self, u): while u != self.par[u]: u = self.par[u] return u def unite(self, u, v): u, v = self.find(u), self.find(v) if u == v: return False if self.ext[u] < self.ext[v]: u, v = v, u self.ext[u] += self.ext[v] self.par[v] = u self.cnt -= 1 return True def main(): n = int(input()) a = list(map(int, input().split())) dsu = UnionFind(n) tab = dict() for i in range(n): tab[a[i]] = i a.sort() for i in range(n): dsu.unite(i, tab[a[i]]) ans = dict() for i in range(n): root = dsu.find(i) if not ans.get(root, None): ans[root] = [] ans[root].append(i+1) print(dsu.cnt) for i in range(n): if not ans.get(i, None): continue print(len(ans[i]), *ans[i]) pypyin = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline cpyin = sys.stdin.readline input = pypyin if 'PyPy' in sys.version else cpyin def strput(): return input().decode() if 'PyPy' in sys.version else input() if __name__ == "__main__": main()
AIM Tech Round 4 (Div. 1)
CF
2,017
1
256
Sorting by Subsequences
You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order. Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places. Every element of the sequence must appear in exactly one subsequence.
The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence. The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements. In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence. Indices could be printed in any order. Every index from 1 to n must appear in output exactly once. If there are several possible answers, print any of them.
null
In the first sample output: After sorting the first subsequence we will get sequence 1 2 3 6 5 4. Sorting the second subsequence changes nothing. After sorting the third subsequence we will get sequence 1 2 3 4 5 6. Sorting the last subsequence changes nothing.
[{"input": "6\n3 2 1 6 5 4", "output": "4\n2 1 3\n1 2\n2 4 6\n1 5"}, {"input": "6\n83 -75 -49 11 37 62", "output": "1\n6 1 2 3 4 5 6"}]
1,400
["dfs and similar", "dsu", "implementation", "math", "sortings"]
71
[{"input": "6\r\n3 2 1 6 5 4\r\n", "output": "4\r\n2 1 3\r\n1 2\r\n2 4 6\r\n1 5\r\n"}, {"input": "6\r\n83 -75 -49 11 37 62\r\n", "output": "1\r\n6 1 2 3 4 5 6\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n1 1\r\n"}, {"input": "2\r\n1 2\r\n", "output": "2\r\n1 1\r\n1 2\r\n"}, {"input": "2\r\n2 1\r\n", "output": "1\r\n2 1 2\r\n"}, {"input": "3\r\n1 2 3\r\n", "output": "3\r\n1 1\r\n1 2\r\n1 3\r\n"}, {"input": "3\r\n3 2 1\r\n", "output": "2\r\n2 1 3\r\n1 2\r\n"}, {"input": "3\r\n3 1 2\r\n", "output": "1\r\n3 1 2 3\r\n"}, {"input": "10\r\n3 7 10 1 9 5 4 8 6 2\r\n", "output": "3\r\n6 1 4 7 2 10 3\r\n3 5 6 9\r\n1 8\r\n"}, {"input": "20\r\n363756450 -204491568 95834122 -840249197 -49687658 470958158 -445130206 189801569 802780784 -790013317 -192321079 586260100 -751917965 -354684803 418379342 -253230108 193944314 712662868 853829789 735867677\r\n", "output": "3\r\n7 1 4 7 2 10 3 13\r\n11 5 14 15 6 16 12 17 18 20 19 9\r\n2 8 11\r\n"}, {"input": "50\r\n39 7 45 25 31 26 50 11 19 37 8 16 22 33 14 6 12 46 49 48 29 27 41 15 34 24 3 13 20 47 9 36 5 43 40 21 2 38 35 42 23 28 1 32 10 17 30 18 44 4\r\n", "output": "6\r\n20 1 43 34 25 4 50 7 2 37 10 45 3 27 22 13 28 42 40 35 39\r\n23 5 33 14 15 24 26 6 16 12 17 46 18 48 20 29 21 36 32 44 49 19 9 31\r\n2 8 11\r\n2 23 41\r\n2 30 47\r\n1 38\r\n"}, {"input": "100\r\n39 77 67 25 81 26 50 11 73 95 86 16 90 33 14 79 12 100 68 64 60 27 41 15 34 24 3 61 83 47 57 65 99 43 40 21 94 72 82 85 23 71 76 32 10 17 30 18 44 59 35 89 6 63 7 69 62 70 4 29 92 87 31 48 36 28 45 97 93 98 56 38 58 80 8 1 74 91 53 55 54 51 96 5 42 52 9 22 78 88 75 13 66 2 37 20 49 19 84 46\r\n", "output": "6\r\n41 1 76 43 34 25 4 59 50 7 55 80 74 77 2 94 37 95 10 45 67 3 27 22 88 90 13 92 61 28 66 93 69 56 71 42 85 40 35 51 82 39\r\n45 5 84 99 33 14 15 24 26 6 53 79 16 12 17 46 100 18 48 64 20 96 83 29 60 21 36 65 32 44 49 97 68 19 98 70 58 73 9 87 62 57 31 63 54 81\r\n8 8 75 91 78 89 52 86 11\r\n2 23 41\r\n2 30 47\r\n2 38 72\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] with open(input_path) as f: n = int(f.readline()) a = list(map(int, f.readline().split())) sorted_a = sorted(a) pos_in_sorted = {v: i+1 for i, v in enumerate(sorted_a)} # 1-based P = {i+1: pos_in_sorted[v] for i, v in enumerate(a)} # 1-based indices visited = set() cycles = [] for node in range(1, n+1): if node not in visited: current = node cycle = [] while current not in visited: visited.add(current) cycle.append(current) current = P[current] cycles.append(cycle) correct_k = len(cycles) computed_cycles = [set(cycle) for cycle in cycles] with open(submission_path) as f: lines = [line.strip() for line in f.readlines() if line.strip()] if not lines: print(0) return try: k = int(lines[0]) if k != correct_k: print(0) return submission_cycles = [] all_indices = set() for line in lines[1:1 + k]: parts = list(map(int, line.split())) if not parts: print(0) return cnt = parts[0] if cnt <= 0 or cnt > n: print(0) return indices = parts[1:1 + cnt] if len(indices) != cnt: print(0) return for idx in indices: if not (1 <= idx <= n): print(0) return s_indices = set(indices) if len(s_indices) != cnt: print(0) return if s_indices & all_indices: print(0) return all_indices.update(s_indices) submission_cycles.append(s_indices) if all_indices != set(range(1, n+1)): print(0) return except: print(0) return for s_cycle in submission_cycles: start = next(iter(s_cycle)) current = start collected = set() while True: collected.add(current) current = P[current] if current == start: break if collected != s_cycle: print(0) return print(100) if __name__ == "__main__": main()
true
24/A
24
A
PyPy 3
TESTS
4
248
0
95306811
arr = {} arr2 = {} cost = 0 cost2 = 0 n = int(input()) for _ in range(n): a, b, c = list(map(int, input().split())) cost2 += c if a in arr: arr[a].append(b) else: arr[a] = [b] arr2[a, b] = c #print(arr) #print(arr2) count = 0 for k in arr: if len(arr[k]) > 1: for kk in arr[k]: if kk not in arr: #print(kk) cost += arr2[k, kk] print(min(cost, cost2-cost))
21
92
0
216797067
n=int(input()) ar=[] br=[] p,y=0,0 for _ in range(n): a,b,c=map(int,input().split()) if (a in ar or b in br): a,b=b,a y+=c p+=c ar.append(a) br.append(b) print(min(y,p-y))
Codeforces Beta Round 24
ICPC
2,010
2
256
Ring road
Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Berland decided to keep up with new trends. Formerly all n cities of Berland were connected by n two-way roads in the ring, i. e. each city was connected directly to exactly two other cities, and from each city it was possible to get to any other city. Government of Berland introduced one-way traffic on all n roads, but it soon became clear that it's impossible to get from some of the cities to some others. Now for each road is known in which direction the traffic is directed at it, and the cost of redirecting the traffic. What is the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other?
The first line contains integer n (3 ≤ n ≤ 100) — amount of cities (and roads) in Berland. Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 100) — road is directed from city ai to city bi, redirecting the traffic costs ci.
Output single integer — the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other.
null
null
[{"input": "3\n1 3 1\n1 2 1\n3 2 1", "output": "1"}, {"input": "3\n1 3 1\n1 2 5\n3 2 1", "output": "2"}, {"input": "6\n1 5 4\n5 3 8\n2 4 15\n1 6 16\n2 3 23\n4 6 42", "output": "39"}, {"input": "4\n1 2 9\n2 3 8\n3 4 7\n4 1 5", "output": "0"}]
1,400
["graphs"]
21
[{"input": "3\r\n1 3 1\r\n1 2 1\r\n3 2 1\r\n", "output": "1\r\n"}, {"input": "3\r\n1 3 1\r\n1 2 5\r\n3 2 1\r\n", "output": "2\r\n"}, {"input": "6\r\n1 5 4\r\n5 3 8\r\n2 4 15\r\n1 6 16\r\n2 3 23\r\n4 6 42\r\n", "output": "39\r\n"}, {"input": "4\r\n1 2 9\r\n2 3 8\r\n3 4 7\r\n4 1 5\r\n", "output": "0\r\n"}, {"input": "5\r\n5 3 89\r\n2 3 43\r\n4 2 50\r\n1 4 69\r\n1 5 54\r\n", "output": "143\r\n"}, {"input": "10\r\n1 8 16\r\n6 1 80\r\n6 5 27\r\n5 7 86\r\n7 9 72\r\n4 9 20\r\n4 3 54\r\n3 2 57\r\n10 2 61\r\n8 10 90\r\n", "output": "267\r\n"}, {"input": "17\r\n8 12 43\r\n13 12 70\r\n7 13 68\r\n11 7 19\r\n5 11 24\r\n5 1 100\r\n4 1 10\r\n3 4 68\r\n2 3 46\r\n15 2 58\r\n15 6 38\r\n6 9 91\r\n9 10 72\r\n14 10 32\r\n14 17 97\r\n17 16 67\r\n8 16 40\r\n", "output": "435\r\n"}, {"input": "22\r\n18 22 46\r\n18 21 87\r\n5 21 17\r\n5 10 82\r\n10 12 81\r\n17 12 98\r\n16 17 17\r\n16 13 93\r\n4 13 64\r\n4 11 65\r\n15 11 18\r\n6 15 35\r\n6 7 61\r\n7 19 12\r\n19 1 65\r\n8 1 32\r\n8 2 46\r\n9 2 19\r\n9 3 58\r\n3 14 65\r\n20 14 67\r\n20 22 2\r\n", "output": "413\r\n"}, {"input": "3\r\n3 1 1\r\n2 1 1\r\n2 3 1\r\n", "output": "1\r\n"}]
false
stdio
null
true
43/E
43
E
PyPy 3-64
TESTS
2
186
70,451,200
153601015
# @Chukamin ICPC_TRAIN def main(): n, s = map(int, input().split()) Nnum = 0 func = [[0 for i in range(40010)] for j in range(210)] for i in range(1, n + 1): data = list(map(int, input().split())) j = 0 cnt = data[0] for _ in range(cnt): v = data[_ * 2 + 1] t = data[_ * 2 + 2] for i in range(t): func[i][j + 1] = func[i][j] + v j += 1 Nnum = max(Nnum, j) ans = 0 for k in range(2, Nnum): for i in range(1, n + 1): if func[i][k]: for j in range(1, n + 1): if i != j and func[j][k]: if func[i][k] == func[j][k]: if func[i][k - 1] < func[j][k - 1] and func[i][k + 1] > func[j][k + 1]: ans += 1 else: if func[i][k - 1] < func[j][k - 1] and func[i][k] > func[j][k]: ans += 1 print(ans) if __name__ == '__main__': main()
48
716
8,396,800
216866689
import sys readline = sys.stdin.readline N, S = 0, 0 cars = [] # input def read(): global N, S N, S = list(map(int, readline().split())) for i in range(N): nums = list(map(int, readline().split())) cars.append([]) carid = len(cars) t0 = 0 for i in range(1, len(nums), 2): v, t = nums[i], nums[i + 1] cars[-1].append((t0, v, carid)) t0 += t cars[-1].append((t0, 0, carid)) # solve def solve(cars:list) -> int: def calc(car1:list, car2:list) -> int: result = 0 car1_id = car1[0][-1] v1, v2 = car1[0][1], car2[0][1] ahead = -1 t = p1 = p2 = 0 que = car1[1:] + car2[1:] que.sort() for t0, v, i in que: p2 += (t0 - t) * v2 p1 += (t0 - t) * v1 t = t0 if i == car1_id: v1 = v else: v2 = v if p1 > p2: result += int(ahead == 2) ahead = 1 elif p1 < p2: result += int(ahead == 1) ahead = 2 return result result = 0 for i in range(N): for j in range(i + 1, N): result += calc(cars[i], cars[j]) return result read() print(solve(cars))
Codeforces Beta Round 42 (Div. 2)
CF
2,010
2
256
Race
Today s kilometer long auto race takes place in Berland. The track is represented by a straight line as long as s kilometers. There are n cars taking part in the race, all of them start simultaneously at the very beginning of the track. For every car is known its behavior — the system of segments on each of which the speed of the car is constant. The j-th segment of the i-th car is pair (vi, j, ti, j), where vi, j is the car's speed on the whole segment in kilometers per hour and ti, j is for how many hours the car had been driving at that speed. The segments are given in the order in which they are "being driven on" by the cars. Your task is to find out how many times during the race some car managed to have a lead over another car. A lead is considered a situation when one car appears in front of another car. It is known, that all the leads happen instantly, i. e. there are no such time segment of positive length, during which some two cars drive "together". At one moment of time on one and the same point several leads may appear. In this case all of them should be taken individually. Meetings of cars at the start and finish are not considered to be counted as leads.
The first line contains two integers n and s (2 ≤ n ≤ 100, 1 ≤ s ≤ 106) — the number of cars and the length of the track in kilometers. Then follow n lines — the description of the system of segments for each car. Every description starts with integer k (1 ≤ k ≤ 100) — the number of segments in the system. Then k space-separated pairs of integers are written. Each pair is the speed and time of the segment. These integers are positive and don't exceed 1000. It is guaranteed, that the sum of lengths of all segments (in kilometers) for each car equals to s; and all the leads happen instantly.
Print the single number — the number of times some car managed to take the lead over another car during the race.
null
null
[{"input": "2 33\n2 5 1 2 14\n1 3 11", "output": "1"}, {"input": "2 33\n2 1 3 10 3\n1 11 3", "output": "0"}, {"input": "5 33\n2 1 3 3 10\n1 11 3\n2 5 3 3 6\n2 3 1 10 3\n2 6 3 3 5", "output": "2"}]
2,300
["brute force", "implementation", "two pointers"]
48
[{"input": "2 33\r\n2 5 1 2 14\r\n1 3 11\r\n", "output": "1\r\n"}, {"input": "2 33\r\n2 1 3 10 3\r\n1 11 3\r\n", "output": "0\r\n"}, {"input": "5 33\r\n2 1 3 3 10\r\n1 11 3\r\n2 5 3 3 6\r\n2 3 1 10 3\r\n2 6 3 3 5\r\n", "output": "2\r\n"}, {"input": "2 166755\r\n2 733 187 362 82\r\n3 813 147 565 57 557 27\r\n", "output": "0\r\n"}, {"input": "3 228385\r\n2 307 733 43 78\r\n2 252 801 157 169\r\n3 86 346 133 886 467 173\r\n", "output": "0\r\n"}, {"input": "4 773663\r\n9 277 398 57 73 62 736 625 393 186 761 129 716 329 179 54 223 554 114\r\n4 463 333 547 696 33 89 505 467\r\n2 527 792 661 539\r\n2 643 976 479 305\r\n", "output": "0\r\n"}, {"input": "5 835293\r\n2 421 965 758 566\r\n3 357 337 956 745 4 691\r\n2 433 925 464 937\r\n5 67 581 109 375 463 71 499 819 589 533\r\n2 918 828 353 213\r\n", "output": "4\r\n"}, {"input": "6 896922\r\n8 295 313 551 122 299 965 189 619 139 566 311 427 47 541 411 231\r\n5 743 210 82 451 921 124 792 397 742 371\r\n7 173 247 608 603 615 383 307 10 112 670 991 103 361 199\r\n2 190 209 961 892\r\n2 821 870 186 982\r\n5 563 456 293 568 247 955 134 787 151 877\r\n", "output": "13\r\n"}, {"input": "7 958552\r\n4 773 315 702 379 382 277 411 835\r\n3 365 416 554 861 921 358\r\n9 137 278 394 557 233 404 653 77 114 527 117 790 338 507 107 353 557 350\r\n3 776 928 43 258 895 254\r\n2 613 684 590 914\r\n4 568 326 917 201 379 173 698 750\r\n2 536 687 785 752\r\n", "output": "10\r\n"}, {"input": "8 394115\r\n8 350 64 117 509 217 451 393 118 99 454 136 37 240 183 937 79\r\n5 222 43 727 39 724 318 281 281 797 59\r\n4 440 139 367 155 415 250 359 480\r\n6 191 480 653 202 367 291 241 167 13 123 706 31\r\n2 410 369 883 275\r\n2 205 307 571 580\r\n2 469 211 452 653\r\n2 822 431 61 653\r\n", "output": "15\r\n"}, {"input": "9 81812\r\n8 31 410 547 18 22 77 449 5 491 8 10 382 746 21 61 523\r\n1 452 181\r\n1 724 113\r\n1 113 724\r\n1 226 362\r\n46 5 257 2 126 373 6 6 491 9 7 137 23 93 73 163 13 17 106 3 100 5 415 270 2 7 723 597 4 176 3 274 18 1 852 334 14 7 25 163 1 3 199 29 140 32 32 191 2 583 3 23 11 22 23 250 1 79 3 33 83 8 433 59 11 2 466 7 761 1 386 6 2 12 68 79 13 4 346 455 1 21 194 58 1 154 12 49 23 7 79 64 87\r\n12 449 11 21 192 328 9 35 381 5 492 361 9 604 11 47 239 543 22 40 265 9 105 27 351\r\n1 181 452\r\n1 362 226\r\n", "output": "5\r\n"}, {"input": "10 746595\r\n4 361 446 717 421 143 532 404 514\r\n2 327 337 724 879\r\n6 733 80 2 994 396 774 841 35 159 15 361 963\r\n5 283 973 43 731 633 521 335 269 173 115\r\n2 727 587 886 361\r\n6 223 683 98 367 80 293 612 584 128 991 224 226\r\n2 911 468 783 409\r\n2 308 983 529 839\r\n2 698 639 367 819\r\n2 275 397 785 812\r\n", "output": "21\r\n"}, {"input": "2 5\r\n3 2 1 1 1 2 1\r\n3 1 1 2 1 1 2\r\n", "output": "0\r\n"}, {"input": "2 6\r\n3 1 2 2 1 1 2\r\n3 2 1 1 2 2 1\r\n", "output": "0\r\n"}]
false
stdio
null
true
653/F
653
F
PyPy 3-64
TESTS
2
62
1,126,400
189179950
import sys, random input = lambda : sys.stdin.readline().rstrip() write = lambda x: sys.stdout.write(x+"\n"); writef = lambda x: print("{:.12f}".format(x)) debug = lambda x: sys.stderr.write(x+"\n") YES="Yes"; NO="No"; pans = lambda v: print(YES if v else NO); INF=10**18 LI = lambda : list(map(int, input().split())); II=lambda : int(input()); SI=lambda : [ord(c)-ord("a") for c in input()] def debug(_l_): for s in _l_.split(): print(f"{s}={eval(s)}", end=" ") print() def dlist(*l, fill=0): if len(l)==1: return [fill]*l[0] ll = l[1:] return [dlist(*ll, fill=fill) for _ in range(l[0])] def sa_naive(s): n = len(s) sa = list(range(n)) sa.sort(key=lambda x: s[x:]) return sa def sa_doubling(s): n = len(s) sa = list(range(n)) rnk = s[:] tmp = [0] * n k = 1 while k < n: sa.sort(key=lambda x: (rnk[x], rnk[x + k]) if x + k < n else (rnk[x], -1)) tmp[sa[0]] = 0 for i in range(1, n): tmp[sa[i]] = tmp[sa[i - 1]] if sa[i - 1] + k < n: x = (rnk[sa[i - 1]], rnk[sa[i - 1] + k]) else: x = (rnk[sa[i - 1]], -1) if sa[i] + k < n: y = (rnk[sa[i]], rnk[sa[i] + k]) else: y = (rnk[sa[i]], -1) if x < y: tmp[sa[i]] += 1 k *= 2 tmp, rnk = rnk, tmp return sa def sa_is(s, upper): n = len(s) if n == 0: return [] if n == 1: return [0] if n == 2: if s[0] < s[1]: return [0, 1] else: return [1, 0] if n < 10: return sa_naive(s) if n < 50: return sa_doubling(s) ls = [0] * n for i in range(n-2, -1, -1): ls[i] = ls[i + 1] if s[i] == s[i + 1] else s[i] < s[i + 1] sum_l = [0] * (upper + 1) sum_s = [0] * (upper + 1) for i in range(n): if ls[i]: sum_l[s[i] + 1] += 1 else: sum_s[s[i]] += 1 for i in range(upper): sum_s[i] += sum_l[i] if i < upper: sum_l[i + 1] += sum_s[i] lms_map = [-1] * (n + 1) m = 0 for i in range(1, n): if not ls[i - 1] and ls[i]: lms_map[i] = m m += 1 lms = [] for i in range(1, n): if not ls[i - 1] and ls[i]: lms.append(i) sa = [-1] * n buf = sum_s.copy() for d in lms: if d == n: continue sa[buf[s[d]]] = d buf[s[d]] += 1 buf = sum_l.copy() sa[buf[s[n - 1]]] = n - 1 buf[s[n - 1]] += 1 for i in range(n): v = sa[i] if v >= 1 and not ls[v - 1]: sa[buf[s[v - 1]]] = v - 1 buf[s[v - 1]] += 1 buf = sum_l.copy() for i in range(n-1, -1, -1): v = sa[i] if v >= 1 and ls[v - 1]: buf[s[v - 1] + 1] -= 1 sa[buf[s[v - 1] + 1]] = v - 1 if m: sorted_lms = [] for v in sa: if lms_map[v] != -1: sorted_lms.append(v) rec_s = [0] * m rec_upper = 0 rec_s[lms_map[sorted_lms[0]]] = 0 for i in range(1, m): l = sorted_lms[i - 1] r = sorted_lms[i] end_l = lms[lms_map[l] + 1] if lms_map[l] + 1 < m else n end_r = lms[lms_map[r] + 1] if lms_map[r] + 1 < m else n same = True if end_l - l != end_r - r: same = False else: while l < end_l: if s[l] != s[r]: break l += 1 r += 1 if l == n or s[l] != s[r]: same = False if not same: rec_upper += 1 rec_s[lms_map[sorted_lms[i]]] = rec_upper rec_sa = sa_is(rec_s, rec_upper) for i in range(m): sorted_lms[i] = lms[rec_sa[i]] sa = [-1] * n buf = sum_s.copy() for d in sorted_lms: if d == n: continue sa[buf[s[d]]] = d buf[s[d]] += 1 buf = sum_l.copy() sa[buf[s[n - 1]]] = n - 1 buf[s[n - 1]] += 1 for i in range(n): v = sa[i] if v >= 1 and not ls[v - 1]: sa[buf[s[v - 1]]] = v - 1 buf[s[v - 1]] += 1 buf = sum_l.copy() for i in range(n-1, -1, -1): v = sa[i] if v >= 1 and ls[v - 1]: buf[s[v - 1] + 1] -= 1 sa[buf[s[v - 1] + 1]] = v - 1 return sa def suffix_array(s): if type(s) is str: s = [ord(c) for c in s] assert min(s)>=0 return sa_is(s[:], max(s)) def lcp_array(s, sa): n = len(s) rnk = [0] * n for i in range(n): rnk[sa[i]] = i lcp = [0] * (n - 1) h = 0 for i in range(n): if h > 0: h -= 1 if rnk[i] == 0: continue j = sa[rnk[i] - 1] while j + h < n and i + h < n: if s[j + h] != s[i + h]: break h += 1 lcp[rnk[i] - 1] = h return lcp ### セグメント木(はやい) class SG: def __init__(self, n, v=None): self._n = n self.geta = 0 x = 0 while (1 << x) < n: x += 1 self._log = x self._size = 1 << self._log self._d = [ninf] * (2 * self._size) if v is not None: for i in range(self._n): self._d[self._size + i] = v[i] for i in range(self._size - 1, 0, -1): self._update(i) def _update(self, k): self._d[k] = op(self._d[2 * k], self._d[2 * k + 1]) def update(self, p, x): assert 0 <= p < self._n # x -= self.geta p += self._size self._d[p] = x for i in range(1, self._log + 1): # self._update(p >> i) k = p>>i self._d[k] = op(self._d[2 * k], self._d[2 * k + 1]) def get(self, p): assert 0 <= p < self._n return self._d[p + self._size] # + self.geta def check(self): return [self.get(p) for p in range(self._n)] def query(self, left, right): # [l,r)の総和 assert 0 <= left <= right <= self._n sml = ninf smr = ninf left += self._size right += self._size # 外側から計算していく(lは小さい側から, rは大きい側から) while left < right: if left & 1: sml = op(sml, self._d[left]) left += 1 if right & 1: right -= 1 smr = op(self._d[right], smr) left >>= 1 right >>= 1 return op(sml, smr) # + self.geta # def update_all(self, v): # # 全体加算 # self.geta += v def query_all(self): return self._d[1] # + self.geta def max_right(self, left, f): """f(op(a[l], a[l + 1], ..., a[r - 1])) = true となる最大の r -> rはf(op(a[l], ..., a[r]))がFalseになる最小のr """ # assert 0 <= left <= self._n # assert f(ninf) if left == self._n: return self._n left += self._size sm = ninf first = True while first or (left & -left) != left: first = False while left % 2 == 0: left >>= 1 if not f(op(sm, self._d[left])): while left < self._size: left *= 2 if f(op(sm, self._d[left])): sm = op(sm, self._d[left]) left += 1 return left - self._size sm = op(sm, self._d[left]) left += 1 return self._n def min_left(self, right, f): """f(op(a[l], a[l + 1], ..., a[r - 1])) = true となる最小の l -> l は f(op(a[l-1] ,..., a[r-1])) が false になる最大の l """ # assert 0 <= right <= self._n # assert f(ninf) if right == 0: return 0 right += self._size sm = ninf first = True while first or (right & -right) != right: first = False right -= 1 while right > 1 and right % 2: right >>= 1 if not f(op(self._d[right], sm)): while right < self._size: right = 2 * right + 1 if f(op(self._d[right], sm)): sm = op(self._d[right], sm) right -= 1 return right + 1 - self._size sm = op(self._d[right], sm) return 0 op = lambda i,j: i+j ninf = 0 # sa[i] : 辞書順i番目のsuffixであるインデックス # rank[i] : インデックスiの順位 # lcp[i] : 辞書順i と i+1 の suffix の共通接頭辞長 # sa = suffix_array(s) # rank = [0]*len(sa) # for i,v in enumerate(sa): # rank[v] = i # lcp = lcp_array(s, sa) n = int(input()) s = input() vs = [] index = [[] for _ in range(n+1)] cum = [0] for i in range(n): v = -1 if s[i]=="(" else 1 vs.append(v+1) cum.append(cum[-1]+v) for i in range(n+1): index[cum[i]].append(i) sg = SG(n, [v-1 for v in vs]) sa = suffix_array(vs) rank = [0]*len(sa) for i,v in enumerate(sa): rank[v] = i lcp = lcp_array(s, sa) ans = 0 from bisect import bisect_left as bl for l in range(n): r = sg.max_right(l, lambda val: val<=0) if rank[l]>0: start = l + lcp[rank[l]-1] else: start = l if r>start: val = bl(index[cum[l]], r) - bl(index[cum[l]], start) else: val = 0 # print(l, start, r, val) if s[l]=="(": ans += val print(ans)
68
904
55,910,400
169979803
import bisect import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline def suffix_array(s): l = len(s) x = [[] for _ in range(222)] for i in range(l): x[s[i]].append(i) y = [] z = [0] for x0 in x: for i in x0: y.append(i) z.append(len(y)) u = list(z) p = 1 for _ in range((l - 1).bit_length()): y0, s0, z0 = [0] * l, [0] * l, [0] for i in range(len(z) - 1): z1, z2 = z[i], z[i + 1] k = z1 for j in range(z1, z2): if y[j] - p >= 0: k = j break for _ in range(z2 - z1): j = s[(y[k] - p) % l] y0[u[j]] = (y[k] - p) % l u[j] += 1 k += 1 if k == z2: k = z1 i, u, v, c = 0, s[y0[0]], s[(y0[0] + p) % l], 0 for j in y0: if u ^ s[j] or v ^ s[(j + p) % l]: i += 1 u, v = s[j], s[(j + p) % l] z0.append(c) c += 1 s0[j] = i z0.append(l) u = list(z0) p *= 2 y, s, z = y0, s0, z0 if len(z) > l: break return y, s def lcp_array(s, sa): l = len(s) lcp, rank = [0] * l, [0] * l for i in range(l): rank[sa[i]] = i v = 0 for i in range(l): if not rank[i]: continue j = sa[rank[i] - 1] u = min(i, j) while u + v < l and s[i + v] == s[j + v]: v += 1 lcp[rank[i]] = v v = max(v - 1, 0) return lcp n = int(input()) s = list(input().rstrip()) + [0] s0 = list(s) sa, p = suffix_array(s0) lcp = lcp_array(s, sa) x = [0] for i in s: if not i & 1: x.append(x[-1] + 1) else: x.append(x[-1] - 1) x.pop() y = [[] for _ in range(n + 5)] for i in range(n + 1): y[x[i]].append(i) for i in range(n + 5): y[i].append(n + 5) z = [0] * n for i in range(1, n + 1): z[sa[i]] = lcp[i] ans = 0 for i in range(n - 1): if s[i] & 1: continue xi = x[i] u = y[xi - 1][bisect.bisect_left(y[xi - 1], i)] ans += max(bisect.bisect_left(y[xi], u) - bisect.bisect_right(y[xi], i + z[i]), 0) print(ans)
IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)
CF
2,016
3
512
Paper task
Alex was programming while Valentina (his toddler daughter) got there and started asking many questions about the round brackets (or parenthesis) in the code. He explained her a bit and when she got it he gave her a task in order to finish his code on time. For the purpose of this problem we consider only strings consisting of opening and closing round brackets, that is characters '(' and ')'. The sequence of brackets is called correct if: 1. it's empty; 2. it's a correct sequence of brackets, enclosed in a pair of opening and closing brackets; 3. it's a concatenation of two correct sequences of brackets. For example, the sequences "()()" and "((()))(())" are correct, while ")(()", "(((((" and "())" are not. Alex took a piece of paper, wrote a string s consisting of brackets and asked Valentina to count the number of distinct non-empty substrings of s that are correct sequences of brackets. In other words, her task is to count the number of non-empty correct sequences of brackets that occur in a string s as a substring (don't mix up with subsequences). When Valentina finished the task, Alex noticed he doesn't know the answer. Help him don't loose face in front of Valentina and solve the problem!
The first line of the input contains an integer n (1 ≤ n ≤ 500 000) — the length of the string s. The second line contains a string s of length n consisting of only '(' and ')'.
Print the number of distinct non-empty correct sequences that occur in s as substring.
null
In the first sample, there are 5 distinct substrings we should count: "()", "()()", "()()()", "()()()()" and "()()()()()". In the second sample, there are 3 distinct substrings we should count: "()", "(())" and "(())()".
[{"input": "10\n()()()()()", "output": "5"}, {"input": "7\n)(())()", "output": "3"}]
2,600
["data structures", "string suffix structures", "strings"]
68
[{"input": "10\r\n()()()()()\r\n", "output": "5\r\n"}, {"input": "7\r\n)(())()\r\n", "output": "3\r\n"}, {"input": "1\r\n(\r\n", "output": "0\r\n"}, {"input": "2\r\n))\r\n", "output": "0\r\n"}, {"input": "15\r\n(())(()())(()()\r\n", "output": "5\r\n"}, {"input": "30\r\n()(())(())(())()(())()()()(()(\r\n", "output": "34\r\n"}, {"input": "100\r\n(((((((((((((((((((((((((((((((((((((((((((((((((())))))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "50\r\n"}, {"input": "1\r\n)\r\n", "output": "0\r\n"}, {"input": "2\r\n)(\r\n", "output": "0\r\n"}, {"input": "2\r\n((\r\n", "output": "0\r\n"}, {"input": "3\r\n(()\r\n", "output": "1\r\n"}, {"input": "3\r\n()(\r\n", "output": "1\r\n"}, {"input": "5\r\n()(()\r\n", "output": "1\r\n"}, {"input": "7\r\n(()(())\r\n", "output": "3\r\n"}, {"input": "10\r\n()()((())(\r\n", "output": "3\r\n"}, {"input": "10\r\n((((()))))\r\n", "output": "5\r\n"}, {"input": "20\r\n()()((()())())()()((\r\n", "output": "13\r\n"}, {"input": "20\r\n(((((((((())))))))))\r\n", "output": "10\r\n"}, {"input": "20\r\n))(())))))((((()))((\r\n", "output": "3\r\n"}, {"input": "50\r\n()(())()()()(()())()(())()()()()(())()()(())()()()\r\n", "output": "142\r\n"}, {"input": "50\r\n((((((((((((((((((((((((()))))))))))))))))))))))))\r\n", "output": "25\r\n"}, {"input": "50\r\n)))(()(())())())(())()())(())()((()()))(()(()(((()\r\n", "output": "13\r\n"}, {"input": "100\r\n)(()))))))(())()))())())(())())))))))()())))()())()((()))())((((()()((()((()((((((((()(((()())(((()(\r\n", "output": "9\r\n"}, {"input": "500\r\n()(()()()()()(())())((()(()))()()(())())()(())(()()())()()(())(())(())(())(())()()(())((()))()()()()()(())()()()()((())())()(())(())(())()((())(()))(())(()()()()(())())((()()(())()))(())(()()()((()))())((())(())(())())()((()(()))())()(())((())()()(()))()(()(())()((()))()())(())()((()))()()(())(((()())())()())((()()(())))()(())((())())()()()((())()(())()())((()))(())(())()()((())()()()(())())()()(()()((())()))()()()((())()()()(()))()(()()())()()()()((())())()()(())((()))()()(()()()()()()())()()((\r\n", "output": "4051\r\n"}, {"input": "500\r\n(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((())))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))\r\n", "output": "250\r\n"}, {"input": "500\r\n)()))))()())))))(())))))))()))))))())))))))()))))))))))()))()))()))(())(()))()())()(()))))()))))))))()))))())))))))))))))(()()))))))))()())))))))()))))()))))))()))())))))))))()))))()))))))))()))))()))))()()))))())))))))())(()()))))))))())))))()())))))((()((((())(()(((((()((((((((((((((()(()((((((((((((((()((((()()))(((((((()((()()(((((((()(((()((((()(((()(((((((((((((((()((()(()((())(()))()(()(((()((((((((()((()(((((((()(((((((()(((((())()())()((((((()(((()(())))((((((()(()((((())(((()((((((()()\r\n", "output": "17\r\n"}]
false
stdio
null
true
652/F
652
F
PyPy 3
TESTS
3
77
0
198407499
import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline n, m, t = map(int, input().split()) j = 0 p, q = [], [] for i in range(n): s, d = input().rstrip().decode().split() s = int(s) % m x = 2 * s + 1 p.append(s * n + i) if d == "R": q.append((s + t) % m) u = 2 * m - x if u < 2 * t: j -= (2 * t - u) // (2 * m) + 1 else: q.append((s - t) % m) u = x if u < 2 * t: j += (2 * t - u) // (2 * m) + 1 j %= n p.sort() q.sort() for i in range(n): p[i] %= n ans = [0] * n for i in range(n): ans[p[(i + j) % n]] = q[i] sys.stdout.write(" ".join(map(str, ans)))
30
1,076
40,243,200
181254094
import sys input = sys.stdin.readline n, m, t = [int(i) for i in input().split()] 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, m) 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
583/A
583
A
Python 3
TESTS
3
46
0
201876395
n = int(input()) ar = [] ans = [] for i in range(n * n): x, y = map(int, input().split()) if x not in ar and y not in ar: ans += [i + 1] ar += [x, y] print(*ans)
39
46
0
177339118
import os,sys,io,math from array import array from math import * I=lambda:[*map(int,sys.stdin.readline().split())] IS=lambda:input() IN=lambda:int(input()) IF=lambda:float(input()) n=IN() a,b,c=[0]*n,[0]*n,[] for i in range(n**2): h,v=map(int,input().split()) if a[h-1]==0 and b[v-1]==0: c.append(i+1) a[h-1]=1 b[v-1]=1 print(*c)
Codeforces Round 323 (Div. 2)
CF
2,015
1
256
Asphalting Roads
City X consists of n vertical and n horizontal infinite roads, forming n × n intersections. Roads (both vertical and horizontal) are numbered from 1 to n, and the intersections are indicated by the numbers of the roads that form them. Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted. Road repairs are planned for n2 days. On the i-th day of the team arrives at the i-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads. According to the schedule of road works tell in which days at least one road will be asphalted.
The first line contains integer n (1 ≤ n ≤ 50) — the number of vertical and horizontal roads in the city. Next n2 lines contain the order of intersections in the schedule. The i-th of them contains two numbers hi, vi (1 ≤ hi, vi ≤ n), separated by a space, and meaning that the intersection that goes i-th in the timetable is at the intersection of the hi-th horizontal and vi-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
null
In the sample the brigade acts like that: 1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 2. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 3. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 4. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.
[{"input": "2\n1 1\n1 2\n2 1\n2 2", "output": "1 4"}, {"input": "1\n1 1", "output": "1"}]
1,000
["implementation"]
39
[{"input": "2\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n", "output": "1 4 \r\n"}, {"input": "1\r\n1 1\r\n", "output": "1 \r\n"}, {"input": "2\r\n1 1\r\n2 2\r\n1 2\r\n2 1\r\n", "output": "1 2 \r\n"}, {"input": "2\r\n1 2\r\n2 2\r\n2 1\r\n1 1\r\n", "output": "1 3 \r\n"}, {"input": "3\r\n2 2\r\n1 2\r\n3 2\r\n3 3\r\n1 1\r\n2 3\r\n1 3\r\n3 1\r\n2 1\r\n", "output": "1 4 5 \r\n"}, {"input": "3\r\n1 3\r\n3 1\r\n2 1\r\n1 1\r\n1 2\r\n2 2\r\n3 2\r\n3 3\r\n2 3\r\n", "output": "1 2 6 \r\n"}, {"input": "4\r\n1 3\r\n2 3\r\n2 4\r\n4 4\r\n3 1\r\n1 1\r\n3 4\r\n2 1\r\n1 4\r\n4 3\r\n4 1\r\n3 2\r\n1 2\r\n4 2\r\n2 2\r\n3 3\r\n", "output": "1 3 5 14 \r\n"}, {"input": "4\r\n3 3\r\n4 2\r\n2 3\r\n3 4\r\n4 4\r\n1 2\r\n3 2\r\n2 2\r\n1 4\r\n3 1\r\n4 1\r\n2 1\r\n1 3\r\n1 1\r\n4 3\r\n2 4\r\n", "output": "1 2 9 12 \r\n"}, {"input": "9\r\n4 5\r\n2 3\r\n8 3\r\n5 6\r\n9 3\r\n4 4\r\n5 4\r\n4 7\r\n1 7\r\n8 4\r\n1 4\r\n1 5\r\n5 7\r\n7 8\r\n7 1\r\n9 9\r\n8 7\r\n7 5\r\n3 7\r\n6 6\r\n7 3\r\n5 2\r\n3 6\r\n7 4\r\n9 6\r\n5 8\r\n9 7\r\n6 3\r\n7 9\r\n1 2\r\n1 1\r\n6 2\r\n5 3\r\n7 2\r\n1 6\r\n4 1\r\n6 1\r\n8 9\r\n2 2\r\n3 9\r\n2 9\r\n7 7\r\n2 8\r\n9 4\r\n2 5\r\n8 6\r\n3 4\r\n2 1\r\n2 7\r\n6 5\r\n9 1\r\n3 3\r\n3 8\r\n5 5\r\n4 3\r\n3 1\r\n1 9\r\n6 4\r\n3 2\r\n6 8\r\n2 6\r\n5 9\r\n8 5\r\n8 8\r\n9 5\r\n6 9\r\n9 2\r\n3 5\r\n4 9\r\n4 8\r\n2 4\r\n5 1\r\n4 6\r\n7 6\r\n9 8\r\n1 3\r\n4 2\r\n8 1\r\n8 2\r\n6 7\r\n1 8\r\n", "output": "1 2 4 9 10 14 16 32 56 \r\n"}, {"input": "8\r\n1 1\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n8 6\r\n1 7\r\n1 8\r\n2 1\r\n8 5\r\n2 3\r\n2 4\r\n2 5\r\n2 6\r\n4 3\r\n2 2\r\n3 1\r\n3 2\r\n3 3\r\n3 4\r\n3 5\r\n3 6\r\n5 6\r\n3 8\r\n4 1\r\n4 2\r\n2 7\r\n4 4\r\n8 8\r\n4 6\r\n4 7\r\n4 8\r\n5 1\r\n5 2\r\n5 3\r\n6 5\r\n5 5\r\n3 7\r\n5 7\r\n5 8\r\n6 1\r\n6 2\r\n6 3\r\n6 4\r\n5 4\r\n6 6\r\n6 7\r\n6 8\r\n7 1\r\n7 2\r\n7 3\r\n7 4\r\n7 5\r\n7 6\r\n7 7\r\n7 8\r\n8 1\r\n8 2\r\n8 3\r\n8 4\r\n2 8\r\n1 6\r\n8 7\r\n4 5\r\n", "output": "1 6 11 18 28 36 39 56 \r\n"}, {"input": "9\r\n9 9\r\n5 5\r\n8 8\r\n3 3\r\n2 2\r\n6 6\r\n4 4\r\n1 1\r\n7 7\r\n8 4\r\n1 4\r\n1 5\r\n5 7\r\n7 8\r\n7 1\r\n1 7\r\n8 7\r\n7 5\r\n3 7\r\n5 6\r\n7 3\r\n5 2\r\n3 6\r\n7 4\r\n9 6\r\n5 8\r\n9 7\r\n6 3\r\n7 9\r\n1 2\r\n4 5\r\n6 2\r\n5 3\r\n7 2\r\n1 6\r\n4 1\r\n6 1\r\n8 9\r\n2 3\r\n3 9\r\n2 9\r\n5 4\r\n2 8\r\n9 4\r\n2 5\r\n8 6\r\n3 4\r\n2 1\r\n2 7\r\n6 5\r\n9 1\r\n8 3\r\n3 8\r\n9 3\r\n4 3\r\n3 1\r\n1 9\r\n6 4\r\n3 2\r\n6 8\r\n2 6\r\n5 9\r\n8 5\r\n4 7\r\n9 5\r\n6 9\r\n9 2\r\n3 5\r\n4 9\r\n4 8\r\n2 4\r\n5 1\r\n4 6\r\n7 6\r\n9 8\r\n1 3\r\n4 2\r\n8 1\r\n8 2\r\n6 7\r\n1 8\r\n", "output": "1 2 3 4 5 6 7 8 9 \r\n"}]
false
stdio
null
true
364/D
364
D
PyPy 3
TESTS
2
3,556
125,030,400
71102642
from typing import List from random import randrange from sys import stdin from functools import reduce def read_integers(): return list(map(int, stdin.readline().strip().split())) def decompose(n): return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))), reverse=True) def ghd(arr: List[int]) -> int: visited = set() n_trials = 15 max_key = 1 i = 0 while i < n_trials: rd = randrange(0, len(arr)) factors = decompose(arr[rd]) for f in factors: if any(f % x == 0 for x in visited): visited.add(f) continue if f not in visited: visited.add(f) cnt = sum(num % f == 0 for num in arr) if cnt * 2 >= len(arr): max_key = max(max_key, f) i += 1 return max_key if True: _, = read_integers() input_arr = read_integers() print(ghd(input_arr))
115
3,900
133,734,400
179459001
import math from random import randint from bisect import bisect_left test = False if test: n = randint(100000,100000) a = [randint(1,100000000) for _ in range(n)] #print('n =', n) #print('a =', a) else: n = int(input()) a=[*map(int,input().split())] dr = 1 for x in range(9): r = randint(0, n-1) ar = a[r] m = int(math.sqrt(ar)) dells = [] dells1 = [] for i in range(1, m+1): if (ar%i == 0): if ar//i > dr: dells1.insert(0, ar//i) if i > dr: dells.append(i) dells.extend(dells1) num_dells = len(dells) s = [0]*num_dells for i in range(n): k = math.gcd(ar, a[i]) lo = bisect_left(dells, k) if lo < num_dells and dells[lo] == k: s[lo] += 1 for i in range(num_dells): for j in range(i+1, num_dells): if dells[j] % dells[i]==0: s[i] += s[j] if (s[i]*2>=n and dr < dells[i]): dr = dells[i] print(dr)
Codeforces Round 213 (Div. 1)
CF
2,013
4
256
Ghd
John Doe offered his sister Jane Doe find the gcd of some set of numbers a. Gcd is a positive integer g, such that all number from the set are evenly divisible by g and there isn't such g' (g' > g), that all numbers of the set are evenly divisible by g'. Unfortunately Jane couldn't cope with the task and John offered her to find the ghd of the same subset of numbers. Ghd is a positive integer g, such that at least half of numbers from the set are evenly divisible by g and there isn't such g' (g' > g) that at least half of the numbers from the set are evenly divisible by g'. Jane coped with the task for two hours. Please try it, too.
The first line contains an integer n (1 ≤ n ≤ 106) showing how many numbers are in set a. The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1012). Please note, that given set can contain equal numbers. Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the %I64d specifier.
Print a single integer g — the Ghd of set a.
null
null
[{"input": "6\n6 2 3 4 5 6", "output": "3"}, {"input": "5\n5 5 6 10 15", "output": "5"}]
2,900
["brute force", "math", "probabilities"]
115
[{"input": "6\r\n6 2 3 4 5 6\r\n", "output": "3\r\n"}, {"input": "5\r\n5 5 6 10 15\r\n", "output": "5\r\n"}, {"input": "100\r\n32 40 7 3 7560 21 7560 7560 10 12 3 7560 7560 7560 7560 5 7560 7560 6 7560 7560 7560 35 7560 18 7560 7560 7560 7560 7560 48 2 7 25 7560 2 2 49 7560 7560 15 16 7560 7560 2 7560 27 7560 7560 7560 7560 3 5 7560 8 7560 42 45 5 7560 5 7560 4 7 3 7560 7 3 7560 7 2 7560 7560 5 3 7560 7560 28 7560 7560 14 7560 5 7560 20 7560 24 7560 2 9 36 7 7560 7560 7560 7560 7560 30 7560 50\r\n", "output": "7560\r\n"}, {"input": "1\r\n3\r\n", "output": "3\r\n"}, {"input": "1\r\n7\r\n", "output": "7\r\n"}, {"input": "2\r\n1 7\r\n", "output": "7\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}]
false
stdio
null
true
831/B
831
B
Python 3
TESTS
2
30
0
136741375
c1 = list(input()) c2 = list(input()) c3 = list(input()) c4 = list() c5 = list() for i in range(len(c3)): if 64 < ord(c3[i]) < 91: c4.append(i) c3 = [x.lower() for x in c3 if isinstance(x,str)] a = len(c3) for i in range(a): n = 0 for b in range(len(c1)): if c3[i] == c1[b]: c5.append(c2[b]) n = n + 1 break else: n = n + 1 if n == 26: c5.append(c3[i]) for i in range(len(c5)): h = 0 for z in range(len(c4)): if i == c4[z]: print(c5[i].upper(),end='') h = h + 1 break else: h = h + 1 if h == len(c4): print(c5[i],end='')
19
31
0
144988702
p1 = input() p2=input() s = input() res = '' for i in s: if(i.lower() in p1): l= p2[p1.index(i.lower())] if(i.isupper()): res+=l.upper() else: res+=l else: res+=i print(res)
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
65/C
65
C
Python 3
TESTS
0
60
102,400
217458268
import math import sys eps = 1e-8 n = int(input()) al = [list(map(int, input().split())) for _ in range(n + 1)] vp, vs = map(int, input().split()) px, py, pz = p0 = map(int, input().split()) al = [(x - px, y - py, z - pz) for x, y, z in al] d3 = lambda x, y, z: x * x + y * y + z * z t0, ts = 0, 0 rt = None for i in range(n): c = [y - x for x, y in zip(al[i], al[i + 1])] ll = d3(*c) l = ll ** 0.5 ts += l te = ts / vs v = [vs * x for x in c] s = [l * x - a * t0 for x, a in zip(al[i], v)] a = d3(*v) - vp * vp * ll b = 2 * sum(x * i for x, i in zip(s, v)) c = d3(*s) d = b * b - 4 * a * c fa = abs(a) < eps f = lambda t: (t0 - eps < t < te + eps) if fa: if abs(b) > eps and f(-c / b): rt = -c / b break elif d > -eps: if d < eps: d = 0 a *= 2.0 d **= 0.5 tl = [t for t in ((-b + d) / a, (-b - d) / a) if f(t)] if tl: rt = min(tl) break t0 = te if rt is None: print("NO") else: print("YES") print("%.9f" % rt) print(" ".join(["%.9f" % ((x + a * rt) / l + p) for x, a, p in zip(s, v, p0)]))
45
248
3,891,200
217460027
import math, sys eps = 1e-8 n = int(input()) al = [list(map(int, input().split())) for _ in range(n + 1)] vp, vs = map(int, input().split()) px, py, pz = map(int, input().split()) al = [(x - px, y - py, z - pz) for x, y, z in al] d3 = lambda x, y, z: x*x + y*y + z*z t0 = 0 rt, pt = None, 0 ts = 0 def tsol(t): global rt, pt if t0 - eps < t < te + eps and (rt is None or rt > t): rt, pt = t, [(x + a * t) / l for x, a in zip(s, v)] for i in range(n): c = [y - x for x, y in zip(al[i], al[i + 1])] ll = d3(*c) l = ll ** 0.5 ts += l te = ts / vs v = [vs * x for x in c] s = [l * x - a * t0 for x, a in zip(al[i], v)] a = d3(*v) - vp * vp * ll b = 2 * sum(x * i for x, i in zip(s, v)) c = d3(*s) d = b * b - 4 * a * c fa = abs(a) < eps if fa: if abs(b) > eps: tsol(-c / b) elif d > -eps: if d < eps: d = 0 a *= 2.0 d **= 0.5 tsol((-b + d) / a) tsol((-b - d) / a) t0 = te if rt is None: print("NO") else: print("YES") print("%.10f" % rt) print("%.10f" % (pt[0] + px), "%.10f" % (pt[1] + py), "%.10f" % (pt[2] + pz))
Codeforces Beta Round 60
CF
2,011
2
256
Harry Potter and the Golden Snitch
Brothers Fred and George Weasley once got into the sporting goods store and opened a box of Quidditch balls. After long and painful experiments they found out that the Golden Snitch is not enchanted at all. It is simply a programmed device. It always moves along the same trajectory, which is a polyline with vertices at the points (x0, y0, z0), (x1, y1, z1), ..., (xn, yn, zn). At the beginning of the game the snitch is positioned at the point (x0, y0, z0), and then moves along the polyline at the constant speed vs. The twins have not yet found out how the snitch behaves then. Nevertheless, they hope that the retrieved information will help Harry Potter and his team in the upcoming match against Slytherin. Harry Potter learned that at the beginning the game he will be at the point (Px, Py, Pz) and his super fast Nimbus 2011 broom allows him to move at the constant speed vp in any direction or remain idle. vp is not less than the speed of the snitch vs. Harry Potter, of course, wants to catch the snitch as soon as possible. Or, if catching the snitch while it is moving along the polyline is impossible, he wants to hurry the Weasley brothers with their experiments. Harry Potter catches the snitch at the time when they are at the same point. Help Harry.
The first line contains a single integer n (1 ≤ n ≤ 10000). The following n + 1 lines contain the coordinates xi, yi, zi, separated by single spaces. The coordinates of any two consecutive points do not coincide. The next line contains the velocities vp and vs, the last line contains Px, Py, Pz, separated by single spaces. All the numbers in the input are integers, their absolute value does not exceed 104. The speeds are strictly positive. It is guaranteed that vs ≤ vp.
If Harry Potter can catch the snitch while it is moving along the polyline (including the end (xn, yn, zn)), print "YES" in the first line (without the quotes). Print in the second line t, which is the earliest moment of time, when Harry will be able to catch the snitch. On the third line print three numbers X, Y, Z, the coordinates of the point at which this happens. The absolute or relative error in the answer should not exceed 10 - 6. If Harry is not able to catch the snitch during its moving along the described polyline, print "NO".
null
null
[{"input": "4\n0 0 0\n0 10 0\n10 10 0\n10 0 0\n0 0 0\n1 1\n5 5 25", "output": "YES\n25.5000000000\n10.0000000000 4.5000000000 0.0000000000"}, {"input": "4\n0 0 0\n0 10 0\n10 10 0\n10 0 0\n0 0 0\n1 1\n5 5 50", "output": "NO"}, {"input": "1\n1 2 3\n4 5 6\n20 10\n1 2 3", "output": "YES\n0.0000000000\n1.0000000000 2.0000000000 3.0000000000"}]
2,100
["binary search", "geometry"]
45
[{"input": "4\r\n0 0 0\r\n0 10 0\r\n10 10 0\r\n10 0 0\r\n0 0 0\r\n1 1\r\n5 5 25\r\n", "output": "YES\r\n25.5000000000\r\n10.0000000000 4.5000000000 0.0000000000\r\n"}, {"input": "4\r\n0 0 0\r\n0 10 0\r\n10 10 0\r\n10 0 0\r\n0 0 0\r\n1 1\r\n5 5 50\r\n", "output": "NO\r\n"}, {"input": "1\r\n1 2 3\r\n4 5 6\r\n20 10\r\n1 2 3\r\n", "output": "YES\r\n0.0000000000\r\n1.0000000000 2.0000000000 3.0000000000\r\n"}, {"input": "4\r\n0 0 0\r\n0 1 0\r\n1 1 0\r\n1 0 0\r\n0 0 0\r\n10 5\r\n0 0 8\r\n", "output": "YES\r\n0.8000000000\r\n0.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "4\r\n1 0 0\r\n0 1 0\r\n-1 0 0\r\n0 -1 0\r\n1 0 0\r\n10 5\r\n9 0 -8\r\n", "output": "YES\r\n1.1313708499\r\n1.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "5\r\n32 -5 -42\r\n-25 -38 -6\r\n-13 41 25\r\n21 -25 -32\r\n43 35 -19\r\n-38 -12 -48\r\n3 2\r\n182 -210 32\r\n", "output": "YES\r\n97.5061769956\r\n-0.5611252637 16.8539490414 4.1465923539\r\n"}, {"input": "10\r\n-20 28 4\r\n-12 -34 49\r\n3 -11 25\r\n-35 -46 25\r\n4 29 -15\r\n17 16 -10\r\n40 -35 16\r\n-15 -25 10\r\n-2 40 20\r\n-26 18 -49\r\n14 8 -44\r\n3 1\r\n-877 450 899\r\n", "output": "YES\r\n437.7804049730\r\n-6.8291526407 15.8542367965 16.2852671995\r\n"}, {"input": "1\r\n5 -22 -3\r\n31 -41 -35\r\n4 4\r\n139 -86 -115\r\n", "output": "NO\r\n"}, {"input": "2\r\n-34 37 40\r\n24 -28 7\r\n-20 -14 -25\r\n1 1\r\n-69 -28 -70\r\n", "output": "YES\r\n107.2130636667\r\n12.9900466281 -24.4968330180 -1.0072388159\r\n"}, {"input": "3\r\n-38 -39 -19\r\n-49 -16 50\r\n-3 -7 5\r\n28 -15 41\r\n1 1\r\n-100 -139 -33\r\n", "output": "NO\r\n"}, {"input": "15\r\n-17 -8 7\r\n-50 -28 8\r\n13 -38 -17\r\n27 -49 15\r\n34 49 17\r\n-17 36 25\r\n-10 -15 28\r\n-15 -36 32\r\n-8 47 26\r\n-19 18 -25\r\n44 36 -16\r\n4 -46 49\r\n46 20 -13\r\n21 -37 -8\r\n35 -38 -26\r\n-26 46 12\r\n4 1\r\n-1693 1363 2149\r\n", "output": "YES\r\n768.5953048926\r\n37.0198725921 5.8883712161 0.2563785546\r\n"}, {"input": "1\r\n0 0 0\r\n0 0 1\r\n10000 10000\r\n0 0 1\r\n", "output": "YES\r\n0.0000500000\r\n0.0000000000 0.0000000000 0.5000000000\r\n"}, {"input": "1\r\n10000 -10000 10000\r\n-10000 10000 -10000\r\n1 1\r\n10000 10000 10000\r\n", "output": "YES\r\n17320.5080756888\r\n0.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "1\r\n10000 -10000 10000\r\n-10000 10000 -10000\r\n10000 1\r\n10000 10000 10000\r\n", "output": "YES\r\n1.9998845433\r\n9998.8453661206 -9998.8453661206 9998.8453661206\r\n"}, {"input": "1\r\n0 0 -1\r\n0 0 1\r\n10000 1\r\n0 0 10000\r\n", "output": "YES\r\n1.0000000000\r\n0.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "1\r\n0 0 0\r\n-1 0 0\r\n10000 1\r\n10000 0 0\r\n", "output": "NO\r\n"}, {"input": "2\r\n10000 10000 10000\r\n10000 10000 -10000\r\n10000 -10000 -10000\r\n1 1\r\n-10000 -10000 10000\r\n", "output": "YES\r\n30000.0000000000\r\n10000.0000000000 0.0000000000 -10000.0000000000\r\n"}, {"input": "4\r\n10000 9999 10000\r\n10000 9999 9999\r\n10000 10000 9999\r\n10000 10000 10000\r\n10000 9999 10000\r\n10000 1\r\n-10000 -10000 -10000\r\n", "output": "YES\r\n3.4640748220\r\n10000.0000000000 9999.5359251780 10000.0000000000\r\n"}, {"input": "3\r\n10000 9999 10000\r\n10000 9999 9999\r\n10000 10000 9999\r\n10000 10000 10000\r\n10000 1\r\n-10000 -10000 -10000\r\n", "output": "NO\r\n"}]
false
stdio
import sys import math def read_lines(path): with open(path, 'r') as f: return [line.strip() for line in f.readlines()] def main(): input_path = sys.argv[1] output_path = sys.argv[2] submission_path = sys.argv[3] ref_lines = read_lines(output_path) sub_lines = read_lines(submission_path) if len(ref_lines) < 1 or len(sub_lines) < 1: print(0) return ref_yesno = ref_lines[0] sub_yesno = sub_lines[0] if ref_yesno != sub_yesno: print(0) return if ref_yesno == "NO": if len(sub_lines) == 1: print(1) else: print(0) return if len(ref_lines) < 3 or len(sub_lines) < 3: print(0) return try: t_ref = float(ref_lines[1]) t_sub = float(sub_lines[1]) except: print(0) return if not math.isclose(t_ref, t_sub, rel_tol=1e-6, abs_tol=1e-6): print(0) return ref_coords = ref_lines[2].split() sub_coords = sub_lines[2].split() if len(ref_coords) != 3 or len(sub_coords) != 3: print(0) return try: x_ref, y_ref, z_ref = map(float, ref_coords) x_sub, y_sub, z_sub = map(float, sub_coords) except: print(0) return if not (math.isclose(x_ref, x_sub, rel_tol=1e-6, abs_tol=1e-6) and math.isclose(y_ref, y_sub, rel_tol=1e-6, abs_tol=1e-6) and math.isclose(z_ref, z_sub, rel_tol=1e-6, abs_tol=1e-6)): print(0) return print(1) if __name__ == "__main__": main()
true
65/C
65
C
Python 3
TESTS
0
62
102,400
217457935
import sys eps = 1e-8 n = int(input()) al = [list(map(int, input().split())) for _ in range(n+1)] vp, vs = map(int, input().split()) px, py, pz = p0 = map(int, input().split()) al = [(x-px, y-py, z-pz) for x, y, z in al] d3 = lambda x, y, z: x*x + y*y + z*z t0, ts = 0, 0 rt = None for i in range(n): c = [y-x for x, y in zip(al[i], al[i+1])] ll = d3(*c) l = ll**0.5 ts += l te = ts / vs v = [vs*x for x in c] s = [l*x - a*t0 for x, a in zip(al[i], v)] a = d3(*v) - vp*vp*ll b = 2 * sum(x*i for x, i in zip(s, v)) c = d3(*s) d = b*b - 4*a*c fa = abs(a) < eps f = lambda t: (t0 - eps < t < te + eps) if fa: if abs(b) > eps and f(-c/b): rt = -c/b break elif d > -eps: if d < eps: d = 0 a *= 2.0 d **= 0.5 tl = [t for t in ((-b+d) / a, (-b-d) / a) if f(t)] if tl: rt = min(tl) break t0 = te if rt is None: print("NO") else: print("YES") print("%.9f" % rt) print(" ".join(["%.9f" % ((x+a*rt) / l+p) for x, a, p in zip(s, v, p0)]))
45
248
3,891,200
217460027
import math, sys eps = 1e-8 n = int(input()) al = [list(map(int, input().split())) for _ in range(n + 1)] vp, vs = map(int, input().split()) px, py, pz = map(int, input().split()) al = [(x - px, y - py, z - pz) for x, y, z in al] d3 = lambda x, y, z: x*x + y*y + z*z t0 = 0 rt, pt = None, 0 ts = 0 def tsol(t): global rt, pt if t0 - eps < t < te + eps and (rt is None or rt > t): rt, pt = t, [(x + a * t) / l for x, a in zip(s, v)] for i in range(n): c = [y - x for x, y in zip(al[i], al[i + 1])] ll = d3(*c) l = ll ** 0.5 ts += l te = ts / vs v = [vs * x for x in c] s = [l * x - a * t0 for x, a in zip(al[i], v)] a = d3(*v) - vp * vp * ll b = 2 * sum(x * i for x, i in zip(s, v)) c = d3(*s) d = b * b - 4 * a * c fa = abs(a) < eps if fa: if abs(b) > eps: tsol(-c / b) elif d > -eps: if d < eps: d = 0 a *= 2.0 d **= 0.5 tsol((-b + d) / a) tsol((-b - d) / a) t0 = te if rt is None: print("NO") else: print("YES") print("%.10f" % rt) print("%.10f" % (pt[0] + px), "%.10f" % (pt[1] + py), "%.10f" % (pt[2] + pz))
Codeforces Beta Round 60
CF
2,011
2
256
Harry Potter and the Golden Snitch
Brothers Fred and George Weasley once got into the sporting goods store and opened a box of Quidditch balls. After long and painful experiments they found out that the Golden Snitch is not enchanted at all. It is simply a programmed device. It always moves along the same trajectory, which is a polyline with vertices at the points (x0, y0, z0), (x1, y1, z1), ..., (xn, yn, zn). At the beginning of the game the snitch is positioned at the point (x0, y0, z0), and then moves along the polyline at the constant speed vs. The twins have not yet found out how the snitch behaves then. Nevertheless, they hope that the retrieved information will help Harry Potter and his team in the upcoming match against Slytherin. Harry Potter learned that at the beginning the game he will be at the point (Px, Py, Pz) and his super fast Nimbus 2011 broom allows him to move at the constant speed vp in any direction or remain idle. vp is not less than the speed of the snitch vs. Harry Potter, of course, wants to catch the snitch as soon as possible. Or, if catching the snitch while it is moving along the polyline is impossible, he wants to hurry the Weasley brothers with their experiments. Harry Potter catches the snitch at the time when they are at the same point. Help Harry.
The first line contains a single integer n (1 ≤ n ≤ 10000). The following n + 1 lines contain the coordinates xi, yi, zi, separated by single spaces. The coordinates of any two consecutive points do not coincide. The next line contains the velocities vp and vs, the last line contains Px, Py, Pz, separated by single spaces. All the numbers in the input are integers, their absolute value does not exceed 104. The speeds are strictly positive. It is guaranteed that vs ≤ vp.
If Harry Potter can catch the snitch while it is moving along the polyline (including the end (xn, yn, zn)), print "YES" in the first line (without the quotes). Print in the second line t, which is the earliest moment of time, when Harry will be able to catch the snitch. On the third line print three numbers X, Y, Z, the coordinates of the point at which this happens. The absolute or relative error in the answer should not exceed 10 - 6. If Harry is not able to catch the snitch during its moving along the described polyline, print "NO".
null
null
[{"input": "4\n0 0 0\n0 10 0\n10 10 0\n10 0 0\n0 0 0\n1 1\n5 5 25", "output": "YES\n25.5000000000\n10.0000000000 4.5000000000 0.0000000000"}, {"input": "4\n0 0 0\n0 10 0\n10 10 0\n10 0 0\n0 0 0\n1 1\n5 5 50", "output": "NO"}, {"input": "1\n1 2 3\n4 5 6\n20 10\n1 2 3", "output": "YES\n0.0000000000\n1.0000000000 2.0000000000 3.0000000000"}]
2,100
["binary search", "geometry"]
45
[{"input": "4\r\n0 0 0\r\n0 10 0\r\n10 10 0\r\n10 0 0\r\n0 0 0\r\n1 1\r\n5 5 25\r\n", "output": "YES\r\n25.5000000000\r\n10.0000000000 4.5000000000 0.0000000000\r\n"}, {"input": "4\r\n0 0 0\r\n0 10 0\r\n10 10 0\r\n10 0 0\r\n0 0 0\r\n1 1\r\n5 5 50\r\n", "output": "NO\r\n"}, {"input": "1\r\n1 2 3\r\n4 5 6\r\n20 10\r\n1 2 3\r\n", "output": "YES\r\n0.0000000000\r\n1.0000000000 2.0000000000 3.0000000000\r\n"}, {"input": "4\r\n0 0 0\r\n0 1 0\r\n1 1 0\r\n1 0 0\r\n0 0 0\r\n10 5\r\n0 0 8\r\n", "output": "YES\r\n0.8000000000\r\n0.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "4\r\n1 0 0\r\n0 1 0\r\n-1 0 0\r\n0 -1 0\r\n1 0 0\r\n10 5\r\n9 0 -8\r\n", "output": "YES\r\n1.1313708499\r\n1.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "5\r\n32 -5 -42\r\n-25 -38 -6\r\n-13 41 25\r\n21 -25 -32\r\n43 35 -19\r\n-38 -12 -48\r\n3 2\r\n182 -210 32\r\n", "output": "YES\r\n97.5061769956\r\n-0.5611252637 16.8539490414 4.1465923539\r\n"}, {"input": "10\r\n-20 28 4\r\n-12 -34 49\r\n3 -11 25\r\n-35 -46 25\r\n4 29 -15\r\n17 16 -10\r\n40 -35 16\r\n-15 -25 10\r\n-2 40 20\r\n-26 18 -49\r\n14 8 -44\r\n3 1\r\n-877 450 899\r\n", "output": "YES\r\n437.7804049730\r\n-6.8291526407 15.8542367965 16.2852671995\r\n"}, {"input": "1\r\n5 -22 -3\r\n31 -41 -35\r\n4 4\r\n139 -86 -115\r\n", "output": "NO\r\n"}, {"input": "2\r\n-34 37 40\r\n24 -28 7\r\n-20 -14 -25\r\n1 1\r\n-69 -28 -70\r\n", "output": "YES\r\n107.2130636667\r\n12.9900466281 -24.4968330180 -1.0072388159\r\n"}, {"input": "3\r\n-38 -39 -19\r\n-49 -16 50\r\n-3 -7 5\r\n28 -15 41\r\n1 1\r\n-100 -139 -33\r\n", "output": "NO\r\n"}, {"input": "15\r\n-17 -8 7\r\n-50 -28 8\r\n13 -38 -17\r\n27 -49 15\r\n34 49 17\r\n-17 36 25\r\n-10 -15 28\r\n-15 -36 32\r\n-8 47 26\r\n-19 18 -25\r\n44 36 -16\r\n4 -46 49\r\n46 20 -13\r\n21 -37 -8\r\n35 -38 -26\r\n-26 46 12\r\n4 1\r\n-1693 1363 2149\r\n", "output": "YES\r\n768.5953048926\r\n37.0198725921 5.8883712161 0.2563785546\r\n"}, {"input": "1\r\n0 0 0\r\n0 0 1\r\n10000 10000\r\n0 0 1\r\n", "output": "YES\r\n0.0000500000\r\n0.0000000000 0.0000000000 0.5000000000\r\n"}, {"input": "1\r\n10000 -10000 10000\r\n-10000 10000 -10000\r\n1 1\r\n10000 10000 10000\r\n", "output": "YES\r\n17320.5080756888\r\n0.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "1\r\n10000 -10000 10000\r\n-10000 10000 -10000\r\n10000 1\r\n10000 10000 10000\r\n", "output": "YES\r\n1.9998845433\r\n9998.8453661206 -9998.8453661206 9998.8453661206\r\n"}, {"input": "1\r\n0 0 -1\r\n0 0 1\r\n10000 1\r\n0 0 10000\r\n", "output": "YES\r\n1.0000000000\r\n0.0000000000 0.0000000000 0.0000000000\r\n"}, {"input": "1\r\n0 0 0\r\n-1 0 0\r\n10000 1\r\n10000 0 0\r\n", "output": "NO\r\n"}, {"input": "2\r\n10000 10000 10000\r\n10000 10000 -10000\r\n10000 -10000 -10000\r\n1 1\r\n-10000 -10000 10000\r\n", "output": "YES\r\n30000.0000000000\r\n10000.0000000000 0.0000000000 -10000.0000000000\r\n"}, {"input": "4\r\n10000 9999 10000\r\n10000 9999 9999\r\n10000 10000 9999\r\n10000 10000 10000\r\n10000 9999 10000\r\n10000 1\r\n-10000 -10000 -10000\r\n", "output": "YES\r\n3.4640748220\r\n10000.0000000000 9999.5359251780 10000.0000000000\r\n"}, {"input": "3\r\n10000 9999 10000\r\n10000 9999 9999\r\n10000 10000 9999\r\n10000 10000 10000\r\n10000 1\r\n-10000 -10000 -10000\r\n", "output": "NO\r\n"}]
false
stdio
import sys import math def read_lines(path): with open(path, 'r') as f: return [line.strip() for line in f.readlines()] def main(): input_path = sys.argv[1] output_path = sys.argv[2] submission_path = sys.argv[3] ref_lines = read_lines(output_path) sub_lines = read_lines(submission_path) if len(ref_lines) < 1 or len(sub_lines) < 1: print(0) return ref_yesno = ref_lines[0] sub_yesno = sub_lines[0] if ref_yesno != sub_yesno: print(0) return if ref_yesno == "NO": if len(sub_lines) == 1: print(1) else: print(0) return if len(ref_lines) < 3 or len(sub_lines) < 3: print(0) return try: t_ref = float(ref_lines[1]) t_sub = float(sub_lines[1]) except: print(0) return if not math.isclose(t_ref, t_sub, rel_tol=1e-6, abs_tol=1e-6): print(0) return ref_coords = ref_lines[2].split() sub_coords = sub_lines[2].split() if len(ref_coords) != 3 or len(sub_coords) != 3: print(0) return try: x_ref, y_ref, z_ref = map(float, ref_coords) x_sub, y_sub, z_sub = map(float, sub_coords) except: print(0) return if not (math.isclose(x_ref, x_sub, rel_tol=1e-6, abs_tol=1e-6) and math.isclose(y_ref, y_sub, rel_tol=1e-6, abs_tol=1e-6) and math.isclose(z_ref, z_sub, rel_tol=1e-6, abs_tol=1e-6)): print(0) return print(1) if __name__ == "__main__": main()
true
30/C
30
C
PyPy 3
TESTS
5
186
0
140566719
def r(x1, y1, x2, y2): return ((x1 - x2) ** 2 + (y1 - y2) ** 2) ** 0.5 n = int(input()) f = [[] for i in range(n)] for i in range(n): x, y, t, p = map(float, input().split()) f[i] = [x, y, t, p] f.sort(key=lambda x: x[0]) dp = [f[i][3] for i in range(n)] for i in range(1, n): for j in range(0, i): if f[i][2] - f[j][2] >= r(f[i][0], f[i][1], f[j][0], f[j][1]): dp[i] = max(dp[i], dp[j] + f[i][3]) print("%.10f" % max(dp))
50
466
2,150,400
117763382
n=int(input()) b=[] for i in range(n): x,y,t,p=map(float,input().split()) b.append([t,x,y,p]) b.sort() dp=[0 for j in range(n)] dp[0]=b[0][3] ans=0 for i in range(1,n): t,x,y,p=b[i] j=i-1 while(j>=0): t1,x1,y1,p1=b[j] if ((x1-x)**2+(y1-y)**2)**0.5<=(t-t1): dp[i]=max(dp[i],dp[j]) j+=-1 dp[i]+=p print(max(dp))
Codeforces Beta Round 30 (Codeforces format)
CF
2,010
2
256
Shooting Gallery
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize — big pink plush panda. The king is not good at shooting, so he invited you to help him. The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it. A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
The first line contains integer n (1 ≤ n ≤ 1000) — amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti — integers, - 1000 ≤ xi, yi ≤ 1000, 0 ≤ ti ≤ 109, real number pi is given with no more than 6 digits after the decimal point, 0 ≤ pi ≤ 1). No two targets may be at the same point.
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
null
null
[{"input": "1\n0 0 0 0.5", "output": "0.5000000000"}, {"input": "2\n0 0 0 0.6\n5 0 5 0.7", "output": "1.3000000000"}]
1,800
["dp", "probabilities"]
50
[{"input": "1\r\n0 0 0 0.5\r\n", "output": "0.5000000000\r\n"}, {"input": "2\r\n0 0 0 0.6\r\n5 0 5 0.7\r\n", "output": "1.3000000000\r\n"}, {"input": "1\r\n-5 2 3 0.886986\r\n", "output": "0.8869860000\r\n"}, {"input": "4\r\n10 -7 14 0.926305\r\n-7 -8 12 0.121809\r\n-7 7 14 0.413446\r\n3 -8 6 0.859061\r\n", "output": "1.7853660000\r\n"}, {"input": "5\r\n-2 -2 34 0.127276\r\n5 -5 4 0.459998\r\n10 3 15 0.293766\r\n1 -3 7 0.089869\r\n-4 -7 11 0.772515\r\n", "output": "0.8997910000\r\n"}, {"input": "5\r\n2 5 1 0.955925\r\n9 -9 14 0.299977\r\n0 1 97 0.114582\r\n-4 -2 66 0.561033\r\n0 -10 75 0.135937\r\n", "output": "1.7674770000\r\n"}, {"input": "10\r\n-4 7 39 0.921693\r\n3 -1 50 0.111185\r\n-2 -8 27 0.976475\r\n-9 -2 25 0.541029\r\n6 -4 21 0.526054\r\n-7 2 19 0.488637\r\n-6 -5 50 0.819011\r\n-7 3 39 0.987596\r\n-3 -8 16 0.685997\r\n4 10 1 0.246686\r\n", "output": "3.0829590000\r\n"}]
false
stdio
import sys def main(): input_path = sys.argv[1] output_path = sys.argv[2] submission_path = sys.argv[3] with open(output_path, 'r') as f: correct_line = f.readline().strip() try: correct = float(correct_line) except: print(0) return with open(submission_path, 'r') as f: sub_line = f.readline().strip() try: sub = float(sub_line) except: print(0) return if abs(correct - sub) <= 1e-6 + 1e-12: # Adding small epsilon for floating point errors print(1) else: print(0) if __name__ == "__main__": main()
true
30/C
30
C
PyPy 3
TESTS
5
248
0
61558376
if __name__ == '__main__': n = int(input()) targets = [input() for _ in range(n)] targets = [line.split() for line in targets] targets = [(int(x), int(y), int(t), float(p)) for x, y, t, p in targets] targets = sorted(targets, key=lambda x: x[2]) max_scores = [0] * n for i in range(n - 1, -1, -1): for j in range(i + 1, n): if targets[j][2] - targets[i][2] >= (abs(targets[j][0] - targets[i][0]) + abs(targets[j][1] - targets[i][1])): max_scores[i] = max(max_scores[i], max_scores[j]) max_scores[i] += targets[i][3] print(max(max_scores))
50
684
2,662,400
168603661
n=int(input()) class Target: def __init__(self,x,y,t,p): self.x=x self.y=y self.t=t self.p=p self.best=0 targets=[0]*(n) for i in range(n): p=[el for el in input().split()] targets[i]=Target(int(p[0]),int(p[1]),int(p[2]),float(p[3])) targets.sort(key=lambda x:(x.t)) best=0 for i in range(n): a=targets[i] a.best+=a.p best=max(best,a.best) for j in range(i+1,n): b=targets[j] if (b.t-a.t)**2<(b.x-a.x)**2+(b.y-a.y)**2: continue b.best=max(b.best,a.best) print(best)
Codeforces Beta Round 30 (Codeforces format)
CF
2,010
2
256
Shooting Gallery
One warm and sunny day king Copa decided to visit the shooting gallery, located at the Central Park, and try to win the main prize — big pink plush panda. The king is not good at shooting, so he invited you to help him. The shooting gallery is an infinite vertical plane with Cartesian coordinate system on it. The targets are points on this plane. Each target is described by it's coordinates xi, and yi, by the time of it's appearance ti and by the number pi, which gives the probability that Copa hits this target if he aims at it. A target appears and disappears instantly, so Copa can hit the target only if at the moment ti his gun sight aimed at (xi, yi). Speed of movement of the gun sight on the plane is equal to 1. Copa knows all the information about the targets beforehand (remember, he is a king!). He wants to play in the optimal way, which maximizes the expected value of the amount of hit targets. He can aim at any target at the moment 0.
The first line contains integer n (1 ≤ n ≤ 1000) — amount of targets in the shooting gallery. Then n lines follow, each describing one target. Each description consists of four numbers xi, yi, ti, pi (where xi, yi, ti — integers, - 1000 ≤ xi, yi ≤ 1000, 0 ≤ ti ≤ 109, real number pi is given with no more than 6 digits after the decimal point, 0 ≤ pi ≤ 1). No two targets may be at the same point.
Output the maximum expected value of the amount of targets that was shot by the king. Your answer will be accepted if it differs from the correct answer by not more than 10 - 6.
null
null
[{"input": "1\n0 0 0 0.5", "output": "0.5000000000"}, {"input": "2\n0 0 0 0.6\n5 0 5 0.7", "output": "1.3000000000"}]
1,800
["dp", "probabilities"]
50
[{"input": "1\r\n0 0 0 0.5\r\n", "output": "0.5000000000\r\n"}, {"input": "2\r\n0 0 0 0.6\r\n5 0 5 0.7\r\n", "output": "1.3000000000\r\n"}, {"input": "1\r\n-5 2 3 0.886986\r\n", "output": "0.8869860000\r\n"}, {"input": "4\r\n10 -7 14 0.926305\r\n-7 -8 12 0.121809\r\n-7 7 14 0.413446\r\n3 -8 6 0.859061\r\n", "output": "1.7853660000\r\n"}, {"input": "5\r\n-2 -2 34 0.127276\r\n5 -5 4 0.459998\r\n10 3 15 0.293766\r\n1 -3 7 0.089869\r\n-4 -7 11 0.772515\r\n", "output": "0.8997910000\r\n"}, {"input": "5\r\n2 5 1 0.955925\r\n9 -9 14 0.299977\r\n0 1 97 0.114582\r\n-4 -2 66 0.561033\r\n0 -10 75 0.135937\r\n", "output": "1.7674770000\r\n"}, {"input": "10\r\n-4 7 39 0.921693\r\n3 -1 50 0.111185\r\n-2 -8 27 0.976475\r\n-9 -2 25 0.541029\r\n6 -4 21 0.526054\r\n-7 2 19 0.488637\r\n-6 -5 50 0.819011\r\n-7 3 39 0.987596\r\n-3 -8 16 0.685997\r\n4 10 1 0.246686\r\n", "output": "3.0829590000\r\n"}]
false
stdio
import sys def main(): input_path = sys.argv[1] output_path = sys.argv[2] submission_path = sys.argv[3] with open(output_path, 'r') as f: correct_line = f.readline().strip() try: correct = float(correct_line) except: print(0) return with open(submission_path, 'r') as f: sub_line = f.readline().strip() try: sub = float(sub_line) except: print(0) return if abs(correct - sub) <= 1e-6 + 1e-12: # Adding small epsilon for floating point errors print(1) else: print(0) if __name__ == "__main__": main()
true
1000/D
1000
D
PyPy 3-64
TESTS
2
77
3,584,000
148239548
def fact(be, en): res, res2 = [1], [1] for i in range(be, en + 1): res.append(mult(res[-1], i)) res2.append(inv(res[-1])) return res, res2 def ncr(n, r): if n < r: return 0 return mult(mult(facs[n], invs[n - r]), invs[r]) mod = 998244353 add = lambda a, b: (a % mod + b % mod) % mod mult = lambda a, b: (a % mod * b % mod) % mod div = lambda a, b: mult(a, inv(b)) inv = lambda a: pow(a, mod - 2, mod) n, a = int(input()), [int(x) for x in input().split()] facs, invs = fact(1, 1000) mem = [0] * n for i in range(n): for j in range(i - 1, -1, -1): if a[j] > 0 and i - j >= a[j]: mem[i] = add(mem[i], ncr(i - j, a[j])) if a[i] > 0: for j in range(i - 1, 0, -1): if a[j] > 0 and i - j >= a[j]: mem[i] = add(mem[i], mult(mem[j - 1], ncr(i - j, a[j]))) else: mem[i] = mem[i - 1] print(mem[-1])
22
124
6,246,400
169757071
N = 10**5 mod = 998244353 fac = [1]*(N+1) finv = [1]*(N+1) for i in range(N): fac[i+1] = fac[i] * (i+1) % mod finv[-1] = pow(fac[-1], mod-2, mod) for i in reversed(range(N)): finv[i] = finv[i+1] * (i+1) % mod def cmb1(n, r): if r <0 or r > n: return 0 r = min(r, n-r) return fac[n] * finv[r] * finv[n-r] % mod n = int(input()) A = list(map(int, input().split())) dp = [0]*(n+1) dp[n] = 1 for i in range(n-1, -1, -1): if A[i] <= 0: continue if i+A[i]+1 > n: continue for j in range(i+A[i]+1, n+1): dp[i] += dp[j]*cmb1(j-i-1, A[i]) dp[i] %= mod ans = 0 for i in range(n): ans += dp[i] ans %= mod print(ans)
Educational Codeforces Round 46 (Rated for Div. 2)
ICPC
2,018
2
256
Yet Another Problem On a Subsequence
The sequence of integers $$$a_1, a_2, \dots, a_k$$$ is called a good array if $$$a_1 = k - 1$$$ and $$$a_1 > 0$$$. For example, the sequences $$$[3, -1, 44, 0], [1, -99]$$$ are good arrays, and the sequences $$$[3, 7, 8], [2, 5, 4, 1], [0]$$$ — are not. A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences $$$[2, -3, 0, 1, 4]$$$, $$$[1, 2, 3, -3, -9, 4]$$$ are good, and the sequences $$$[2, -3, 0, 1]$$$, $$$[1, 2, 3, -3 -9, 4, 1]$$$ — are not. For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.
The first line contains the number $$$n~(1 \le n \le 10^3)$$$ — the length of the initial sequence. The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n~(-10^9 \le a_i \le 10^9)$$$ — the sequence itself.
In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.
null
In the first test case, two good subsequences — $$$[a_1, a_2, a_3]$$$ and $$$[a_2, a_3]$$$. In the second test case, seven good subsequences — $$$[a_1, a_2, a_3, a_4], [a_1, a_2], [a_1, a_3], [a_1, a_4], [a_2, a_3], [a_2, a_4]$$$ and $$$[a_3, a_4]$$$.
[{"input": "3\n2 1 1", "output": "2"}, {"input": "4\n1 1 1 1", "output": "7"}]
1,900
["combinatorics", "dp"]
22
[{"input": "3\r\n2 1 1\r\n", "output": "2\r\n"}, {"input": "4\r\n1 1 1 1\r\n", "output": "7\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
false
stdio
null
true
1000/D
1000
D
PyPy 3-64
TESTS
2
1,388
10,035,200
188233846
def Fact(N): global factorials return factorials[N] def C(N, K): global mod f1, f2, f3 = Fact(N), Fact(K), Fact(N - K) return (f1 * pow((f2 * f3) % mod, mod - 2, mod)) % mod n = int(input()) a = list(map(int, input().split())) mod = 998244353 pref, zero, dp, factorials = [0] * (n + 1), [0] * (n + 1), [0] * (n + 1), [1] * (n + 1) zero[n - 1] = 1 * (not a[n - 1]) factorials[0] = 1 for i in range(1, n + 1): factorials[i] = factorials[i - 1] * i for i in range(n - 2, -1, -1): pref[i], zero[i] = pref[i + 1], zero[i + 1] if a[i] > 0: for j in range(i + a[i], n): dp[i] += ((pref[j + 1] + zero[j + 1] + 1) % mod) * C(j - i - 1, a[i] - 1) dp[i] %= mod elif not a[i]: dp[i] = (zero[i + 1] + pref[i + 1]) % mod zero[i] += 1 pref[i] += dp[i] pref[i] %= mod print(pref[0] % mod)
22
171
4,300,800
193699512
from heapq import heappush, heappop from collections import defaultdict, Counter, deque import threading import sys import bisect input = sys.stdin.readline def ri(): return int(input()) def rs(): return input() def rl(): return list(map(int, input().split())) def rls(): return list(input().split()) # threading.stack_size(10**8) # sys.setrecursionlimit(10**6) class Factorial: def __init__(self, N, mod) -> None: N += 1 self.mod = mod self.f = [1 for _ in range(N)] self.g = [1 for _ in range(N)] for i in range(1, N): self.f[i] = self.f[i - 1] * i % self.mod self.g[-1] = pow(self.f[-1], mod - 2, mod) for i in range(N - 2, -1, -1): self.g[i] = self.g[i + 1] * (i + 1) % self.mod def fac(self, n): return self.f[n] def fac_inv(self, n): return self.g[n] def comb(self, n, m): if n < m: return 0 return self.f[n] * self.g[m] % self.mod * self.g[n - m] % self.mod def perm(self, n, m): return self.f[n] * self.g[n - m] % self.mod def catalan(self, n): return (self.comb(2 * n, n) - self.comb(2 * n, n - 1)) % self.mod def inv(self, n): return self.f[n-1] * self.g[n] % mod def main(): mod = 998244353 n = ri() a = rl() f = Factorial(10**3, mod) dp = [0]*(n+1) dp[-1] = 1 for i in range(n-1, -1, -1): if a[i] > 0 and i+a[i] < n: for j in range(i+a[i]+1, n+1): dp[i] += f.comb(j-i-1, a[i])*dp[j] dp[i] %= mod print((sum(dp)-1) % mod) pass main() # threading.Thread(target=main).start()
Educational Codeforces Round 46 (Rated for Div. 2)
ICPC
2,018
2
256
Yet Another Problem On a Subsequence
The sequence of integers $$$a_1, a_2, \dots, a_k$$$ is called a good array if $$$a_1 = k - 1$$$ and $$$a_1 > 0$$$. For example, the sequences $$$[3, -1, 44, 0], [1, -99]$$$ are good arrays, and the sequences $$$[3, 7, 8], [2, 5, 4, 1], [0]$$$ — are not. A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences $$$[2, -3, 0, 1, 4]$$$, $$$[1, 2, 3, -3, -9, 4]$$$ are good, and the sequences $$$[2, -3, 0, 1]$$$, $$$[1, 2, 3, -3 -9, 4, 1]$$$ — are not. For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.
The first line contains the number $$$n~(1 \le n \le 10^3)$$$ — the length of the initial sequence. The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n~(-10^9 \le a_i \le 10^9)$$$ — the sequence itself.
In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.
null
In the first test case, two good subsequences — $$$[a_1, a_2, a_3]$$$ and $$$[a_2, a_3]$$$. In the second test case, seven good subsequences — $$$[a_1, a_2, a_3, a_4], [a_1, a_2], [a_1, a_3], [a_1, a_4], [a_2, a_3], [a_2, a_4]$$$ and $$$[a_3, a_4]$$$.
[{"input": "3\n2 1 1", "output": "2"}, {"input": "4\n1 1 1 1", "output": "7"}]
1,900
["combinatorics", "dp"]
22
[{"input": "3\r\n2 1 1\r\n", "output": "2\r\n"}, {"input": "4\r\n1 1 1 1\r\n", "output": "7\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "1\r\n1\r\n", "output": "0\r\n"}]
false
stdio
null
true
990/F
990
F
Python 3
TESTS
1
62
0
39187965
def i_ints(): return list(map(int, input().split())) ############# n, = i_ints() s = [0] + i_ints() m, = i_ints() vs = [i_ints() for _ in range(m)] used_pipe = [0] * (n + 1) # 0 is dummy pipe to have x, y be indices into array unseen = set(range(1, n+1)) for i, (x, y) in enumerate(vs): if x in unseen: unseen.discard(x) used_pipe[x] = i if y in unseen: unseen.discard(y) used_pipe[y] = i f = [0] * m order = sorted(range(1, n+1), key=used_pipe.__getitem__, reverse=True) for i in order[:-1]: p = used_pipe[i] x, y = vs[p] if x == i: f[p] = -s[x] s[y] += s[x] else: f[p] = s[y] s[x] += s[y] if s[order[-1]] == 0: print("Possible") for i in f: print(i) else: print("Imposible")
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
786/B
786
B
Python 3
TESTS
3
61
4,915,200
25805460
plans = [] helpl = [] matrix = [] minprice = [] now = 0 res = '' (n,q,se) = map(int,input().split()) for i in range(0,q): s = input().split() helpl = [int(c) for c in s] plans.append([]) for j in range(0,len(helpl)): plans[i].append(helpl[j]) for i in range(0,n): matrix.append([]) for j in range(0,n): matrix[i].append(1000000) #print(matrix[i]) for i in range(0,q): #print(plans[i]) if plans[i][0] == 1: if matrix[plans[i][1]-1][plans[i][2]-1] > plans[i][3] or matrix[plans[i][1]-1][plans[i][2]-1] == 0: matrix[plans[i][1]-1][plans[i][2]-1] = plans[i][3] elif plans[i][0] == 2: for j in range(plans[i][2]-1,plans[i][3]): if matrix[plans[i][1] - 1][j] > plans[i][4] or matrix[plans[i][1] - 1][j] == 0: matrix[plans[i][1] - 1][j] = plans[i][4] elif plans[i][0] == 3: for j in range(plans[i][2]-1,plans[i][3]): if matrix[j][plans[i][1]-1] > plans[i][4] or matrix[j][plans[i][1]-1] == 0: matrix[j][plans[i][1]-1] = plans[i][4] #for i in range(0,n): #print(matrix[i]) d = matrix.copy() for i in range(0,n): d[i][i] = 0 #print(d[i]) for k in range(0,n): for i in range(0,n): for j in range(0,n): d[i][j] = min(d[i][j], d[i][k] + d[k][j]) for i in range(0,n): for j in range(0,n): if d[i][j] == 1000000: d[i][j] = -1 #print(d[i]) for i in range(0,n): res = res + str(d[se-1][i]) + ' ' print(res)
49
1,981
182,374,400
205981387
from sys import stdin input=lambda :stdin.readline()[:-1] from heapq import heappush, heappop INF=10**18 def dijkstra(start,n): dist=[INF]*n hq=[(0,start)] dist[start]=0 seen=[False]*n while hq: w,v=heappop(hq) if dist[v]<w: continue seen[v]=True for to,cost in edge[v]: if seen[to]==False and dist[v]+cost<dist[to]: dist[to]=dist[v]+cost heappush(hq,(dist[to],to)) return dist n,q,s=map(int,input().split()) m=0 while n>(1<<m): m+=1 def gen(l,r): l+=1<<m r+=1<<m res=[] while l<r: if l&1: res.append(l) l+=1 if r&1: r-=1 res.append(r) l>>=1 r>>=1 return res N=1<<m edge=[[] for i in range(4*N)] for _ in range(q): query=list(map(int,input().split())) if query[0]==1: v,u,w=query[1:] query=[2,v,u,u,w] if query[0]==2: v,l,r,w=query[1:] for x in gen(l-1,r): edge[N+v-1].append((x,w)) else: v,l,r,w=query[1:] for x in gen(l-1,r): edge[2*N+x].append((3*N+v-1,w)) for i in range(N): edge[N+i].append((3*N+i,0)) edge[3*N+i].append((N+i,0)) for i in range(1,N): l,r=2*i,2*i+1 edge[i].append((l,0)) edge[i].append((r,0)) edge[2*N+l].append((2*N+i,0)) edge[2*N+r].append((2*N+i,0)) dist=dijkstra(N+s-1,4*N) ans=dist[N:N+n] for i in range(n): if ans[i]==INF: ans[i]=-1 print(*ans)
Codeforces Round 406 (Div. 1)
CF
2,017
2
256
Legacy
Rick and his co-workers have made a new radioactive formula and a lot of bad guys are after them. So Rick wants to give his legacy to Morty before bad guys catch them. There are n planets in their universe numbered from 1 to n. Rick is in planet number s (the earth) and he doesn't know where Morty is. As we all know, Rick owns a portal gun. With this gun he can open one-way portal from a planet he is in to any other planet (including that planet). But there are limits on this gun because he's still using its free trial. By default he can not open any portal by this gun. There are q plans in the website that sells these guns. Every time you purchase a plan you can only use it once but you can purchase it again if you want to use it more. Plans on the website have three types: 1. With a plan of this type you can open a portal from planet v to planet u. 2. With a plan of this type you can open a portal from planet v to any planet with index in range [l, r]. 3. With a plan of this type you can open a portal from any planet with index in range [l, r] to planet v. Rick doesn't known where Morty is, but Unity is going to inform him and he wants to be prepared for when he finds and start his journey immediately. So for each planet (including earth itself) he wants to know the minimum amount of money he needs to get from earth to that planet.
The first line of input contains three integers n, q and s (1 ≤ n, q ≤ 105, 1 ≤ s ≤ n) — number of planets, number of plans and index of earth respectively. The next q lines contain the plans. Each line starts with a number t, type of that plan (1 ≤ t ≤ 3). If t = 1 then it is followed by three integers v, u and w where w is the cost of that plan (1 ≤ v, u ≤ n, 1 ≤ w ≤ 109). Otherwise it is followed by four integers v, l, r and w where w is the cost of that plan (1 ≤ v ≤ n, 1 ≤ l ≤ r ≤ n, 1 ≤ w ≤ 109).
In the first and only line of output print n integers separated by spaces. i-th of them should be minimum money to get from earth to i-th planet, or - 1 if it's impossible to get to that planet.
null
In the first sample testcase, Rick can purchase 4th plan once and then 2nd plan in order to get to get to planet number 2.
[{"input": "3 5 1\n2 3 2 3 17\n2 3 2 2 16\n2 2 2 3 3\n3 3 1 1 12\n1 3 3 17", "output": "0 28 12"}, {"input": "4 3 1\n3 4 1 3 12\n2 2 3 4 10\n1 2 4 16", "output": "0 -1 -1 12"}]
2,300
["data structures", "graphs", "shortest paths"]
49
[{"input": "3 5 1\r\n2 3 2 3 17\r\n2 3 2 2 16\r\n2 2 2 3 3\r\n3 3 1 1 12\r\n1 3 3 17\r\n", "output": "0 28 12 \r\n"}, {"input": "4 3 1\r\n3 4 1 3 12\r\n2 2 3 4 10\r\n1 2 4 16\r\n", "output": "0 -1 -1 12 \r\n"}, {"input": "6 1 5\r\n1 3 6 80612370\r\n", "output": "-1 -1 -1 -1 0 -1 \r\n"}, {"input": "10 8 7\r\n1 10 7 366692903\r\n1 4 8 920363557\r\n2 7 5 10 423509459\r\n2 2 5 7 431247033\r\n2 7 3 5 288617239\r\n2 7 3 3 175870925\r\n3 9 3 8 651538651\r\n3 4 2 5 826387883\r\n", "output": "-1 -1 175870925 288617239 288617239 423509459 0 423509459 423509459 423509459 \r\n"}, {"input": "1 1 1\r\n1 1 1 692142678\r\n", "output": "0 \r\n"}, {"input": "2 4 2\r\n3 2 1 2 227350719\r\n2 2 1 1 111798664\r\n1 2 2 972457508\r\n2 2 2 2 973058334\r\n", "output": "111798664 0 \r\n"}, {"input": "8 8 1\r\n3 7 2 5 267967223\r\n1 6 7 611402069\r\n3 7 2 8 567233748\r\n2 2 1 8 28643141\r\n3 3 3 8 79260103\r\n1 6 8 252844388\r\n2 1 4 4 827261673\r\n3 4 4 5 54569367\r\n", "output": "0 -1 906521776 827261673 -1 -1 1095228896 -1 \r\n"}, {"input": "100000 1 63256\r\n3 15441 33869 86113 433920134\r\n", "output": "-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 -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 -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 -1 -1 -1 -1 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-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 -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 -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 \n"}]
false
stdio
null
true
226/D
226
D
PyPy 3
TESTS
3
278
0
84116366
import math import sys input = sys.stdin.readline n, m = map(int, input().split()) a = [[] for _ in range(n)] for i in range(n): a[i] = [int(_) for _ in input().split()] row = [] col = [] while True: bad = False for i in range(n): total = 0 for j in range(m): total += a[i][j] if total < 0: bad = True row.append(i + 1) for j in range(m): a[i][j] *= -1 for i in range(m): total = 0 for j in range(n): total += a[j][i] if total < 0: bad = True col.append(i + 1) for j in range(n): a[j][i] *= -1 if not bad: break print(len(row), ' '.join(map(str, row))) print(len(col), ' '.join(map(str, col)))
95
372
2,048,000
84116428
import math import sys input = sys.stdin.readline n, m = map(int, input().split()) a = [[] for _ in range(n)] for i in range(n): a[i] = [int(_) for _ in input().split()] row = [] col = [] rowStat = [1] * n colStat = [1] * m while True: bad = False for i in range(n): total = 0 for j in range(m): total += a[i][j] if total < 0: bad = True rowStat[i] *= -1 for j in range(m): a[i][j] *= -1 for i in range(m): total = 0 for j in range(n): total += a[j][i] if total < 0: bad = True colStat[i] *= -1 for j in range(n): a[j][i] *= -1 if not bad: break for i in range(n): if rowStat[i] == -1: row.append(i + 1) for i in range(m): if colStat[i] == -1: col.append(i + 1) print(len(row), ' '.join(map(str, row))) print(len(col), ' '.join(map(str, col)))
Codeforces Round 140 (Div. 1)
CF
2,012
2
256
The table
Harry Potter has a difficult homework. Given a rectangular table, consisting of n × m cells. Each cell of the table contains the integer. Harry knows how to use two spells: the first spell change the sign of the integers in the selected row, the second — in the selected column. Harry's task is to make non-negative the sum of the numbers in each row and each column using these spells. Alone, the boy can not cope. Help the young magician!
The first line contains two integers n and m (1 ≤ n,  m ≤ 100) — the number of rows and the number of columns. Next n lines follow, each contains m integers: j-th integer in the i-th line is ai, j (|ai, j| ≤ 100), the number in the i-th row and j-th column of the table. The rows of the table numbered from 1 to n. The columns of the table numbered from 1 to m.
In the first line print the number a — the number of required applications of the first spell. Next print a space-separated integers — the row numbers, you want to apply a spell. These row numbers must be distinct! In the second line print the number b — the number of required applications of the second spell. Next print b space-separated integers — the column numbers, you want to apply a spell. These column numbers must be distinct! If there are several solutions are allowed to print any of them.
null
null
[{"input": "4 1\n-1\n-1\n-1\n-1", "output": "4 1 2 3 4\n0"}, {"input": "2 4\n-1 -1 -1 2\n1 1 1 1", "output": "1 1\n1 4"}]
2,100
["constructive algorithms", "greedy"]
95
[{"input": "4 1\r\n-1\r\n-1\r\n-1\r\n-1\r\n", "output": "4 1 2 3 4 \r\n0 \r\n"}, {"input": "2 4\r\n-1 -1 -1 2\r\n1 1 1 1\r\n", "output": "1 1 \r\n1 4 \r\n"}, {"input": "10 5\r\n1 7 1 6 -3\r\n8 -8 0 -7 -8\r\n7 -10 -8 -3 6\r\n-3 0 -9 0 -3\r\n-1 5 -2 -9 10\r\n-2 9 2 0 7\r\n5 0 -1 -10 6\r\n7 -8 -3 -9 1\r\n-5 10 -10 5 9\r\n-7 4 -8 0 -4\r\n", "output": "6 2 3 4 7 8 10 \r\n1 1 \r\n"}, {"input": "5 10\r\n-2 -7 -10 -9 5 -9 -3 8 -8 5\r\n3 0 9 8 -4 -3 -8 1 8 1\r\n2 3 7 5 -8 -3 0 -9 -7 -2\r\n-6 -7 0 0 6 9 -8 6 -8 3\r\n7 9 -4 -5 -9 -3 8 6 -5 6\r\n", "output": "2 1 4 \r\n4 5 6 8 10 \r\n"}, {"input": "1 1\r\n-10\r\n", "output": "1 1 \r\n0 \r\n"}, {"input": "1 70\r\n98 66 2 43 -22 -31 29 -19 -42 89 -70 7 -41 33 42 -23 67 -4 23 -67 93 77 83 91 5 94 -12 37 -32 -9 69 24 79 54 40 -2 -25 50 2 -19 65 73 77 2 -34 -64 -43 93 28 86 67 -54 61 88 -3 72 63 38 40 4 98 21 31 -35 -38 84 43 62 50 84\r\n", "output": "0 \r\n22 5 6 8 9 11 13 16 18 20 27 29 30 36 37 40 45 46 47 52 55 64 65 \r\n"}]
false
stdio
null
true
95/C
95
C
PyPy 3-64
TESTS
1
92
0
174810199
''' # Submitted By M7moud Ala3rj Don't Copy This Code, CopyRight . [email protected] © 2022-2023 :) ''' # Problem Name = "Volleyball" # Class: C import sys from heapq import * #sys.setrecursionlimit(2147483647) input = sys.stdin.readline def print(*args, end='\n', sep=' ') -> None: sys.stdout.write(sep.join(map(str, args)) + end) def Solve(): n, m = list(map(int, input().split())) graph = {x:set() for x in range(1, n+1)} x, y = list(map(int, input().split())) for _ in range(m): u, v, w = list(map(int, input().split())) graph[u].add((w, v)) ; graph[v].add((w, u)) taxis = [[0, 0]] * (n+1) for i in range(1, n+1): taxis[i] = list(map(int, input().split())) seen = [0] * (n + 1) pq = [[0, x, 0, []]] while pq: cost, node, ctaxi, tt = heappop(pq) if seen[node]: continue if node == y: print(cost) return seen[node] = 1 for cd, child in graph[node]: childcost = cost ; childctaxi = ctaxi ; childtt = tt if ctaxi-cd<=0 and node not in childtt: cc, dd = taxis[node] childcost+=cc ; childctaxi+=dd ; childtt+=[node] if childctaxi-cd<0: continue heappush(pq, [childcost, child, childctaxi-cd, childtt]) if __name__ == "__main__": Solve()
65
966
28,057,600
144034839
import sys def input(): return sys.stdin.readline()[:-1] N,M = list(map(int,input().split())) e_list = [[] for i in range(N)] X,Y = list(map(int,input().split())) X-=1 Y-=1 for i in range(M): u,v,w = list(map(int,input().split())) u-=1 v-=1 e_list[u].append((v,w)) e_list[v].append((u,w)) DP = [0]*N from heapq import heappop, heappush def dijkstra(vi): for i in range(N): DP[i] = 10**18 DP[vi] = 0 q = [] heappush(q, vi) while q: p = heappop(q) d = p>>10 v = p&((1<<10)-1) if DP[v]<d: continue for v1,d1 in e_list[v]: if d+d1<DP[v1]: DP[v1] = d+d1 heappush(q, ((d+d1)<<10)+v1) new_edges = [[] for i in range(N)] for i in range(N): t,c = list(map(int,input().split())) dijkstra(i) for j,d in enumerate(DP): if i!=j and d<=t: new_edges[i].append((j,c)) e_list = new_edges dijkstra(X) if DP[Y]<10**18: print(DP[Y]) else: print(-1)
Codeforces Beta Round 77 (Div. 1 Only)
CF
2,011
2
256
Volleyball
Petya loves volleyball very much. One day he was running late for a volleyball match. Petya hasn't bought his own car yet, that's why he had to take a taxi. The city has n junctions, some of which are connected by two-way roads. The length of each road is defined by some positive integer number of meters; the roads can have different lengths. Initially each junction has exactly one taxi standing there. The taxi driver from the i-th junction agrees to drive Petya (perhaps through several intermediate junctions) to some other junction if the travel distance is not more than ti meters. Also, the cost of the ride doesn't depend on the distance and is equal to ci bourles. Taxis can't stop in the middle of a road. Each taxi can be used no more than once. Petya can catch taxi only in the junction, where it stands initially. At the moment Petya is located on the junction x and the volleyball stadium is on the junction y. Determine the minimum amount of money Petya will need to drive to the stadium.
The first line contains two integers n and m (1 ≤ n ≤ 1000, 0 ≤ m ≤ 1000). They are the number of junctions and roads in the city correspondingly. The junctions are numbered from 1 to n, inclusive. The next line contains two integers x and y (1 ≤ x, y ≤ n). They are the numbers of the initial and final junctions correspondingly. Next m lines contain the roads' description. Each road is described by a group of three integers ui, vi, wi (1 ≤ ui, vi ≤ n, 1 ≤ wi ≤ 109) — they are the numbers of the junctions connected by the road and the length of the road, correspondingly. The next n lines contain n pairs of integers ti and ci (1 ≤ ti, ci ≤ 109), which describe the taxi driver that waits at the i-th junction — the maximum distance he can drive and the drive's cost. The road can't connect the junction with itself, but between a pair of junctions there can be more than one road. All consecutive numbers in each line are separated by exactly one space character.
If taxis can't drive Petya to the destination point, print "-1" (without the quotes). Otherwise, print the drive's minimum cost. Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
null
An optimal way — ride from the junction 1 to 2 (via junction 4), then from 2 to 3. It costs 7+2=9 bourles.
[{"input": "4 4\n1 3\n1 2 3\n1 4 1\n2 4 1\n2 3 5\n2 7\n7 2\n1 2\n7 7", "output": "9"}]
1,900
["shortest paths"]
65
[{"input": "4 4\r\n1 3\r\n1 2 3\r\n1 4 1\r\n2 4 1\r\n2 3 5\r\n2 7\r\n7 2\r\n1 2\r\n7 7\r\n", "output": "9\r\n"}, {"input": "3 3\r\n1 3\r\n1 2 2\r\n1 3 3\r\n3 2 1\r\n2 7\r\n2 7\r\n3 6\r\n", "output": "14\r\n"}, {"input": "3 1\r\n1 3\r\n1 2 2\r\n2 7\r\n2 7\r\n3 6\r\n", "output": "-1\r\n"}, {"input": "3 2\r\n3 3\r\n1 2 2\r\n1 3 3\r\n2 7\r\n2 7\r\n3 7\r\n", "output": "0\r\n"}, {"input": "2 2\r\n1 2\r\n1 2 3\r\n1 2 2\r\n2 7\r\n3 2\r\n", "output": "7\r\n"}, {"input": "1 0\r\n1 1\r\n74 47\r\n", "output": "0\r\n"}, {"input": "5 5\r\n1 3\r\n1 3 3\r\n5 1 6\r\n4 3 8\r\n1 3 3\r\n5 2 4\r\n1 2\r\n4 1\r\n2 5\r\n5 2\r\n1 2\r\n", "output": "-1\r\n"}, {"input": "7 4\r\n3 4\r\n6 2 7\r\n6 1 4\r\n4 3 5\r\n3 6 4\r\n1 6\r\n7 3\r\n3 6\r\n6 5\r\n3 7\r\n4 4\r\n1 4\r\n", "output": "-1\r\n"}, {"input": "5 5\r\n4 5\r\n1 4 8\r\n4 2 4\r\n4 1 3\r\n3 1 9\r\n4 5 4\r\n2 7\r\n2 7\r\n5 1\r\n6 3\r\n3 2\r\n", "output": "3\r\n"}, {"input": "5 5\r\n5 4\r\n2 4 10\r\n2 4 7\r\n3 1 7\r\n2 4 2\r\n5 3 9\r\n6 17\r\n2 4\r\n3 12\r\n7 18\r\n2 5\r\n", "output": "-1\r\n"}, {"input": "4 7\r\n3 4\r\n2 3 5\r\n3 2 9\r\n4 1 9\r\n3 2 1\r\n3 1 2\r\n2 3 6\r\n1 2 8\r\n2 2\r\n5 3\r\n2 1\r\n1 5\r\n", "output": "-1\r\n"}, {"input": "7 14\r\n7 5\r\n1 3 15\r\n2 1 10\r\n1 3 5\r\n2 4 9\r\n5 4 19\r\n1 3 11\r\n5 1 1\r\n2 4 5\r\n2 3 11\r\n3 2 10\r\n3 4 18\r\n5 1 18\r\n6 2 5\r\n5 6 2\r\n3 6\r\n6 7\r\n9 1\r\n3 6\r\n1 1\r\n9 4\r\n9 8\r\n", "output": "-1\r\n"}, {"input": "7 15\r\n5 5\r\n3 4 6\r\n7 4 3\r\n7 2 8\r\n2 5 2\r\n7 2 8\r\n5 2 9\r\n3 1 7\r\n1 2 4\r\n7 1 8\r\n7 5 7\r\n2 4 2\r\n4 3 9\r\n7 4 2\r\n5 4 8\r\n7 2 8\r\n15 4\r\n18 18\r\n6 8\r\n16 5\r\n11 1\r\n5 3\r\n18 4\r\n", "output": "0\r\n"}, {"input": "8 20\r\n8 4\r\n6 3 1\r\n3 4 4\r\n5 2 2\r\n3 6 3\r\n5 8 7\r\n6 2 7\r\n8 6 4\r\n6 4 5\r\n4 2 5\r\n5 3 3\r\n5 7 3\r\n8 1 6\r\n2 4 3\r\n6 8 5\r\n1 8 6\r\n8 2 7\r\n8 2 3\r\n1 6 7\r\n8 7 3\r\n6 3 6\r\n2 2\r\n5 9\r\n1 9\r\n4 7\r\n1 8\r\n4 8\r\n9 7\r\n9 3\r\n", "output": "3\r\n"}, {"input": "8 20\r\n8 2\r\n1 7 5\r\n3 2 3\r\n2 7 6\r\n6 5 6\r\n4 8 5\r\n7 8 4\r\n1 6 2\r\n7 4 3\r\n4 3 1\r\n6 7 5\r\n4 2 4\r\n2 8 7\r\n6 2 2\r\n2 3 4\r\n3 7 3\r\n7 8 4\r\n5 4 2\r\n7 1 1\r\n5 7 3\r\n4 3 7\r\n4 4\r\n2 7\r\n3 5\r\n3 1\r\n3 5\r\n1 5\r\n11 4\r\n10 5\r\n", "output": "5\r\n"}, {"input": "9 20\r\n5 1\r\n8 9 3\r\n1 8 6\r\n5 6 3\r\n2 1 4\r\n7 1 6\r\n1 4 4\r\n3 2 4\r\n5 6 4\r\n3 9 6\r\n6 2 3\r\n9 1 7\r\n1 7 1\r\n1 3 3\r\n8 4 7\r\n7 1 7\r\n6 9 3\r\n5 8 3\r\n9 4 5\r\n6 9 1\r\n6 2 6\r\n1 7\r\n1 3\r\n6 1\r\n1 2\r\n6 1\r\n2 2\r\n4 7\r\n4 5\r\n4 1\r\n", "output": "-1\r\n"}, {"input": "10 21\r\n9 5\r\n5 2 6\r\n1 9 7\r\n6 2 7\r\n8 10 2\r\n7 2 1\r\n6 9 6\r\n10 9 4\r\n2 10 2\r\n10 8 4\r\n10 1 7\r\n9 1 7\r\n1 8 5\r\n10 9 7\r\n7 5 3\r\n2 10 6\r\n4 7 3\r\n10 5 6\r\n5 10 4\r\n6 9 2\r\n2 3 6\r\n1 9 3\r\n10 6\r\n7 12\r\n13 3\r\n17 4\r\n18 17\r\n1 9\r\n16 16\r\n12 13\r\n1 10\r\n6 15\r\n", "output": "-1\r\n"}, {"input": "14 20\r\n7 2\r\n12 2 17\r\n13 3 8\r\n6 8 3\r\n14 4 16\r\n13 5 17\r\n7 14 7\r\n11 10 6\r\n12 4 16\r\n6 11 7\r\n2 13 12\r\n13 2 18\r\n2 10 12\r\n13 12 1\r\n12 5 4\r\n9 4 16\r\n7 6 7\r\n2 3 15\r\n4 14 1\r\n13 5 3\r\n10 9 3\r\n4 6\r\n4 5\r\n5 3\r\n2 6\r\n2 4\r\n1 2\r\n4 7\r\n2 2\r\n6 1\r\n1 1\r\n6 5\r\n7 7\r\n3 4\r\n2 6\r\n", "output": "-1\r\n"}, {"input": "5 0\r\n2 4\r\n1 2\r\n6 9\r\n4 585\r\n6 9\r\n7 4\r\n", "output": "-1\r\n"}, {"input": "5 8\r\n1 3\r\n1 2 1\r\n1 3 3\r\n1 5 1\r\n1 4 2\r\n5 4 3\r\n3 5 2\r\n2 3 8\r\n1 3 2\r\n2 7\r\n10 3\r\n4 7\r\n2 1\r\n1 1\r\n", "output": "7\r\n"}, {"input": "5 7\r\n1 3\r\n1 2 1\r\n1 3 3\r\n1 5 1\r\n1 4 2\r\n5 4 3\r\n3 5 2\r\n2 3 8\r\n2 7\r\n10 3\r\n4 7\r\n2 1\r\n1 1\r\n", "output": "10\r\n"}, {"input": "7 5\r\n6 7\r\n1 3 1000000000\r\n2 7 999999999\r\n5 7 123456789\r\n1 5 148879589\r\n5 4 1000000000\r\n1000000000 1000000000\r\n1000000000 1000000000\r\n999999999 145785965\r\n1000000000 1000000000\r\n1000000000 1\r\n123456789 123568591\r\n1000000000 1000000000\r\n", "output": "-1\r\n"}, {"input": "2 1\r\n1 2\r\n1 2 1\r\n1 999999998\r\n1 999999998\r\n", "output": "999999998\r\n"}, {"input": "3 3\r\n1 3\r\n2 1 1\r\n1 2 1000000000\r\n1 2 1000000000\r\n1000000000 1000000000\r\n1000000000 1000000000\r\n1000000000 1000000000\r\n", "output": "-1\r\n"}, {"input": "3 3\r\n1 2\r\n2 1 1\r\n1 2 1000000000\r\n1 2 1000000000\r\n10000000 1000000000\r\n10000000 1000000000\r\n10000000 1000000000\r\n", "output": "1000000000\r\n"}, {"input": "3 3\r\n1 2\r\n2 1 1000000000\r\n1 2 1000000000\r\n1 2 1000000000\r\n10000000 1000000000\r\n10000000 1000000000\r\n10000000 1000000000\r\n", "output": "-1\r\n"}]
false
stdio
null
true
272/D
272
D
PyPy 3
TESTS
2
184
0
117478825
n=int(input()) a=sorted(list(map(int,input().split()))+list(map(int,input().split()))) m=int(input()) b=[] c=1 for i in range(1,n+1): c*=i c%=m b.append(c) d=0 e=1 for i in range(n): if a[i]!=a[i-1]: e*=b[i-d-1] e%=m d=i e*=b[n-d-1] print(e%m)
51
872
81,203,200
190272675
import abc import itertools import math from math import gcd as gcd import sys import queue import itertools from heapq import heappop, heappush import random def solve(): n = int(sys.stdin.readline()) a = list(map(int, sys.stdin.readline().split())) b = list(map(int, sys.stdin.readline().split())) m = int(sys.stdin.readline()) c = {} def fill(a): for i in range(n): if a[i] not in c: c[a[i]] = {i: 1} else: if i not in c[a[i]]: c[a[i]][i] = 1 else: c[a[i]][i] += 1 fill(a) fill(b) facts = [1 for i in range(2 * 10 ** 5 + 10)] d2 = [0 for i in range(2 * 10 ** 5 + 10)] for i in range(1, 2 * 10 ** 5 + 10): r = 0 k = i while k % 2 == 0: k //= 2 r += 1 d2[i] = d2[i - 1] + r facts[i] = (facts[i - 1] * k) % m res = 1 e = 0 for key in c: dd = c[key] rem = sum(list(dd.values())) e += d2[rem] res = (res * facts[rem]) % m for el in dd: e -= dd[el] - 1 #print(e) print((res * pow(2, e, m)) % m) if __name__ == '__main__': multi_test = 0 if multi_test == 1: t = int(sys.stdin.readline()) for _ in range(t): solve() else: solve()
Codeforces Round 167 (Div. 2)
CF
2,013
2
256
Dima and Two Sequences
Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n). Now Dima wants to count the number of distinct sequences of points of length 2·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence. Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2·n, q2·n) and (x1, y1), (x2, y2), ..., (x2·n, y2·n) distinct, if there is such i (1 ≤ i ≤ 2·n), that (pi, qi) ≠ (xi, yi). As the answer can be rather large, print the remainder from dividing the answer by number m.
The first line contains integer n (1 ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 109). The numbers in the lines are separated by spaces. The last line contains integer m (2 ≤ m ≤ 109 + 7).
In the single line print the remainder after dividing the answer to the problem by number m.
null
In the first sample you can get only one sequence: (1, 1), (2, 1). In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
[{"input": "1\n1\n2\n7", "output": "1"}, {"input": "2\n1 2\n2 3\n11", "output": "2"}]
1,600
["combinatorics", "math", "sortings"]
51
[{"input": "1\r\n1\r\n2\r\n7\r\n", "output": "1\r\n"}, {"input": "2\r\n1 2\r\n2 3\r\n11\r\n", "output": "2\r\n"}, {"input": "2\r\n1 2\r\n1 2\r\n4\r\n", "output": "1\r\n"}, {"input": "4\r\n1 2 3 4\r\n4 3 2 1\r\n1009\r\n", "output": "16\r\n"}, {"input": "5\r\n1 2 3 3 5\r\n1 2 3 5 3\r\n12\r\n", "output": "0\r\n"}, {"input": "1\r\n1000000000\r\n1000000000\r\n2\r\n", "output": "1\r\n"}, {"input": "2\r\n1 2\r\n2 2\r\n4\r\n", "output": "3\r\n"}]
false
stdio
null
true
272/D
272
D
Python 3
TESTS
2
154
0
32560588
f = lambda: map(int, input().split()) n = int(input()) c, d = {}, {} for x, y in zip(f(), f()): c[x] = c.get(x, 0) + 1 c[y] = c.get(y, 0) + 1 if x == y: d[x] = d.get(x, 0) + 2 s, m = 1, int(input()) for k, v in c.items(): u = d.get(k, 0) for i in range(v - u, v, 2): s = s * (i * i + i) // 2 % m for i in range(v - u): s = s * (i + 1) % m print(s)
51
902
20,070,400
121344663
n = input() c_n = {} d_n = {} for a_i, b_i in zip(input().split(), input().split()): if a_i == b_i: d_n[a_i] = d_n.get(a_i,0) + 2 c_n[a_i] = c_n.get(a_i,0) + 1 c_n[b_i] = c_n.get(b_i,0) + 1 result = 1 k = int(input()) for a_i, cant in c_n.items(): cant_rep = d_n.get(a_i,0) for i in range(cant - cant_rep + 1, cant, 2): result = (result * i * (i + 1) // 2) % k for i in range(1, cant - cant_rep + 1): result = (result * i) % k print(result)
Codeforces Round 167 (Div. 2)
CF
2,013
2
256
Dima and Two Sequences
Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n). Now Dima wants to count the number of distinct sequences of points of length 2·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence. Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2·n, q2·n) and (x1, y1), (x2, y2), ..., (x2·n, y2·n) distinct, if there is such i (1 ≤ i ≤ 2·n), that (pi, qi) ≠ (xi, yi). As the answer can be rather large, print the remainder from dividing the answer by number m.
The first line contains integer n (1 ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 109). The numbers in the lines are separated by spaces. The last line contains integer m (2 ≤ m ≤ 109 + 7).
In the single line print the remainder after dividing the answer to the problem by number m.
null
In the first sample you can get only one sequence: (1, 1), (2, 1). In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
[{"input": "1\n1\n2\n7", "output": "1"}, {"input": "2\n1 2\n2 3\n11", "output": "2"}]
1,600
["combinatorics", "math", "sortings"]
51
[{"input": "1\r\n1\r\n2\r\n7\r\n", "output": "1\r\n"}, {"input": "2\r\n1 2\r\n2 3\r\n11\r\n", "output": "2\r\n"}, {"input": "2\r\n1 2\r\n1 2\r\n4\r\n", "output": "1\r\n"}, {"input": "4\r\n1 2 3 4\r\n4 3 2 1\r\n1009\r\n", "output": "16\r\n"}, {"input": "5\r\n1 2 3 3 5\r\n1 2 3 5 3\r\n12\r\n", "output": "0\r\n"}, {"input": "1\r\n1000000000\r\n1000000000\r\n2\r\n", "output": "1\r\n"}, {"input": "2\r\n1 2\r\n2 2\r\n4\r\n", "output": "3\r\n"}]
false
stdio
null
true
383/E
383
E
Python 3
TESTS
1
124
409,600
45697746
import collections, functools M = 10 n = int(input()) f = collections.defaultdict(int) for _ in range(n): f[functools.reduce(lambda x, y: x | y, [1<<(ord(x)-ord('a')) for x in input()])] += 1 ans = 0 for i in range(M): for mask in range(1<<M): if(mask & (1<<i)): f[mask] += f[mask^(1<<i)]; for mask in range(1<<M): ans ^= (n-f[mask])*(n-f[mask]) print(ans)
68
3,026
69,324,800
155280550
import sys readline=sys.stdin.readline N=int(readline()) dp=[0]*(1<<24) dp[-1]=N for _ in range(N): bit=(1<<24)-1 for a in readline().rstrip(): a=ord(a)-97 if bit&1<<a: bit^=1<<a dp[bit]-=1 for i in range(24): for bit in range((1<<24)-1,-1,-1): if not bit&1<<i: dp[bit]+=dp[bit^1<<i] ans=0 for x in dp: ans^=x**2 print(ans)
Codeforces Round 225 (Div. 1)
CF
2,014
4
256
Vowels
Iahubina is tired of so many complicated languages, so she decided to invent a new, simple language. She already made a dictionary consisting of n 3-words. A 3-word is a sequence of exactly 3 lowercase letters of the first 24 letters of the English alphabet (a to x). She decided that some of the letters are vowels, and all the others are consonants. The whole language is based on a simple rule: any word that contains at least one vowel is correct. Iahubina forgot which letters are the vowels, and wants to find some possible correct sets of vowels. She asks Iahub questions. In each question, she will give Iahub a set of letters considered vowels (in this question). For each question she wants to know how many words of the dictionary are correct, considering the given set of vowels. Iahubina wants to know the xor of the squared answers to all the possible questions. There are 224 different questions, they are all subsets of the set of the first 24 letters of the English alphabet. Help Iahub find that number.
The first line contains one integer, n (1 ≤ n ≤ 104). Each of the next n lines contains a 3-word consisting of 3 lowercase letters. There will be no two identical 3-words.
Print one number, the xor of the squared answers to the queries.
null
null
[{"input": "5\nabc\naaa\nada\nbcd\ndef", "output": "0"}]
2,700
["combinatorics", "divide and conquer", "dp"]
68
[{"input": "5\r\nabc\r\naaa\r\nada\r\nbcd\r\ndef\r\n", "output": "0\r\n"}, {"input": "100\r\namd\r\namj\r\natr\r\nbcp\r\nbjm\r\ncna\r\ncpj\r\ncse\r\ndij\r\ndjp\r\ndlv\r\nebk\r\nedf\r\nelw\r\nfbr\r\nfcl\r\nfhs\r\nflo\r\nfmj\r\ngcg\r\ngen\r\nghg\r\ngvb\r\ngxx\r\nhbe\r\nhbf\r\nhgu\r\nhlv\r\nhqa\r\nibg\r\nifp\r\nima\r\nitt\r\nivl\r\nixu\r\njle\r\njli\r\nket\r\nkit\r\nkws\r\nlep\r\nles\r\nleu\r\nmbp\r\nmci\r\nmdv\r\nmhf\r\nmih\r\nmll\r\nmop\r\nndp\r\nnfs\r\nngl\r\nnng\r\noic\r\nomo\r\nooj\r\noti\r\npax\r\npfo\r\npjd\r\npup\r\nqer\r\nrad\r\nrdg\r\nrfq\r\nrvt\r\nrwa\r\nrxj\r\nshc\r\nsjv\r\nswx\r\ntcu\r\ntlm\r\ntmb\r\ntml\r\ntmw\r\ntvr\r\ntvx\r\nuid\r\nuir\r\nukf\r\nulg\r\nvce\r\nves\r\nvfb\r\nvok\r\nvut\r\nvvi\r\nvwb\r\nwab\r\nwba\r\nwdf\r\nweq\r\nwog\r\nwsl\r\nxbk\r\nxiq\r\nxop\r\nxpp\r\n", "output": "13888\r\n"}, {"input": 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false
stdio
null
true
526/B
526
B
Python 3
PRETESTS
0
46
0
10582995
# fin = open("input.txt") # n = int(fin.readline()) # A = [0] + list(map(int, fin.readline().split())) n = int(input()) A = list(map(int, input().split())) C = 0 for i in range(n - 1, -1, -1): C += abs(A[i * 2] - A[i * 2 + 1]) A[i] += max(A[i * 2], A[i * 2 + 1]) print(C)
38
61
0
10586802
from math import floor def main(): n = int(input()) a = list(map(int, input().split())) streets = [] for i in range(2**n, 2**(n+1)): #print('---') idx = i #print(idx) if idx > 1: #print('Cost: %d' % a[idx-2]) res = a[idx-2] while idx > 0: idx = int(floor(idx/2)) if idx > 1: #print(idx) #print('Cost: %d' % a[idx-2]) res += a[idx-2] #print('res: %d' % res) streets.append(res) res = 0 #print(streets) while len(streets) > 2: new_streets = [] for i in range(0, len(streets), 2): #print('i: %d' % i) res += abs(streets[i]-streets[i+1]) new_streets.append(max(streets[i], streets[i+1])) #print(new_streets, cur_diff) streets = new_streets print(res+abs(streets[0]-streets[1])) if __name__ == '__main__': main()
ZeptoLab Code Rush 2015
CF
2,015
1
256
Om Nom and Dark Park
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him. The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square $$\frac{i}{2}$$. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads. Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe. He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit. The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and $$\frac{i}{2}$$. All numbers ai are positive integers, not exceeding 100.
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
null
Picture for the sample test. Green color denotes the additional street lights.
[{"input": "2\n1 2 3 4 5 6", "output": "5"}]
1,400
["dfs and similar", "greedy", "implementation"]
38
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false
stdio
null
true
832/D
832
D
PyPy 3
TESTS
3
61
409,600
149775433
import sys input = sys.stdin.readline from collections import deque def solve(n, q, graph, ABC): for a, b, c in ABC: a, b, c = sorted([a, b, c]) if a == b == c: print(1) else: q = deque([(a, 0)]) vis = {a} ab, ac = -1, -1 while q: node, dis = q.popleft() if node == b: ab = dis+1 if node == c: ac = dis+1 for nei in graph[node]: if nei in vis: continue q.append((nei, dis+1)) vis.add(nei) q = deque([(b, 0)]) vis = {b} bc = -1 while q: node, dis = q.popleft() if node == c: bc = dis+1 break for nei in graph[node]: if nei in vis: continue vis.add(nei) q.append((nei, dis+1)) print(sorted([ab, ac, bc])[1]) n, q = map(int, input().split()) P = list(map(int, input().split())) graph = [[] for _ in range(n+1)] for i in range(2, n+1): graph[i].append(P[i-2]) graph[P[i-2]].append(i) ABC = [tuple(map(int, input().split())) for _ in range(q)] solve(n, q, graph, ABC)
94
1,325
40,550,400
110042385
# by the authority of GOD author: manhar singh sachdev # import os,sys from io import BytesIO, IOBase def pre(n,path,lim): # lim = n.bit_length() up = [[-1]*(lim+1) for _ in range(n)] st,visi,height = [0],[1]+[0]*(n-1),[0]*n start,finish,time = [0]*n,[0]*n,0 while len(st): x = st[-1] y,j = up[x][0],0 while y != -1: up[x][j] = y y = up[y][j] j += 1 while len(path[x]) and visi[path[x][-1]]: path[x].pop() if not len(path[x]): time += 1 finish[x] = time st.pop() else: i = path[x].pop() st.append(i) time += 1 visi[i],start[i],up[i][0],height[i] = 1,time,x,height[x]+1 return start,finish,up,height def is_ancestor(u,v,start,finish): return start[u] <= start[v] and finish[u] >= finish[v] def lca(u,v,up,start,finish,lim): if is_ancestor(u,v,start,finish): return u if is_ancestor(v,u,start,finish): return v for i in range(lim,-1,-1): if up[u][i] != -1 and not is_ancestor(up[u][i],v,start,finish): u = up[u][i] return up[u][0] def solve(s,f,t,up,start,finish,lim,height): a = lca(s,f,up,start,finish,lim) b = lca(t,f,up,start,finish,lim) ans = height[f]-max(height[a],height[b])+1 if a == b: x = lca(s,t,up,start,finish,lim) ans += height[x]-height[a] return ans def main(): n,q = map(int,input().split()) path = [[] for _ in range(n)] for ind,i in enumerate(map(int,input().split())): path[ind+1].append(i-1) path[i-1].append(ind+1) lim = n.bit_length() start,finish,up,height = pre(n,path,lim) for _ in range(q): a,b,c = map(lambda xx:int(xx)-1,input().split()) print(max(solve(a,b,c,up,start,finish,lim,height), solve(a,c,b,up,start,finish,lim,height), solve(b,a,c,up,start,finish,lim,height))) # Fast IO Region BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") if __name__ == "__main__": main()
Codeforces Round 425 (Div. 2)
CF
2,017
2
256
Misha, Grisha and Underground
Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other. The boys decided to have fun and came up with a plan. Namely, in some day in the morning Misha will ride the underground from station s to station f by the shortest path, and will draw with aerosol an ugly text "Misha was here" on every station he will pass through (including s and f). After that on the same day at evening Grisha will ride from station t to station f by the shortest path and will count stations with Misha's text. After that at night the underground workers will wash the texts out, because the underground should be clean. The boys have already chosen three stations a, b and c for each of several following days, one of them should be station s on that day, another should be station f, and the remaining should be station t. They became interested how they should choose these stations s, f, t so that the number Grisha will count is as large as possible. They asked you for help.
The first line contains two integers n and q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of stations and the number of days. The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi ≤ n). The integer pi means that there is a route between stations pi and i. It is guaranteed that it's possible to reach every station from any other. The next q lines contains three integers a, b and c each (1 ≤ a, b, c ≤ n) — the ids of stations chosen by boys for some day. Note that some of these ids could be same.
Print q lines. In the i-th of these lines print the maximum possible number Grisha can get counting when the stations s, t and f are chosen optimally from the three stations on the i-th day.
null
In the first example on the first day if s = 1, f = 2, t = 3, Misha would go on the route 1 $$\rightarrow$$ 2, and Grisha would go on the route 3 $$\rightarrow$$ 1 $$\rightarrow$$ 2. He would see the text at the stations 1 and 2. On the second day, if s = 3, f = 2, t = 3, both boys would go on the route 3 $$\rightarrow$$ 1 $$\rightarrow$$ 2. Grisha would see the text at 3 stations. In the second examle if s = 1, f = 3, t = 2, Misha would go on the route 1 $$\rightarrow$$ 2 $$\rightarrow$$ 3, and Grisha would go on the route 2 $$\rightarrow$$ 3 and would see the text at both stations.
[{"input": "3 2\n1 1\n1 2 3\n2 3 3", "output": "2\n3"}, {"input": "4 1\n1 2 3\n1 2 3", "output": "2"}]
1,900
["dfs and similar", "graphs", "trees"]
94
[{"input": "3 2\r\n1 1\r\n1 2 3\r\n2 3 3\r\n", "output": "2\r\n3\r\n"}, {"input": "4 1\r\n1 2 3\r\n1 2 3\r\n", "output": "2\r\n"}, {"input": "2 4\r\n1\r\n1 1 1\r\n1 1 2\r\n1 2 2\r\n2 2 2\r\n", "output": "1\r\n2\r\n2\r\n1\r\n"}, {"input": "5 20\r\n4 1 1 4\r\n2 2 5\r\n3 2 5\r\n2 3 4\r\n4 2 5\r\n4 1 2\r\n5 3 1\r\n2 1 2\r\n4 3 2\r\n1 3 3\r\n4 2 5\r\n5 1 4\r\n4 5 4\r\n1 2 4\r\n3 3 1\r\n5 4 5\r\n1 1 1\r\n1 4 4\r\n5 3 2\r\n4 2 1\r\n3 1 4\r\n", "output": "3\r\n3\r\n3\r\n2\r\n2\r\n3\r\n3\r\n3\r\n2\r\n2\r\n2\r\n2\r\n2\r\n2\r\n2\r\n1\r\n2\r\n3\r\n2\r\n2\r\n"}, {"input": "5 20\r\n5 5 1 4\r\n1 4 3\r\n2 4 1\r\n1 5 5\r\n5 1 4\r\n5 1 5\r\n1 5 5\r\n5 4 4\r\n2 3 3\r\n4 4 1\r\n1 4 1\r\n4 5 4\r\n1 4 5\r\n4 1 5\r\n2 4 2\r\n4 3 3\r\n2 5 5\r\n1 5 4\r\n3 3 4\r\n5 5 1\r\n3 4 1\r\n", "output": "3\r\n3\r\n3\r\n2\r\n3\r\n3\r\n2\r\n3\r\n2\r\n2\r\n2\r\n2\r\n2\r\n3\r\n3\r\n2\r\n2\r\n3\r\n3\r\n3\r\n"}]
false
stdio
null
true
379/C
379
C
PyPy 3-64
TESTS
5
670
74,444,800
213215323
from collections import defaultdict as ded d = ded(lambda:None) n = int(input()) a = input().split() for i in a: if d[i]==None: print(i,end= " ") d[i]=int(i)+1 else: j =d[i] while d[str(j)]: j+=1 print(j,end=" ") d[i]=j+1
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
340/E
340
E
PyPy 3-64
TESTS
1
92
0
200984693
def main(): n, *a = map(int, open(0).read().split()) ok = [True] * (n + 1) ok[0] = False for i, x in enumerate(a, 1): if x != -1: ok[i] = False ok[x] = False n = sum(ok) mod = 10 ** 9 + 7 m = Mod(n, mod) ans = m.fact[n] for i in range(1, n + 1): x = m.comb(n, i) * m.fact[n - i] % mod if i & 1: x = -x ans += x ans %= mod print(ans) class Mod: def __init__(self, n, mod): self.mod = mod self.fact = [0] * (n + 1) self.fact_inv = [0] * (n + 1) self.inv = [0] * (n + 1) self.fact[0] = 1 for i in range(1, n + 1): self.fact[i] = self.fact[i - 1] * i % mod self.fact_inv[n] = inv(self.fact[n], self.mod) for i in range(n - 1, -1, -1): self.fact_inv[i] = self.fact_inv[i + 1] * (i + 1) % mod for i in range(1, n + 1): self.inv[i] = self.fact_inv[i] * self.fact[i - 1] % mod def comb(self, n, r): return self.fact[n] * self.fact_inv[r] % self.mod * self.fact_inv[ n - r] % self.mod def inv(a, mod): b = mod s = 1 t = 0 while 0 < b: q = a // b r = a - b * q a = b b = r r = s s = t t = r - t * q return s + (-s + mod + 1) // mod * mod if __name__ == '__main__': main()
18
218
512,000
102160229
MOD = 10**9+7 def solve(values): notUsed = set(range(1, len(values)+1)) chairs = set() for i, a in enumerate(values, 1): if a == -1: chairs.add(i) else: notUsed -= {a} fixed = len(chairs & notUsed) m = len(notUsed) fact, fact_inv = facts(m) ans = fact[m] for k in range(1, fixed+1): ans += nCr(fixed, k, fact, fact_inv)*fact[m-k]*(-1)**k return ans % MOD def facts(m): fact = [0]*(m+1) fact[0] = 1 for i in range(1, m+1): fact[i] = fact[i-1]*i % MOD fact_inv = [0]*(m+1) fact_inv[m] = pow(fact[m], MOD-2, MOD) for i in reversed(range(m)): fact_inv[i] = fact_inv[i + 1]*(i + 1) % MOD return fact, fact_inv def nCr(n, r, fact, fact_inv): if r < 0 or n < r: return 0 return int(fact[n] * fact_inv[r] * fact_inv[n-r]) if __name__ == '__main__': input() values = list(map(int, input().split())) print(solve(values))
Codeforces Round 198 (Div. 2)
CF
2,013
1
256
Iahub and Permutations
Iahub is so happy about inventing bubble sort graphs that he's staying all day long at the office and writing permutations. Iahubina is angry that she is no more important for Iahub. When Iahub goes away, Iahubina comes to his office and sabotage his research work. The girl finds an important permutation for the research. The permutation contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n). She replaces some of permutation elements with -1 value as a revenge. When Iahub finds out his important permutation is broken, he tries to recover it. The only thing he remembers about the permutation is it didn't have any fixed point. A fixed point for a permutation is an element ak which has value equal to k (ak = k). Your job is to proof to Iahub that trying to recover it is not a good idea. Output the number of permutations which could be originally Iahub's important permutation, modulo 1000000007 (109 + 7).
The first line contains integer n (2 ≤ n ≤ 2000). On the second line, there are n integers, representing Iahub's important permutation after Iahubina replaces some values with -1. It's guaranteed that there are no fixed points in the given permutation. Also, the given sequence contains at least two numbers -1 and each positive number occurs in the sequence at most once. It's guaranteed that there is at least one suitable permutation.
Output a single integer, the number of ways Iahub could recover his permutation, modulo 1000000007 (109 + 7).
null
For the first test example there are two permutations with no fixed points are [2, 5, 4, 3, 1] and [5, 1, 4, 3, 2]. Any other permutation would have at least one fixed point.
[{"input": "5\n-1 -1 4 3 -1", "output": "2"}]
2,000
["combinatorics", "math"]
18
[{"input": "5\r\n-1 -1 4 3 -1\r\n", "output": "2\r\n"}, {"input": "8\r\n2 4 5 3 -1 8 -1 6\r\n", "output": "1\r\n"}, {"input": "7\r\n-1 -1 4 -1 7 1 6\r\n", "output": "4\r\n"}, {"input": "6\r\n-1 -1 -1 -1 -1 -1\r\n", "output": "265\r\n"}, {"input": "2\r\n-1 -1\r\n", "output": "1\r\n"}, {"input": "10\r\n4 10 -1 1 6 8 9 2 -1 -1\r\n", "output": "4\r\n"}, {"input": "20\r\n-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1\r\n", "output": "927799753\r\n"}]
false
stdio
null
true
340/E
340
E
Python 3
TESTS
2
124
7,065,600
36718163
from math import factorial as f def der(n): s=0 for i in range(2,n+1): s+=((-1)**i)/f(i) return f(n)*s x=input() l=[i for i in input().split()] print(int(der(l.count("-1"))%(1000000007)))
18
248
7,168,000
85499531
MOD = 10**9+7 n = int(input()) notUsed = set(range(1, n+1)) chairs = set() for i, a in enumerate(map(int, input().split()), 1): if a == -1: chairs.add(i) else: notUsed -= {a} fixed = len(chairs & notUsed) m = len(notUsed) U = m fact = [0]*(U+1) fact[0] = 1 for i in range(1, U+1): fact[i] = fact[i-1]*i % MOD invfact = [0]*(U+1) invfact[U] = pow(fact[U], MOD-2, MOD) for i in reversed(range(U)): invfact[i] = invfact[i+1]*(i+1) % MOD def nCr(n, r): if r < 0 or n < r: return 0 return fact[n]*invfact[r]*invfact[n-r] ans = fact[m] for k in range(1, fixed+1): ans += nCr(fixed, k)*fact[m-k]*(-1)**k ans %= MOD print(ans)
Codeforces Round 198 (Div. 2)
CF
2,013
1
256
Iahub and Permutations
Iahub is so happy about inventing bubble sort graphs that he's staying all day long at the office and writing permutations. Iahubina is angry that she is no more important for Iahub. When Iahub goes away, Iahubina comes to his office and sabotage his research work. The girl finds an important permutation for the research. The permutation contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n). She replaces some of permutation elements with -1 value as a revenge. When Iahub finds out his important permutation is broken, he tries to recover it. The only thing he remembers about the permutation is it didn't have any fixed point. A fixed point for a permutation is an element ak which has value equal to k (ak = k). Your job is to proof to Iahub that trying to recover it is not a good idea. Output the number of permutations which could be originally Iahub's important permutation, modulo 1000000007 (109 + 7).
The first line contains integer n (2 ≤ n ≤ 2000). On the second line, there are n integers, representing Iahub's important permutation after Iahubina replaces some values with -1. It's guaranteed that there are no fixed points in the given permutation. Also, the given sequence contains at least two numbers -1 and each positive number occurs in the sequence at most once. It's guaranteed that there is at least one suitable permutation.
Output a single integer, the number of ways Iahub could recover his permutation, modulo 1000000007 (109 + 7).
null
For the first test example there are two permutations with no fixed points are [2, 5, 4, 3, 1] and [5, 1, 4, 3, 2]. Any other permutation would have at least one fixed point.
[{"input": "5\n-1 -1 4 3 -1", "output": "2"}]
2,000
["combinatorics", "math"]
18
[{"input": "5\r\n-1 -1 4 3 -1\r\n", "output": "2\r\n"}, {"input": "8\r\n2 4 5 3 -1 8 -1 6\r\n", "output": "1\r\n"}, {"input": "7\r\n-1 -1 4 -1 7 1 6\r\n", "output": "4\r\n"}, {"input": "6\r\n-1 -1 -1 -1 -1 -1\r\n", "output": "265\r\n"}, {"input": "2\r\n-1 -1\r\n", "output": "1\r\n"}, {"input": "10\r\n4 10 -1 1 6 8 9 2 -1 -1\r\n", "output": "4\r\n"}, {"input": "20\r\n-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1\r\n", "output": "927799753\r\n"}]
false
stdio
null
true
296/B
296
B
PyPy 3
TESTS
2
124
0
136781572
def special(lst1, lst2, m): a,b,c=1,1,1 ln=len(lst1) for i in range(ln): if lst1[i]=="?" and lst2[i]=="?": b*=10 elif lst1[i]!=lst2[i] and lst1[i]!="?" and lst2[i]!="?": b=0 b%=m for i in range(n): if lst1[i]=="?" and lst2[i]=="?": a*=55 c*=55 elif lst1[i]!="?" and lst2[i]!="?": if int(lst1[i])>int(lst2[i]): a=0 if int(lst1[i])<int(lst2[i]): c=0 else: if lst1[i]=="?" and lst2[i]!="?": a*=(int(lst2[i])+1) c*=10-int(lst2[i]) elif lst1[i]!="?" and lst2[i]=="?": c*=(int(lst2[i])+1) a*=10-int(lst1[i]) ans=a+b+c ans%=m d=1 for i in range(n): if lst1[i]=="?": d*=10 if lst2[i]=="?": d*=10 d%=m ans=(d-ans) return ans%m n=int(input()) lst1=input() lst2=input() print(special(lst1,lst2,1000000007))
38
840
7,065,600
28195911
from functools import reduce n, s1, s2, f1, f2 = int(input()), str(input()), str(input()), lambda x: reduce((lambda a, b: (a * b) % 1000000007), x, 1), lambda x: reduce((lambda a, b: a or b), x, False) print((10 ** sum([(s1[i] == '?') + (s2[i] == '?') for i in range(n)]) - (not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)])) * f1([55 if s1[i] == '?' and s2[i] == '?' else (ord(s2[i]) - ord('0') + 1) if s1[i] == '?' else (10 - ord(s1[i]) + ord('0')) if s2[i] == '?' else 1 for i in range(n)]) - (not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)])) * f1([55 if s1[i] == '?' and s2[i] == '?' else (10 - ord(s2[i]) + ord('0')) if s1[i] == '?' else (ord(s1[i]) - ord('0')) + 1 if s2[i] == '?' else 1 for i in range(n)]) + (not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)]) and not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)])) * f1([10 if s1[i] == '?' and s2[i] == '?' else 1 for i in range(n)])) % 1000000007)
Codeforces Round 179 (Div. 2)
CF
2,013
2
256
Yaroslav and Two Strings
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w. A string's template is a string that consists of digits and question marks ("?"). Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers. Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
null
The first test contains no question marks and both strings are incomparable, so the answer is 1. The second test has no question marks, but the given strings are comparable, so the answer is 0.
[{"input": "2\n90\n09", "output": "1"}, {"input": "2\n11\n55", "output": "0"}, {"input": "5\n?????\n?????", "output": "993531194"}]
2,000
["combinatorics", "dp"]
38
[{"input": "2\r\n90\r\n09\r\n", "output": "1\r\n"}, {"input": "2\r\n11\r\n55\r\n", "output": "0\r\n"}, {"input": "5\r\n?????\r\n?????\r\n", "output": "993531194\r\n"}, {"input": "10\r\n104?3?1??3\r\n?1755?1??7\r\n", "output": "91015750\r\n"}, {"input": "10\r\n6276405116\r\n6787?352?9\r\n", "output": "46\r\n"}, {"input": "10\r\n0844033584\r\n0031021311\r\n", "output": "0\r\n"}, {"input": "10\r\n???0?19?01\r\n957461????\r\n", "output": "983368000\r\n"}, {"input": "10\r\n8703870339\r\n994987934?\r\n", "output": "9\r\n"}, {"input": "10\r\n?8?528?91?\r\n45??06???1\r\n", "output": "980398000\r\n"}, {"input": "10\r\n8030456630\r\n83406?6890\r\n", "output": "5\r\n"}, {"input": "1\r\n?\r\n?\r\n", "output": "0\r\n"}, {"input": "2\r\n12\r\n?9\r\n", "output": "1\r\n"}, {"input": "3\r\n??1\r\n?12\r\n", "output": "890\r\n"}, {"input": "3\r\n?12\r\n??1\r\n", "output": "890\r\n"}, {"input": "5\r\n??15?\r\n?32??\r\n", "output": "939500\r\n"}, {"input": "5\r\n??25?\r\n?32??\r\n", "output": "812550\r\n"}, {"input": "5\r\n??55?\r\n?32??\r\n", "output": "872950\r\n"}, {"input": "5\r\n?32??\r\n??55?\r\n", "output": "872950\r\n"}]
false
stdio
null
true
690/F1
690
F1
Python 3
TESTS
4
46
0
19014662
n = int(input()) l = [] for _ in range(n-1): a,b = map(int,input().split()) l.append([a,b]) c=0 for i in l: for j in l: if i[0] == j[0] and i[1] != j[1]: c+=1 elif i[1] == j[0] and i[0] != j[1]: c+=1 elif i[0] == j[1] and i[1] != j[0]: c+=1 print(c//2)
19
62
102,400
198725128
import math n=int(input()) l=[0]*(n+1) # print(*l) for i in range(n-1): a,b=map(int,input().split()) l[a]+=1 l[b]+=1 ans=0 for i in range(1,n+1): if l[i]>=2: ans+=math.comb(l[i],2) print(ans)
Helvetic Coding Contest 2016 online mirror (teams, unrated)
ICPC
2,016
2
256
Tree of Life (easy)
Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies. On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of n points (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (a sequence of one or more edges). To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!
The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.
Print one integer – the number of lifelines in the tree.
null
In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.
[{"input": "4\n1 2\n1 3\n1 4", "output": "3"}, {"input": "5\n1 2\n2 3\n3 4\n3 5", "output": "4"}]
1,300
[]
19
[{"input": "4\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "3"}, {"input": "5\r\n1 2\r\n2 3\r\n3 4\r\n3 5\r\n", "output": "4"}, {"input": "2\r\n1 2\r\n", "output": "0"}, {"input": "3\r\n2 1\r\n3 2\r\n", "output": "1"}, {"input": "10\r\n5 1\r\n1 2\r\n9 3\r\n10 5\r\n6 3\r\n8 5\r\n2 7\r\n2 3\r\n9 4\r\n", "output": "11"}]
false
stdio
null
true
690/F1
690
F1
PyPy 3
TESTS
4
140
0
92413083
a=int(input());ans=0 z=[list(map(int,input().split())) for _ in " "*(a-1)] z.sort();k=a-1 r =lambda x:(x*(x+1))//2;t=0 for i in range(k-1): if z[i][0]==z[i+1][0]:t+=1 else:ans+=r(t);t=0 ans+=r(t);t=0 for i in range(k-1): if z[i][1]==z[i+1][1]:t+=1 else:ans+=r(t);t=0 ans+=r(t);t=0 for i in range(k-1): if z[i][0]==z[i+1][1]:t+=1 else:ans+=r(t);t=0 ans+=r(t);t=0 for i in range(k-1): if z[i][1]==z[i+1][0]:t+=1 else:ans+=r(t);t=0 ans+=r(t);t=0 print(ans)
19
62
2,252,800
164936319
import sys input = sys.stdin.readline n = int(input()) d = [0 for _ in range(n)] for i in range(n-1): a, b = map(int, input().split()) d[a-1] += 1 d[b-1] += 1 print(sum(i*(i-1)//2 for i in d))
Helvetic Coding Contest 2016 online mirror (teams, unrated)
ICPC
2,016
2
256
Tree of Life (easy)
Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies. On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of n points (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (a sequence of one or more edges). To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!
The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.
Print one integer – the number of lifelines in the tree.
null
In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.
[{"input": "4\n1 2\n1 3\n1 4", "output": "3"}, {"input": "5\n1 2\n2 3\n3 4\n3 5", "output": "4"}]
1,300
[]
19
[{"input": "4\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "3"}, {"input": "5\r\n1 2\r\n2 3\r\n3 4\r\n3 5\r\n", "output": "4"}, {"input": "2\r\n1 2\r\n", "output": "0"}, {"input": "3\r\n2 1\r\n3 2\r\n", "output": "1"}, {"input": "10\r\n5 1\r\n1 2\r\n9 3\r\n10 5\r\n6 3\r\n8 5\r\n2 7\r\n2 3\r\n9 4\r\n", "output": "11"}]
false
stdio
null
true
690/F1
690
F1
PyPy 3
TESTS
4
140
0
92409871
from sys import stdin a=int(stdin.readline());ans=0 z=[list(map(int,stdin.readline().split())) for _ in " "*(a-1)] w=[] for i in range(a-1): s=z[i] for j in range(a-1): if i!=j: if [max(i,j),min(i,j)] in w:continue if s[0]==z[j][1] or s[0]==z[j][0]:ans+=1;w.append([max(i,j),min(i,j)]) print(ans)
19
77
7,884,800
124377070
# Test ##################################################################################################################### def main(): tree_sLength = int(input()) - 1 tree = {} for i in range(tree_sLength): v1, v2 = map(int, input().split()) tree[v1] = tree.get(v1, ()) + (v2, ) tree[v2] = tree.get(v2, ()) + (v1, ) return nLifeLines(tree) def nLifeLines(tree): return sum(subTree_snLifeLines(len(tree[x])) for x in tree)//2 def subTree_snLifeLines(nBranches): return nBranches*(nBranches - 1) if __name__ == '__main__': print(main())
Helvetic Coding Contest 2016 online mirror (teams, unrated)
ICPC
2,016
2
256
Tree of Life (easy)
Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies. On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of n points (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (a sequence of one or more edges). To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!
The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.
Print one integer – the number of lifelines in the tree.
null
In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.
[{"input": "4\n1 2\n1 3\n1 4", "output": "3"}, {"input": "5\n1 2\n2 3\n3 4\n3 5", "output": "4"}]
1,300
[]
19
[{"input": "4\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "3"}, {"input": "5\r\n1 2\r\n2 3\r\n3 4\r\n3 5\r\n", "output": "4"}, {"input": "2\r\n1 2\r\n", "output": "0"}, {"input": "3\r\n2 1\r\n3 2\r\n", "output": "1"}, {"input": "10\r\n5 1\r\n1 2\r\n9 3\r\n10 5\r\n6 3\r\n8 5\r\n2 7\r\n2 3\r\n9 4\r\n", "output": "11"}]
false
stdio
null
true
796/D
796
D
PyPy 3-64
TESTS
4
1,668
195,788,800
195240992
from collections import defaultdict, deque from sys import stdin, stdout def inp(): return int(stdin.readline()) def inlt(): return list(map(int, stdin.readline().strip().split())) def insr(): s = stdin.readline().strip() return list(s[:len(s)]) def invr(): return map(int, stdin.readline().strip().split()) ''' 6 2 4 1 6 1 2 2 3 3 4 4 5 5 6 ''' def main(): graph = defaultdict(list) idx_mapper = {} n, k, d = inlt() vstd = set() redundant = set() def bfs(node): deq.append((node, 0)) currv.add(node) while deq: node, dist = deq.popleft() if dist < d: for i in graph[node]: if i in vstd: redundant.add(idx_mapper[(i, node)]) elif i not in currv and i not in preserve: currv.add(i) deq.append((i, dist + 1)) k_s = inlt() preserve = set(k_s) for i in range(n - 1): a, b = invr() idx_mapper[(a, b)] = i + 1 idx_mapper[(b, a)] = i + 1 graph[a].append(b) graph[b].append(a) for i in k_s: deq = deque() currv = set() bfs(i) for i in currv: vstd.add(i) print(len(redundant)) print(*redundant) main()
135
1,669
108,544,000
152757338
import math from collections import defaultdict, deque, Counter from sys import stdin import heapq inf = int(1e18) input = stdin.readline n,k,d = map(int, input().split()) ps = [int(i) for i in input().split()] adj = [[] for i in range (n+1)] for i in range (n-1): u,v = map(int, input().split()) adj[u].append((v,i+1)) adj[v].append((u,i+1)) visit = [False]*(n+1) edges = set() q = deque() for p in ps: q.append(p) visit[p] = True while len(q)>0: node = q.popleft() for child, i in adj[node]: if not visit[child]: visit[child] = True q.append(child) edges.add(i) print(n-1-len(edges)) ans = [] for i in range (n-1): if i+1 not in edges: ans.append(i+1) print(*ans)
Codeforces Round 408 (Div. 2)
CF
2,017
2
256
Police Stations
Inzane finally found Zane with a lot of money to spare, so they together decided to establish a country of their own. Ruling a country is not an easy job. Thieves and terrorists are always ready to ruin the country's peace. To fight back, Zane and Inzane have enacted a very effective law: from each city it must be possible to reach a police station by traveling at most d kilometers along the roads. There are n cities in the country, numbered from 1 to n, connected only by exactly n - 1 roads. All roads are 1 kilometer long. It is initially possible to travel from a city to any other city using these roads. The country also has k police stations located in some cities. In particular, the city's structure satisfies the requirement enforced by the previously mentioned law. Also note that there can be multiple police stations in one city. However, Zane feels like having as many as n - 1 roads is unnecessary. The country is having financial issues, so it wants to minimize the road maintenance cost by shutting down as many roads as possible. Help Zane find the maximum number of roads that can be shut down without breaking the law. Also, help him determine such roads.
The first line contains three integers n, k, and d (2 ≤ n ≤ 3·105, 1 ≤ k ≤ 3·105, 0 ≤ d ≤ n - 1) — the number of cities, the number of police stations, and the distance limitation in kilometers, respectively. The second line contains k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — each denoting the city each police station is located in. The i-th of the following n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities directly connected by the road with index i. It is guaranteed that it is possible to travel from one city to any other city using only the roads. Also, it is possible from any city to reach a police station within d kilometers.
In the first line, print one integer s that denotes the maximum number of roads that can be shut down. In the second line, print s distinct integers, the indices of such roads, in any order. If there are multiple answers, print any of them.
null
In the first sample, if you shut down road 5, all cities can still reach a police station within k = 4 kilometers. In the second sample, although this is the only largest valid set of roads that can be shut down, you can print either 4 5 or 5 4 in the second line.
[{"input": "6 2 4\n1 6\n1 2\n2 3\n3 4\n4 5\n5 6", "output": "1\n5"}, {"input": "6 3 2\n1 5 6\n1 2\n1 3\n1 4\n1 5\n5 6", "output": "2\n4 5"}]
2,100
["constructive algorithms", "dfs and similar", "dp", "graphs", "shortest paths", "trees"]
135
[{"input": "6 2 4\r\n1 6\r\n1 2\r\n2 3\r\n3 4\r\n4 5\r\n5 6\r\n", "output": "1\r\n3 "}, {"input": "6 3 2\r\n1 5 6\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n5 6\r\n", "output": "2\r\n4 5 "}, {"input": "10 1 5\r\n5\r\n1 2\r\n2 3\r\n3 4\r\n4 5\r\n5 6\r\n6 7\r\n7 8\r\n8 9\r\n9 10\r\n", "output": "0\r\n"}, {"input": "11 1 5\r\n6\r\n1 2\r\n2 3\r\n3 4\r\n4 5\r\n5 6\r\n6 7\r\n7 8\r\n8 9\r\n9 10\r\n10 11\r\n", "output": "0\r\n"}, {"input": "2 1 1\r\n1\r\n1 2\r\n", "output": "0\r\n"}]
false
stdio
null
true
999/D
999
D
PyPy 3-64
TESTS
6
187
34,201,600
219020028
def rl(): return list(map(int,input().split())) def ri(): return int(input()) def rs(): return input() def rm(): return map(int,input().split()) def main(): n,m=rm() a=rl() cnt=[0]*m loc=[[] for i in range(m)] for i in range(n): cnt[a[i]%m]+=1 loc[a[i]%m].append(i) target=n//m ans=0 st=[] for i in range(m): if cnt[i]>target: for j in range(cnt[i]-target): st.append(loc[i].pop()) cnt[i]=target else: for j in range(target-cnt[i]): if len(st)==0: break lc=st.pop() add=(i-a[lc])%m a[lc]+=add ans+=add cnt[i]+=1 for i in range(m): if cnt[i]==target: continue for j in range(target-cnt[i]): lc=st.pop() a[lc]+=(i-lc)%m ans+=(i-lc)%m cnt[i]+=1 print(ans) print(*a) main()
47
389
40,755,200
167327303
from collections import * from heapq import * from bisect import * from itertools import * from functools import * from math import * from string import * import sys input = sys.stdin.readline n, m = map(int, input().split()) A = list(map(int, input().split())) S = sum(A) target = n // m locs = [[] for _ in range(m)] for i, x in enumerate(A): locs[x % m].append(i) i = 0 for k in range(m): while len(locs[k]) > target: while i < k or len(locs[i % m]) >= target: i += 1 pos = locs[k].pop() A[pos] += (i - k) % m locs[A[pos] % m].append(pos) print(sum(A) - S) print(*A)
Codeforces Round 490 (Div. 3)
ICPC
2,018
3
256
Equalize the Remainders
You are given an array consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$, and a positive integer $$$m$$$. It is guaranteed that $$$m$$$ is a divisor of $$$n$$$. In a single move, you can choose any position $$$i$$$ between $$$1$$$ and $$$n$$$ and increase $$$a_i$$$ by $$$1$$$. Let's calculate $$$c_r$$$ ($$$0 \le r \le m-1)$$$ — the number of elements having remainder $$$r$$$ when divided by $$$m$$$. In other words, for each remainder, let's find the number of corresponding elements in $$$a$$$ with that remainder. Your task is to change the array in such a way that $$$c_0 = c_1 = \dots = c_{m-1} = \frac{n}{m}$$$. Find the minimum number of moves to satisfy the above requirement.
The first line of input contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 2 \cdot 10^5, 1 \le m \le n$$$). It is guaranteed that $$$m$$$ is a divisor of $$$n$$$. The second line of input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 10^9$$$), the elements of the array.
In the first line, print a single integer — the minimum number of moves required to satisfy the following condition: for each remainder from $$$0$$$ to $$$m - 1$$$, the number of elements of the array having this remainder equals $$$\frac{n}{m}$$$. In the second line, print any array satisfying the condition and can be obtained from the given array with the minimum number of moves. The values of the elements of the resulting array must not exceed $$$10^{18}$$$.
null
null
[{"input": "6 3\n3 2 0 6 10 12", "output": "3\n3 2 0 7 10 14"}, {"input": "4 2\n0 1 2 3", "output": "0\n0 1 2 3"}]
1,900
["data structures", "greedy", "implementation"]
47
[{"input": "6 3\r\n3 2 0 6 10 12\r\n", "output": "3\r\n3 2 0 7 10 14 \r\n"}, {"input": "4 2\r\n0 1 2 3\r\n", "output": "0\r\n0 1 2 3 \r\n"}, {"input": "1 1\r\n1000000000\r\n", "output": "0\r\n1000000000 \r\n"}, {"input": "6 3\r\n3 2 0 6 10 11\r\n", "output": "1\r\n3 2 0 7 10 11 \r\n"}, {"input": "100 25\r\n6745 2075 7499 7517 1776 5164 2335 2745 4465 1457 7565 2232 2486 9025 8059 9646 8017 7662 9690 3352 2306 366 7422 1073 7169 8966 4506 8225 5614 8628 2908 7452 9625 9332 7097 353 1043 8118 5794 4486 626 971 6731 6618 887 6354 4814 7307 7681 6160 9351 2579 411 3436 5570 2812 2726 4433 3220 577 5891 3861 528 2183 127 5579 6979 4005 9953 5038 9937 4792 3003 9417 8796 1565 11 2596 2486 3494 4464 9568 5512 5565 9822 9820 4848 2889 9527 2249 9860 8236 256 8434 8038 6407 5570 5922 7435 2815\r\n", "output": "88\r\n6745 2075 7499 7517 1776 5164 2335 2745 4465 1457 7565 2232 2486 9025 8059 9646 8017 7662 9690 3352 2306 366 7422 1073 7169 8966 4506 8225 5614 8628 2908 7452 9625 9332 7097 353 1043 8118 5794 4486 626 971 6731 6618 887 6354 4814 7307 7681 6160 9351 2579 411 3436 5570 2812 2726 4433 3220 577 5891 3863 528 2183 127 5579 6979 4005 9953 5038 9937 4792 3005 9417 8796 1565 24 2596 2505 3494 4464 9568 5513 5566 9822 9823 4848 2899 9530 2249 9860 8259 259 8434 8038 6408 5573 5922 7435 2819 \r\n"}]
false
stdio
import sys def main(): input_path = sys.argv[1] ref_output_path = sys.argv[2] sub_output_path = sys.argv[3] # Read input with open(input_path) as f: n, m = map(int, f.readline().split()) a_original = list(map(int, f.readline().split())) # Read reference output to get correct moves with open(ref_output_path) as f: correct_moves = int(f.readline().strip()) # Read submission output with open(sub_output_path) as f: lines = f.readlines() if len(lines) < 2: print(0) return try: submission_moves = int(lines[0].strip()) submission_array = list(map(int, lines[1].split())) except: print(0) return # Check submission moves match reference if submission_moves != correct_moves: print(0) return # Check array length if len(submission_array) != n: print(0) return # Check elements are non-decreasing for sub, orig in zip(submission_array, a_original): if sub < orig: print(0) return # Check total moves sum total_moves = sum(sub - orig for sub, orig in zip(submission_array, a_original)) if total_moves != submission_moves: print(0) return # Check remainder counts k = n // m remainders = [0] * m for num in submission_array: remainders[num % m] += 1 if all(cnt == k for cnt in remainders): print(1) else: print(0) if __name__ == "__main__": main()
true
484/D
484
D
Python 3
TESTS
0
31
0
228199270
def max_sociability(n, charisma): max_diff = 0 current_diff = 0 for i in range(1, n): diff = abs(charisma[i] - charisma[i - 1]) current_diff = max(current_diff + diff, diff) max_diff = max(max_diff, current_diff) return max_diff n = int(input()) charisma = list(map(int, input().split())) result = max_sociability(n, charisma) print(result)
71
1,310
79,462,400
147542209
n = int(input()) if n == 1: print(0) else: a = [0]*(n+1) i = 1 for v in map(int, input().split()): a[i] = v i += 1 dp = [0]*(n+1) dp[2] = abs(a[1]-a[2]) for i in range(3, n+1): # 单调 if (a[i-2] < a[i-1] and a[i-1] < a[i]) or (a[i-2] > a[i-1] and a[i-1] > a[i]): # 转移时a[i-1]造成的贡献会被抵消 dp[i] = dp[i-1]+abs(a[i]-a[i-1]) else: # 此时不单调 多选浪费 dp[i] = max(dp[i-1], dp[i-2]+abs(a[i]-a[i-1])) print(dp[n])
Codeforces Round 276 (Div. 1)
CF
2,014
2
256
Kindergarten
In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated each child with his or her integer charisma value. Each child should go to exactly one group. Each group should be a nonempty segment of consecutive children of a line. A group's sociability is the maximum difference of charisma of two children in the group (in particular, if the group consists of one child, its sociability equals a zero). The teacher wants to divide the children into some number of groups in such way that the total sociability of the groups is maximum. Help him find this value.
The first line contains integer n — the number of children in the line (1 ≤ n ≤ 106). The second line contains n integers ai — the charisma of the i-th child ( - 109 ≤ ai ≤ 109).
Print the maximum possible total sociability of all groups.
null
In the first test sample one of the possible variants of an division is following: the first three children form a group with sociability 2, and the two remaining children form a group with sociability 1. In the second test sample any division leads to the same result, the sociability will be equal to 0 in each group.
[{"input": "5\n1 2 3 1 2", "output": "3"}, {"input": "3\n3 3 3", "output": "0"}]
2,400
["data structures", "dp", "greedy"]
71
[{"input": "5\r\n1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "3\r\n3 3 3\r\n", "output": "0\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "2\r\n-1000000000 1000000000\r\n", "output": "2000000000\r\n"}, {"input": "4\r\n1 4 2 3\r\n", "output": "4\r\n"}, {"input": "4\r\n23 5 7 1\r\n", "output": "24\r\n"}, {"input": "4\r\n23 7 5 1\r\n", "output": "22\r\n"}, {"input": "8\r\n23 2 7 5 15 8 4 10\r\n", "output": "37\r\n"}, {"input": "8\r\n4 5 3 6 2 7 1 8\r\n", "output": "16\r\n"}]
false
stdio
null
true
24/C
24
C
PyPy 3
TESTS
2
280
4,096,000
154968873
n,j=map(int,input().split()) x0,y0=map(int,input().split()) A=[tuple(map(int,input().split())) for i in range(n)] x,y=x0,y0 for z,w in A: x=2*z-x y=2*w-y rep=j//n rest=j%n nx=(x-x0)*rep+x0 ny=(y-y0)*rep+y0 x,y=nx,ny for i in range(rest): z,w=A[i] x=2*z-x y=2*w-y print(x,y)
23
310
14,643,200
212806082
from sys import stdin N, J = [int(w) for w in stdin.readline().split()] J = J % (N + N) points = [] x, y = [int(w) for w in stdin.readline().split()] for _ in range(N): points.append([int(w) for w in stdin.readline().split()]) for i in range(J): i %= N x += (points[i][0] - x) * 2 y += (points[i][1] - y) * 2 print(x, y)
Codeforces Beta Round 24
ICPC
2,010
2
256
Sequence of points
You are given the following points with integer coordinates on the plane: M0, A0, A1, ..., An - 1, where n is odd number. Now we define the following infinite sequence of points Mi: Mi is symmetric to Mi - 1 according $${ A } _ { ( i - 1 ) \mod n }$$ (for every natural number i). Here point B is symmetric to A according M, if M is the center of the line segment AB. Given index j find the point Mj.
On the first line you will be given an integer n (1 ≤ n ≤ 105), which will be odd, and j (1 ≤ j ≤ 1018), where j is the index of the desired point. The next line contains two space separated integers, the coordinates of M0. After that n lines follow, where the i-th line contain the space separated integer coordinates of the point Ai - 1. The absolute values of all input coordinates will not be greater then 1000.
On a single line output the coordinates of Mj, space separated.
null
null
[{"input": "3 4\n0 0\n1 1\n2 3\n-5 3", "output": "14 0"}, {"input": "3 1\n5 5\n1000 1000\n-1000 1000\n3 100", "output": "1995 1995"}]
1,800
["geometry", "implementation", "math"]
23
[{"input": "3 4\r\n0 0\r\n1 1\r\n2 3\r\n-5 3\r\n", "output": "14 0\r\n"}, {"input": "3 1\r\n5 5\r\n1000 1000\r\n-1000 1000\r\n3 100\r\n", "output": "1995 1995\r\n"}, {"input": "1 1\r\n-1000 -1000\r\n1000 1000\r\n", "output": "3000 3000\r\n"}, {"input": "1 1000000000000000000\r\n-1000 1000\r\n1000 -1000\r\n", "output": "-1000 1000\r\n"}, {"input": "1 900000000000000001\r\n-1000 -1000\r\n-1000 -1000\r\n", "output": "-1000 -1000\r\n"}]
false
stdio
null
true
999/D
999
D
Python 3
TESTS
7
919
23,142,400
121169118
n,m = map(int,input().split()) a = list(map(int,input().split())) b = [0]*m x = n//m ans = 0 c = [] for i in range(n): cr = a[i]%m if b[cr]<x: b[cr]+=1 else: c.append([cr,i]) c.sort() i = 0 j = 0 n1 = len(c) while i<m and j<n1: if b[i]>=x: i+=1 else: b[i]+=1 cnt = i-c[j][0] if cnt<0: cnt+=m a[c[j][1]]+=cnt ans+=cnt j+=1 print(ans) for i in a: print(i,end=' ') print()
47
389
53,862,400
167326676
from collections import * from heapq import * from bisect import * from itertools import * from functools import * from math import * from string import * import sys input = sys.stdin.readline def main(): n, m = map(int, input().split()) A = list(map(int, input().split())) target = n // m locs = [[] for _ in range(m)] for i, x in enumerate(A): locs[x % m].append(i) ans = 0 free = [] for _ in range(2): for k in range(m): while len(locs[k]) > target: i = locs[k].pop() free.append([i, k]) while len(locs[k]) < target and free: pos, mod_class = free.pop() dist = (k - mod_class) % m A[pos] += dist ans += dist locs[k].append(pos) print(ans) print(*A) if __name__ == "__main__": main()
Codeforces Round 490 (Div. 3)
ICPC
2,018
3
256
Equalize the Remainders
You are given an array consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$, and a positive integer $$$m$$$. It is guaranteed that $$$m$$$ is a divisor of $$$n$$$. In a single move, you can choose any position $$$i$$$ between $$$1$$$ and $$$n$$$ and increase $$$a_i$$$ by $$$1$$$. Let's calculate $$$c_r$$$ ($$$0 \le r \le m-1)$$$ — the number of elements having remainder $$$r$$$ when divided by $$$m$$$. In other words, for each remainder, let's find the number of corresponding elements in $$$a$$$ with that remainder. Your task is to change the array in such a way that $$$c_0 = c_1 = \dots = c_{m-1} = \frac{n}{m}$$$. Find the minimum number of moves to satisfy the above requirement.
The first line of input contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 2 \cdot 10^5, 1 \le m \le n$$$). It is guaranteed that $$$m$$$ is a divisor of $$$n$$$. The second line of input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 10^9$$$), the elements of the array.
In the first line, print a single integer — the minimum number of moves required to satisfy the following condition: for each remainder from $$$0$$$ to $$$m - 1$$$, the number of elements of the array having this remainder equals $$$\frac{n}{m}$$$. In the second line, print any array satisfying the condition and can be obtained from the given array with the minimum number of moves. The values of the elements of the resulting array must not exceed $$$10^{18}$$$.
null
null
[{"input": "6 3\n3 2 0 6 10 12", "output": "3\n3 2 0 7 10 14"}, {"input": "4 2\n0 1 2 3", "output": "0\n0 1 2 3"}]
1,900
["data structures", "greedy", "implementation"]
47
[{"input": "6 3\r\n3 2 0 6 10 12\r\n", "output": "3\r\n3 2 0 7 10 14 \r\n"}, {"input": "4 2\r\n0 1 2 3\r\n", "output": "0\r\n0 1 2 3 \r\n"}, {"input": "1 1\r\n1000000000\r\n", "output": "0\r\n1000000000 \r\n"}, {"input": "6 3\r\n3 2 0 6 10 11\r\n", "output": "1\r\n3 2 0 7 10 11 \r\n"}, {"input": "100 25\r\n6745 2075 7499 7517 1776 5164 2335 2745 4465 1457 7565 2232 2486 9025 8059 9646 8017 7662 9690 3352 2306 366 7422 1073 7169 8966 4506 8225 5614 8628 2908 7452 9625 9332 7097 353 1043 8118 5794 4486 626 971 6731 6618 887 6354 4814 7307 7681 6160 9351 2579 411 3436 5570 2812 2726 4433 3220 577 5891 3861 528 2183 127 5579 6979 4005 9953 5038 9937 4792 3003 9417 8796 1565 11 2596 2486 3494 4464 9568 5512 5565 9822 9820 4848 2889 9527 2249 9860 8236 256 8434 8038 6407 5570 5922 7435 2815\r\n", "output": "88\r\n6745 2075 7499 7517 1776 5164 2335 2745 4465 1457 7565 2232 2486 9025 8059 9646 8017 7662 9690 3352 2306 366 7422 1073 7169 8966 4506 8225 5614 8628 2908 7452 9625 9332 7097 353 1043 8118 5794 4486 626 971 6731 6618 887 6354 4814 7307 7681 6160 9351 2579 411 3436 5570 2812 2726 4433 3220 577 5891 3863 528 2183 127 5579 6979 4005 9953 5038 9937 4792 3005 9417 8796 1565 24 2596 2505 3494 4464 9568 5513 5566 9822 9823 4848 2899 9530 2249 9860 8259 259 8434 8038 6408 5573 5922 7435 2819 \r\n"}]
false
stdio
import sys def main(): input_path = sys.argv[1] ref_output_path = sys.argv[2] sub_output_path = sys.argv[3] # Read input with open(input_path) as f: n, m = map(int, f.readline().split()) a_original = list(map(int, f.readline().split())) # Read reference output to get correct moves with open(ref_output_path) as f: correct_moves = int(f.readline().strip()) # Read submission output with open(sub_output_path) as f: lines = f.readlines() if len(lines) < 2: print(0) return try: submission_moves = int(lines[0].strip()) submission_array = list(map(int, lines[1].split())) except: print(0) return # Check submission moves match reference if submission_moves != correct_moves: print(0) return # Check array length if len(submission_array) != n: print(0) return # Check elements are non-decreasing for sub, orig in zip(submission_array, a_original): if sub < orig: print(0) return # Check total moves sum total_moves = sum(sub - orig for sub, orig in zip(submission_array, a_original)) if total_moves != submission_moves: print(0) return # Check remainder counts k = n // m remainders = [0] * m for num in submission_array: remainders[num % m] += 1 if all(cnt == k for cnt in remainders): print(1) else: print(0) if __name__ == "__main__": main()
true
216/D
216
D
Python 3
TESTS
0
60
0
206619409
def main(): n = int(input()) bridges = [] for i in range(n): k, *bridge_distances = map(int, input().split()) bridges.append(bridge_distances) unstable_cells = 0 for i in range(n): k = len(bridges[i]) for j in range(k): left = bridges[i][k - 1] if j == 0 else bridges[i][j - 1] right = bridges[i][0] if j == k - 1 else bridges[i][j + 1] if left != right: unstable_cells += 1 print(unstable_cells) if __name__ == "__main__": main()
55
560
30,412,800
125084763
import math import sys def search(arr,l,r,p1,p2): i1=-1 i2=-1 copy_l=l copy_r=r while l<=r: m=(l+r)//2 val=arr[m] if p1<=val and val<=p2: i1=m r=m-1 elif val>p2: r=m-1 else: l=m+1 l=copy_l r=copy_r while l<=r: m=(l+r)//2 val=arr[m] if p1<=val and val<=p2: i2=m l=m+1 elif val>p2: r=m-1 else: l=m+1 if i1==-1 or i2==-1: return 0 else: return i2-i1+1 def main(arr): n=len(arr) for i in range(len(arr)): arr[i].sort() ans=0 for i in range(len(arr)): f_arr=arr[(i-1)%n] s_arr=arr[(i+1)%n] for j in range(len(arr[i])-1): p1=arr[i][j] p2=arr[i][j+1] val1=search(f_arr,0,len(f_arr)-1,p1,p2) val2=search(s_arr,0,len(s_arr)-1,p1,p2) if val1!=val2: ans+=1 return ans n=int(input()) arr=[] for i in range(n): arr2=list(map(int,input().split())) arr2.pop(0) arr.append(arr2) print(main(arr))
Codeforces Round 133 (Div. 2)
CF
2,012
2
256
Spider's Web
Paw the Spider is making a web. Web-making is a real art, Paw has been learning to do it his whole life. Let's consider the structure of the web. There are n main threads going from the center of the web. All main threads are located in one plane and divide it into n equal infinite sectors. The sectors are indexed from 1 to n in the clockwise direction. Sectors i and i + 1 are adjacent for every i, 1 ≤ i < n. In addition, sectors 1 and n are also adjacent. Some sectors have bridge threads. Each bridge connects the two main threads that make up this sector. The points at which the bridge is attached to the main threads will be called attachment points. Both attachment points of a bridge are at the same distance from the center of the web. At each attachment point exactly one bridge is attached. The bridges are adjacent if they are in the same sector, and there are no other bridges between them. A cell of the web is a trapezoid, which is located in one of the sectors and is bounded by two main threads and two adjacent bridges. You can see that the sides of the cell may have the attachment points of bridges from adjacent sectors. If the number of attachment points on one side of the cell is not equal to the number of attachment points on the other side, it creates an imbalance of pulling forces on this cell and this may eventually destroy the entire web. We'll call such a cell unstable. The perfect web does not contain unstable cells. Unstable cells are marked red in the figure. Stable cells are marked green. Paw the Spider isn't a skillful webmaker yet, he is only learning to make perfect webs. Help Paw to determine the number of unstable cells in the web he has just spun.
The first line contains integer n (3 ≤ n ≤ 1000) — the number of main threads. The i-th of following n lines describe the bridges located in the i-th sector: first it contains integer ki (1 ≤ ki ≤ 105) equal to the number of bridges in the given sector. Then follow ki different integers pij (1 ≤ pij ≤ 105; 1 ≤ j ≤ ki). Number pij equals the distance from the attachment points of the j-th bridge of the i-th sector to the center of the web. It is guaranteed that any two bridges between adjacent sectors are attached at a different distance from the center of the web. It is guaranteed that the total number of the bridges doesn't exceed 105.
Print a single integer — the number of unstable cells in Paw the Spider's web.
null
null
[{"input": "7\n3 1 6 7\n4 3 5 2 9\n2 8 1\n4 3 7 6 4\n3 2 5 9\n3 6 3 8\n3 4 2 9", "output": "6"}]
1,700
["binary search", "sortings", "two pointers"]
55
[{"input": "7\r\n3 1 6 7\r\n4 3 5 2 9\r\n2 8 1\r\n4 3 7 6 4\r\n3 2 5 9\r\n3 6 3 8\r\n3 4 2 9\r\n", "output": "6"}, {"input": "3\r\n1 1\r\n1 2\r\n1 3\r\n", "output": "0"}, {"input": "3\r\n2 1 2\r\n2 3 4\r\n2 5 6\r\n", "output": "0"}, {"input": "5\r\n3 2 4 10\r\n2 1 6\r\n2 8 7\r\n3 2 4 10\r\n2 1 6\r\n", "output": "2"}, {"input": "3\r\n5 2 7 6 9 8\r\n4 10 1 5 4\r\n1 3\r\n", "output": "2"}, {"input": "4\r\n5 2 6 10 8 3\r\n1 7\r\n2 10 4\r\n2 9 5\r\n", "output": "5"}, {"input": "5\r\n3 12 10 19\r\n5 1 4 18 11 13\r\n4 17 15 2 6\r\n4 3 12 18 10\r\n4 2 8 5 9\r\n", "output": "8"}, {"input": "10\r\n1 77966\r\n1 79480\r\n1 94920\r\n1 53920\r\n1 15585\r\n1 57339\r\n1 1585\r\n1 91802\r\n1 27934\r\n1 20354\r\n", "output": "0"}]
false
stdio
null
true
463/D
463
D
PyPy 3
TESTS
1
140
1,536,000
118700926
import sys input=sys.stdin.readline n,k=map(int,input().split()) a=[list(map(int,input().split())) for i in range(k)] g=[[] for i in range(n+1)] for i in range(n): x=a[0][i] s1=set(a[0][i+1:]) idx2=a[1].index(x) s2=set(a[1][idx2+1:]) idx3=a[2].index(x) s3=set(a[2][idx3+1:]) s=s1&s2&s3 for y in s: g[x].append(y) MAX=0 for s in range(1,n+1): q=[[0,s]] dist=[-1]*(n+1) dist[s]=0 while q: cost,ver=q.pop() for to in g[ver]: if dist[to]<cost+1: dist[to]=cost+1 q.append([dist[to],to]) MAX=max(MAX,max(dist)) print(MAX+1)
40
124
2,867,200
197562869
n, k = map(int, input().split()) a, b = [[] for row in range(6)], [[0 for col in range(n + 1)] for row in range(6)] for i in range(k): a[i] = list(map(int, input().split())) for j in range(n): b[i][a[i][j]] = j dp = [1] * n for i in range(n): for j in range(i): flag = 1 for x in range(1, k): if b[x][a[0][j]] > b[x][a[0][i]]: flag = 0 break if flag: dp[i] = max(dp[i], dp[j] + 1) print(max(dp))
Codeforces Round 264 (Div. 2)
CF
2,014
2
256
Gargari and Permutations
Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have found k permutations. Each of them consists of numbers 1, 2, ..., n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari? You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
The first line contains two integers n and k (1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the next k lines contains integers 1, 2, ..., n in some order — description of the current permutation.
Print the length of the longest common subsequence.
null
The answer for the first test sample is subsequence [1, 2, 3].
[{"input": "4 3\n1 4 2 3\n4 1 2 3\n1 2 4 3", "output": "3"}]
1,900
["dfs and similar", "dp", "graphs", "implementation"]
40
[{"input": "4 3\r\n1 4 2 3\r\n4 1 2 3\r\n1 2 4 3\r\n", "output": "3\r\n"}, {"input": "6 3\r\n2 5 1 4 6 3\r\n5 1 4 3 2 6\r\n5 4 2 6 3 1\r\n", "output": "3\r\n"}, {"input": "41 4\r\n24 15 17 35 13 41 4 14 23 5 8 16 21 18 30 36 6 22 11 29 26 1 40 31 7 3 32 10 28 38 12 20 39 37 34 19 33 27 2 25 9\r\n22 13 25 24 38 35 29 12 15 8 11 37 3 19 4 23 18 32 30 40 36 21 16 34 27 9 5 41 39 2 14 17 31 33 26 7 1 10 20 6 28\r\n31 27 39 16 22 12 13 32 6 10 19 29 37 7 18 33 24 21 1 9 36 4 34 41 25 28 17 40 30 35 23 14 11 8 2 15 38 20 26 5 3\r\n8 18 39 38 7 34 16 31 15 1 40 20 37 4 25 11 17 19 33 26 6 14 13 41 12 32 2 21 10 35 27 9 28 5 30 24 22 23 29 3 36\r\n", "output": "4\r\n"}, {"input": "1 2\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "28 5\r\n3 14 12 16 13 27 20 8 1 10 24 11 5 9 7 18 17 23 22 25 28 19 4 21 26 6 15 2\r\n7 12 23 27 22 26 16 18 19 5 6 9 11 28 25 4 10 3 1 14 8 17 15 2 20 13 24 21\r\n21 20 2 5 19 15 12 4 18 9 23 16 11 14 8 6 25 27 13 17 10 26 7 24 28 1 3 22\r\n12 2 23 11 20 18 25 21 13 27 14 8 4 6 9 16 7 3 10 1 22 15 26 19 5 17 28 24\r\n13 2 6 19 22 23 4 1 28 10 18 17 21 8 9 3 26 11 12 27 14 20 24 25 15 5 16 7\r\n", "output": "3\r\n"}, {"input": "6 3\r\n2 5 1 4 6 3\r\n5 1 4 6 2 3\r\n5 4 2 6 3 1\r\n", "output": "4\r\n"}, {"input": "41 4\r\n24 15 17 35 13 41 4 14 23 5 8 16 21 18 30 36 6 22 11 29 26 1 40 31 7 3 32 10 28 38 12 20 39 37 34 19 33 27 2 25 9\r\n22 13 25 24 38 35 29 12 15 8 11 37 3 19 4 23 18 32 30 40 36 21 16 34 27 9 5 41 39 2 14 17 31 33 26 7 1 10 20 6 28\r\n31 27 39 16 22 12 13 32 6 10 19 29 37 7 18 33 24 21 1 9 36 4 34 41 25 28 17 40 30 35 23 14 11 8 2 15 38 20 26 5 3\r\n8 18 39 38 7 34 16 31 15 1 40 20 37 4 25 11 17 19 33 26 6 14 13 41 12 32 2 21 10 35 27 9 28 5 30 24 22 23 29 3 36\r\n", "output": "4\r\n"}, {"input": "37 3\r\n6 3 19 20 15 4 1 35 8 24 12 21 34 26 18 14 23 33 28 9 36 11 37 31 25 32 29 22 13 27 16 17 10 7 5 30 2\r\n10 3 35 17 34 21 14 8 26 28 11 19 27 7 4 23 24 22 12 13 16 1 25 29 5 31 30 20 32 18 15 9 2 36 37 33 6\r\n19 9 22 32 26 35 29 23 5 6 14 34 33 10 2 28 15 11 24 4 13 7 8 31 37 36 1 27 3 16 30 25 20 21 18 17 12\r\n", "output": "7\r\n"}]
false
stdio
null
true
960/F
960
F
Python 3
PRETESTS
3
62
7,065,600
37064950
n,m = map(int,input().split()) e=[tuple(map(int,input().split()))for i in range(m)] ans = {} def f(lans,w): left=0 right = len(lans) while True: middle = (left+right)//2 if right==left+1: if lans[left][1]<=w: return lans[left][0] return 0 if lans[middle][1]>w: right=middle else: left=middle for elem in e: a,b,w=elem if a in ans: if b not in ans: ans[b]=set() t=(f(list(ans[a]),w)+1,w) ans[b].add(t) else: if b not in ans: ans[b]=set() ans[b].add((1,w)) print(max([b[0] for a in ans.values() for b in a]))
63
685
44,236,800
206868890
import bisect import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline def f(u, v): return u * z + v def update(i, x): i += l1 tree[i] = x i //= 2 while i: tree[i] = max(tree[2 * i], tree[2 * i + 1]) i //= 2 return def get_max(s, t): s, t = s + l1, t + l1 ans = 0 while s <= t: if s % 2 == 0: s //= 2 else: ans = max(ans, tree[s]) s = (s + 1) // 2 if t % 2 == 1: t //= 2 else: ans = max(ans, tree[t]) t = (t - 1) // 2 return ans n, m = map(int, input().split()) abw = [tuple(map(int, input().split())) for _ in range(m)] z = pow(10, 5) + 5 s = [] for a, b, w in abw: s.append(f(b, w)) s = list(set(sorted(s))) s.sort() l = len(s) d = dict() for i in range(l): d[s[i]] = i l1 = pow(2, (l + 1).bit_length()) l2 = 2 * l1 tree = [0] * l2 for a, b, w in abw: l0 = bisect.bisect_left(s, f(a, 0)) r0 = bisect.bisect_left(s, f(a, w)) - 1 update(d[f(b, w)], max(get_max(l0, r0) + 1, tree[d[f(b, w)] + l1])) ans = tree[1] print(ans)
Divide by Zero 2018 and Codeforces Round 474 (Div. 1 + Div. 2, combined)
CF
2,018
1
256
Pathwalks
You are given a directed graph with n nodes and m edges, with all edges having a certain weight. There might be multiple edges and self loops, and the graph can also be disconnected. You need to choose a path (possibly passing through same vertices multiple times) in the graph such that the weights of the edges are in strictly increasing order, and these edges come in the order of input. Among all such paths, you need to find the the path that has the maximum possible number of edges, and report this value. Please note that the edges picked don't have to be consecutive in the input.
The first line contains two integers n and m (1 ≤ n ≤ 100000,1 ≤ m ≤ 100000) — the number of vertices and edges in the graph, respectively. m lines follows. The i-th of these lines contains three space separated integers ai, bi and wi (1 ≤ ai, bi ≤ n, 0 ≤ wi ≤ 100000), denoting an edge from vertex ai to vertex bi having weight wi
Print one integer in a single line — the maximum number of edges in the path.
null
The answer for the first sample input is 2: $$1 \rightarrow 2 \rightarrow 3$$. Note that you cannot traverse $$1 \rightarrow 2 \rightarrow 3 \rightarrow 1$$ because edge $$\overset{3}{\rightarrow}1$$ appears earlier in the input than the other two edges and hence cannot be picked/traversed after either of the other two edges. In the second sample, it's optimal to pick 1-st, 3-rd and 5-th edges to get the optimal answer: $$1 \rightarrow 3 \rightarrow 4 \rightarrow 5$$.
[{"input": "3 3\n3 1 3\n1 2 1\n2 3 2", "output": "2"}, {"input": "5 5\n1 3 2\n3 2 3\n3 4 5\n5 4 0\n4 5 8", "output": "3"}]
2,100
["data structures", "dp", "graphs"]
63
[{"input": "3 3\r\n3 1 3\r\n1 2 1\r\n2 3 2\r\n", "output": "2"}, {"input": "5 5\r\n1 3 2\r\n3 2 3\r\n3 4 5\r\n5 4 0\r\n4 5 8\r\n", "output": "3"}, {"input": "5 10\r\n3 4 8366\r\n5 1 6059\r\n2 1 72369\r\n2 2 35472\r\n5 3 50268\r\n2 4 98054\r\n5 1 26220\r\n2 3 24841\r\n1 3 42450\r\n3 1 59590\r\n", "output": "3"}, {"input": "1000 10\r\n11 368 48256\r\n192 176 81266\r\n236 360 25346\r\n377 962 3089\r\n486 176 49857\r\n693 810 36660\r\n692 698 35141\r\n879 822 10964\r\n974 439 31998\r\n364 142 62668\r\n", "output": "1"}, {"input": "1 1\r\n1 1 1000\r\n", "output": "1"}, {"input": "6 5\r\n1 2 1\r\n2 3 2\r\n3 4 3\r\n5 4 10\r\n4 6 11\r\n", "output": "4"}]
false
stdio
null
true
690/D1
690
D1
Python 3
TESTS
0
31
4,608,000
25933138
import sys #with open(filename, 'r') as f: with sys.stdin as f: input_list = list(f) for i, line in enumerate(input_list): if i == 0: R = int(line.split(" ")[0]) else: if i == R: print(len(list(filter(lambda x: len(x) > 0, line.split('.')))))
119
62
0
18999968
r, c = map(int, input().split()) for _ in range(r-1): input() bottom = input() sections = 0 looking_for_new_section = True for brick in bottom: if looking_for_new_section: if brick == 'B': looking_for_new_section = False sections += 1 else: if brick == '.': looking_for_new_section = True print(sections)
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
797/E
797
E
PyPy 3-64
TESTS
7
218
72,704,000
223292411
# https://github.com/Tsuzat/fast-IO-for-python/blob/master/fast_IO/__init__.py #!/usr/bin/env python import os import sys from io import BytesIO, IOBase # region fastio BUFSIZE: int = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") def print(*args, **kwargs): """Prints the values to a stream, or to sys.stdout by default.""" sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout) at_start = True for x in args: if not at_start: file.write(sep) file.write(str(x)) at_start = False file.write(kwargs.pop("end", "\n")) if kwargs.pop("flush", False): file.flush() sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # endregion import math,random,collections from collections import deque from math import floor,ceil,sqrt,gcd from random import randint if __name__ == "__main__": pass # prime factorize prime = [] primeset = set() # https://www.geeksforgeeks.org/sieve-eratosthenes-0n-time-complexity/ def getPrimes(N): global prime global primeset isprime = [True for i in range(N+1)] isprime[0] = isprime[1] = False SPF = [None] * (N) for i in range(2, N): if isprime[i]: prime.append(i) SPF[i] = i j = 0 while (j < len(prime) and i * prime[j] < N and prime[j] <= SPF[i]): isprime[i * prime[j]] = False SPF[i * prime[j]] = prime[j] j += 1 primeset = set(prime) def getRidOf(n,p): while n % p == 0: n = n//p return n def factors(n): ans = dict() if len(prime) < math.ceil(math.sqrt(n))+3: getPrimes(math.ceil(math.sqrt(n))+5) for i in prime: if i > math.ceil(math.sqrt(n))+3: break while n % i == 0: if i in ans: ans[i] += 1 else: ans[i] = 1 n = n // i if n > 1: if n in ans: ans[n] += 1 else: ans[n] = 1 return ans def totient(n): totient = n for factor in factors(n): totient -= totient // factor return totient # https://www.geeksforgeeks.org/square-free-number/ def squarefree(n): if n % 2 == 0: n = n / 2 if n % 2 == 0: return False for i in range(3, int(sqrt(n) + 1)): if n % i == 0: n = n / i if n % i == 0: return False return True # https://smsxgz.github.io/post/pe/counting_square_free_numbers/ mobiusList = [] def mobius(n): global mobiusList if n < len(mobiusList): return mobiusList[n] prime = [1] * (n + 1) mobiusList = [1] * (n + 1) for i in range(2, n + 1): if not prime[i]: continue mobiusList[i] = -1 for j in range(2, n // i + 1): prime[i * j] = 0 mobiusList[i * j] *= -1 for j in range(1, n // (i * i) + 1): mobiusList[j * i * i] = 0 return mobiusList # https://smsxgz.github.io/post/pe/counting_square_free_numbers/ def count_squarefree(N): global mobius Imax = int(pow(N//4,1/3)) D = int(sqrt(N / Imax)) mobius(D) # compute S1 s1 = 0 for i in range(1, D + 1): s1 += mobiusList[i] * (N // (i * i)) # compute M(d), d = 1, ..., D M_list = [0] M = 0 for m in mobiusList[1:]: M += m M_list.append(M) # compute M(sqrt(n / i)), i = Imax - 1, ..., 1 Mxi_list = [] Mxi_sum = 0 for i in range(Imax - 1, 0, -1): Mxi = 1 xi = int(sqrt(N // i)) sqd = int(sqrt(xi)) # sqd < D <= xi for j in range(1, xi // (sqd + 1) + 1): Mxi -= (xi // j - xi // (j + 1)) * M_list[j] for j in range(2, sqd + 1): if xi // j <= D: Mxi -= M_list[xi // j] else: Mxi -= Mxi_list[Imax - j * j * i - 1] Mxi_list.append(Mxi) Mxi_sum += Mxi # compute S2 s2 = Mxi_sum - (Imax - 1) * M_list[-1] return s1 + s2 def intArrToStr(a): return " ".join([str(x) for x in a]) def isPrime(n): prime_flag = 0 if(n > 1): for i in range(2, int(sqrt(n)) + 1): if (n % i == 0): prime_flag = 1 break return (prime_flag == 0) else: return False from itertools import compress def primes2(n): """ Input n>=6, Returns a list of primes, 2 <= p < n """ n, correction = n-n%6+6, 2-(n%6>1) sieve = [True] * (n//3) for i in range(1,int(n**0.5)//3+1): if sieve[i]: k=3*i+1|1 sieve[ k*k//3 ::2*k] = [False] * ((n//6-k*k//6-1)//k+1) sieve[k*(k-2*(i&1)+4)//3::2*k] = [False] * ((n//6-k*(k-2*(i&1)+4)//6-1)//k+1) return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]] def dsum(a): if type(a) == type(1): return a else: return sum([dsum(b) for b in a]) # Modify these functions for binary search def binarySearchArrayGet(x): pass def binarySearchArrayLength(): pass def binarySearchCompare(x,y): pass # Exact, Floor, Ceiling def binarySearch(element, style="Exact"): l = 0 r = binarySearchArrayLength()-1 while l+1 < r: m = (l+r+randint(0,1))//2 if binarySearchCompare(binarySearchArrayGet(m), element) <= 0: l = m if m != binarySearchArrayLength()-1 and binarySearchCompare(binarySearchArrayGet(m+1), element) > 0: r = m+1 if style in ["Exact","exact","E","e",0.5]: if binarySearchCompare(binarySearchArrayGet(l), element) == 0: return l if binarySearchCompare(binarySearchArrayGet(r), element) == 0: return r return -1 if style in ["Floor","floor","F","f",0]: if binarySearchCompare(binarySearchArrayGet(l), element) > 0: return -1 if binarySearchCompare(binarySearchArrayGet(r), element) > 0: return l return r if style in ["Ceiling","ceiling","Ceil","ceil","C","c",1]: if binarySearchCompare(binarySearchArrayGet(r), element) < 0: return -1 if binarySearchCompare(binarySearchArrayGet(l), element) < 0: return r return l raise Exception("Binary search style invalid!") def column(arr,col): a = [] for i in arr: a.append(i[col]) return a def removeDuplicates(a): b = [] for i in range(len(a))-1: if a[i] != a[i+1]: b.append(a[i]) b.append(a[-1]) return b def intArrInput(): *a, = map(int,input().split()) return a def digitStringToArr(s): return [int(x) for x in s] def stringSeparatorToArr(s,sep): return [int(x) for x in s.split(sep)] def spacedStringToArr(s): return stringSeparatorToArr(s," ") # https://en.wikipedia.org/wiki/Fast_Walsh%E2%80%93Hadamard_transform def fwht(a) -> None: """In-place Fast Walsh–Hadamard Transform of array a.""" h = 1 while h < len(a): # perform FWHT for i in range(0, len(a), h * 2): for j in range(i, i + h): x = a[j] y = a[j + h] a[j] = x + y a[j + h] = x - y # normalize and increment h *= 2 def xorConvolution(c,d): # assuming a and b already have the same number of elements that is a power of 2 a = c.copy() b = d.copy() fwht(a) fwht(b) hi = [] for i in range(len(a)): hi.append(a[i]*b[i]) fwht(hi) for j in range(len(hi)): hi[j] //= len(a) return hi def mex(a): b = set(a) for i in range(len(a)+2): if i not in b: return i raise Exception("What") def freqDict(a): ans = dict() for i in a: if i in ans: ans[i] += 1 else: ans[i] = 1 return ans from random import getrandbits RANDOM = getrandbits(32) class Wrapper(int): def __init__(self, x): int.__init__(x) def __hash__(self): return super(Wrapper, self).__hash__() ^ RANDOM def dictGet(x,d): if type(x) is int: x = str(x) return d[x] def addTo(x,d): if type(x) is int: x = str(x) if x in d: d[x] += 1 else: d[x] = 1 import random RANDOMFORDICT = random.randrange((1 << 31) - 1) class Dict(dict): def __setitem__(self, __k, __v): return super().__setitem__(__k ^ RANDOMFORDICT, __v) def __delitem__(self, __v): return super().__delitem__(__v ^ RANDOMFORDICT) def __getitem__(self, __k): return super().__getitem__(__k ^ RANDOMFORDICT) def __contains__(self, __o: object) -> bool: return super().__contains__(__o ^ RANDOMFORDICT) # ----------------------------------------------------------------- def binarySearchArrayGet(x): return [1,2,3,5,6][x] def binarySearchArrayLength(): return 5 def binarySearchCompare(x,y): return x-y generatePrimes = False PRIMELIMIT = 10**5+12345 if generatePrimes: prime = primes2(PRIMELIMIT) primeset = set(prime) # ----------------------------------------------------------------- import math n = int(input()) a = intArrInput() q = int(input()) precomp = [[]] for i in range(1,int(math.sqrt(n))): stuff = [0 for i in range(n)] stuff[-1] = 1 for j in range(n-2,-1,-1): lol = a[j] + j + i if lol >= n: stuff[j] = 1 else: stuff[j] = stuff[lol] + 1 precomp.append(stuff) for i in range(n): x,y = intArrInput() x -= 1 c = 0 if y > math.sqrt(n)-3: while x < n: x += a[x] + y c += 1 else: c = precomp[y][x] print(c)
53
576
92,364,800
92972860
import sys n = int(sys.stdin.buffer.readline().decode('utf-8')) a = [0] + list(map(int, sys.stdin.buffer.readline().decode('utf-8').split())) q = int(sys.stdin.buffer.readline().decode('utf-8')) dp_size = 200 dp = [[1]*(n+1) for _ in range(dp_size+1)] for k in range(1, dp_size+1): for i in range(n, 0, -1): if i+a[i]+k <= n: dp[k][i] = dp[k][i+a[i]+k]+1 ans = [0]*q for i, (p, k) in enumerate(map(int, line.decode('utf-8').split()) for line in sys.stdin.buffer): if k <= dp_size: ans[i] = dp[k][p] else: while p <= n: p += a[p] + k ans[i] += 1 sys.stdout.buffer.write('\n'.join(map(str, ans)).encode('utf-8'))
Educational Codeforces Round 19
ICPC
2,017
2
256
Array Queries
a is an array of n positive integers, all of which are not greater than n. You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.
The first line contains one integer n (1 ≤ n ≤ 100000). The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n). The third line containts one integer q (1 ≤ q ≤ 100000). Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).
Print q integers, ith integer must be equal to the answer to ith query.
null
Consider first example: In first query after first operation p = 3, after second operation p = 5. In next two queries p is greater than n after the first operation.
[{"input": "3\n1 1 1\n3\n1 1\n2 1\n3 1", "output": "2\n1\n1"}]
2,000
["brute force", "data structures", "dp"]
53
[{"input": "3\r\n1 1 1\r\n3\r\n1 1\r\n2 1\r\n3 1\r\n", "output": "2\r\n1\r\n1\r\n"}, {"input": "10\r\n3 5 4 3 7 10 6 7 2 3\r\n10\r\n4 5\r\n2 10\r\n4 6\r\n9 9\r\n9 2\r\n5 1\r\n6 4\r\n1 1\r\n5 6\r\n6 4\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n1\r\n"}, {"input": "50\r\n6 2 5 6 10 2 5 8 9 2 9 5 10 4 3 6 10 6 1 1 3 7 2 1 7 8 5 9 6 2 7 6 1 7 2 10 10 2 4 2 8 4 3 10 7 1 7 8 6 3\r\n50\r\n23 8\r\n12 8\r\n3 3\r\n46 3\r\n21 6\r\n7 4\r\n26 4\r\n12 1\r\n25 4\r\n18 7\r\n29 8\r\n27 5\r\n47 1\r\n50 2\r\n46 7\r\n13 6\r\n44 8\r\n12 2\r\n18 3\r\n35 10\r\n38 1\r\n7 10\r\n4 2\r\n22 8\r\n36 3\r\n25 2\r\n47 3\r\n33 5\r\n10 5\r\n12 9\r\n7 4\r\n26 4\r\n19 4\r\n3 8\r\n12 3\r\n35 8\r\n31 4\r\n25 5\r\n3 5\r\n46 10\r\n37 6\r\n8 9\r\n20 5\r\n36 1\r\n41 9\r\n6 7\r\n40 5\r\n24 4\r\n41 10\r\n14 8\r\n", "output": "3\r\n4\r\n6\r\n2\r\n4\r\n5\r\n3\r\n8\r\n3\r\n3\r\n2\r\n2\r\n1\r\n1\r\n1\r\n3\r\n1\r\n6\r\n5\r\n2\r\n3\r\n3\r\n7\r\n2\r\n2\r\n4\r\n1\r\n3\r\n4\r\n3\r\n5\r\n3\r\n5\r\n5\r\n6\r\n2\r\n3\r\n2\r\n4\r\n1\r\n1\r\n3\r\n4\r\n2\r\n1\r\n5\r\n2\r\n4\r\n1\r\n3\r\n"}, {"input": "50\r\n49 21 50 31 3 19 10 40 20 5 47 12 12 30 5 4 5 2 11 23 5 39 3 30 19 3 23 40 4 14 39 50 45 14 33 37 1 15 43 30 45 23 32 3 9 17 9 5 14 12\r\n50\r\n36 29\r\n44 24\r\n38 22\r\n30 43\r\n15 19\r\n39 2\r\n13 8\r\n29 50\r\n37 18\r\n32 19\r\n42 39\r\n20 41\r\n14 11\r\n4 25\r\n14 35\r\n17 23\r\n1 29\r\n3 19\r\n8 6\r\n26 31\r\n9 46\r\n36 31\r\n21 49\r\n17 38\r\n47 27\r\n5 21\r\n42 22\r\n36 34\r\n50 27\r\n11 45\r\n26 41\r\n1 16\r\n21 39\r\n18 43\r\n21 37\r\n12 16\r\n22 32\r\n7 18\r\n10 14\r\n43 37\r\n18 50\r\n21 32\r\n27 32\r\n48 16\r\n5 6\r\n5 11\r\n5 6\r\n39 29\r\n40 13\r\n14 6\r\n", "output": "1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n2\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n2\r\n3\r\n1\r\n1\r\n1\r\n1\r\n1\r\n3\r\n3\r\n3\r\n1\r\n1\r\n2\r\n"}]
false
stdio
null
true
19/A
19
A
Python 3
TESTS
3
92
0
214148780
n = int(input()) teams, index = [], {} for i in range(n): team = input() teams.append([team, [0, 0, 0]]) index[team] = i for i in range(int(n*(n-1)/2)): game = input().split() A, B = game[0].split('-') a, b = map(int, game[1].split(':')) if a > b: teams[index[A]][1][0] += 3 elif a == b: teams[index[A]][1][0] += 1 teams[index[B]][1][0] += 1 else: teams[index[B]][1][0] += 3 teams[index[A]][1][1] += a teams[index[A]][1][2] += b teams[index[B]][1][1] += b teams[index[B]][1][2] += a for i in sorted(teams, key=lambda x: (x[1][0], x[1][1], x[1][2]))[int(len(teams)/2):]: print(i[0])
29
92
0
151243272
class Info: def __init__(self, _teamName, _points, _goalDiff, _scoredGoals): self.teamName = _teamName self.points = _points self.goalDiff = _goalDiff self.scoredGoals = _scoredGoals def __str__(self): return f"[teamName: '{self.teamName}', points: {self.points}, goalDiff: {self.goalDiff}, scoredGoals: {self.scoredGoals}]" def output(cont): for item in cont: print(item) def findIndexByName(cont, searchName): for i in range(len(cont)): if cont[i].teamName == searchName: return i return -1 N = int(input()) cont = [] for i in range(N): # newTeamName = input() # obj = Info(newTeamName,0,0,0) # cont.append(obj) cont.append(Info(input(), 0, 0, 0)) ''' 01234567890123456789 'Barcelona-Real 15:10' ''' for i in range(N * (N - 1) // 2): line = input() dashInd = line.index('-') spaceInd = line.index(' ') colonInd = line.index(':') team1Name = line[:dashInd] team2Name = line[dashInd + 1:spaceInd] goal1 = int(line[spaceInd + 1:colonInd]) goal2 = int(line[colonInd + 1:]) team1Index = findIndexByName(cont,team1Name) team2Index = findIndexByName(cont, team2Name) #update point if goal1 > goal2: cont[team1Index].points += 3 elif goal2 > goal1: cont[team2Index].points += 3 else: cont[team1Index].points += 1 cont[team2Index].points += 1 #update goalDiff cont[team1Index].goalDiff += goal1 - goal2 cont[team2Index].goalDiff += goal2 - goal1 #update scored goals cont[team1Index].scoredGoals += goal1 cont[team2Index].scoredGoals += goal2 cont.sort(key=lambda item:(item.points,item.goalDiff,item.scoredGoals),reverse=True) del cont[N // 2:] cont.sort(key=lambda item:item.teamName) for item in cont: print(item.teamName)
Codeforces Beta Round 19
ICPC
2,010
2
64
World Football Cup
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations: - the final tournament features n teams (n is always even) - the first n / 2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity. You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
The first input line contains the only integer n (1 ≤ n ≤ 50) — amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following n·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 — names of the teams; num1, num2 (0 ≤ num1, num2 ≤ 100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output n / 2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
null
null
[{"input": "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3", "output": "A\nD"}, {"input": "2\na\nA\na-A 2:1", "output": "a"}]
1,400
["implementation"]
29
[{"input": "4\r\nA\r\nB\r\nC\r\nD\r\nA-B 1:1\r\nA-C 2:2\r\nA-D 1:0\r\nB-C 1:0\r\nB-D 0:3\r\nC-D 0:3\r\n", "output": "A\r\nD\r\n"}, {"input": "2\r\na\r\nA\r\na-A 2:1\r\n", "output": "a\r\n"}, {"input": "2\r\nEULEUbCmfrmqxtzvg\r\nuHGRmKUhDcxcfqyruwzen\r\nuHGRmKUhDcxcfqyruwzen-EULEUbCmfrmqxtzvg 13:92\r\n", "output": "EULEUbCmfrmqxtzvg\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 2:2\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "A\r\nB\r\nF\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 7:7\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "B\r\nC\r\nF\r\n"}]
false
stdio
null
true
643/A
643
A
Python 3
TESTS
2
46
0
21002892
def main(): n = int(input()) cnt, winner = ([[0] * i for i in range(1, n + 1)] for _ in (0, 1)) order = [[] for _ in range(n)] for i, t in enumerate(map(int, input().split())): order[t - 1].append(i) for t, l in enumerate(order): while l: x, hi = len(l), l.pop() row = cnt[hi] winner[hi][hi] = t for lo in l: if row[lo] < x: row[lo] = x winner[hi][lo] = t x -= 1 cnt = [0] * n for row in winner: for t in row: cnt[t] += 1 print(*cnt) if __name__ == '__main__': main()
35
265
9,523,200
230903750
n = int(input()) t = [int(q) - 1 for q in input().split()] k = [0] * n for i in range(n): s = [0] * n b = 0 for a in t[i:]: s[a] += 1 if s[a] > s[b] or a < b and s[a] == s[b]: b = a k[b] += 1 print(*k)
VK Cup 2016 - Round 3
CF
2,016
2
256
Bear and Colors
Bear Limak has n colored balls, arranged in one long row. Balls are numbered 1 through n, from left to right. There are n possible colors, also numbered 1 through n. The i-th ball has color ti. For a fixed interval (set of consecutive elements) of balls we can define a dominant color. It's a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant. There are $$\frac{n(n+1)}{2}$$ non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant.
The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — the number of balls. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ n) where ti is the color of the i-th ball.
Print n integers. The i-th of them should be equal to the number of intervals where i is a dominant color.
null
In the first sample, color 2 is dominant in three intervals: - An interval [2, 2] contains one ball. This ball's color is 2 so it's clearly a dominant color. - An interval [4, 4] contains one ball, with color 2 again. - An interval [2, 4] contains two balls of color 2 and one ball of color 1. There are 7 more intervals and color 1 is dominant in all of them.
[{"input": "4\n1 2 1 2", "output": "7 3 0 0"}, {"input": "3\n1 1 1", "output": "6 0 0"}]
1,500
["implementation"]
35
[{"input": "4\r\n1 2 1 2\r\n", "output": "7 3 0 0 \r\n"}, {"input": "3\r\n1 1 1\r\n", "output": "6 0 0 \r\n"}, {"input": "10\r\n9 1 5 2 9 2 9 2 1 1\r\n", "output": "18 30 0 0 1 0 0 0 6 0 \r\n"}, {"input": "50\r\n17 13 19 19 19 34 32 24 24 13 34 17 19 19 7 32 19 13 13 30 19 34 34 28 41 24 24 47 22 34 21 21 30 7 22 21 32 19 34 19 34 22 7 28 6 13 19 30 13 30\r\n", "output": "0 0 0 0 0 22 40 0 0 0 0 0 98 0 0 0 5 0 675 0 165 9 0 61 0 0 0 5 0 6 0 4 0 183 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 \r\n"}, {"input": "150\r\n28 124 138 71 71 18 78 136 138 93 145 93 18 15 71 47 47 64 18 72 138 72 18 150 7 71 109 149 18 115 149 149 15 78 124 27 72 124 28 108 138 109 108 111 148 138 78 27 28 150 138 65 15 145 109 47 102 62 28 7 115 108 102 149 150 27 111 64 149 124 13 21 108 64 7 15 72 72 124 47 102 28 109 18 124 28 111 138 7 13 21 62 136 62 13 64 71 7 130 47 77 65 71 148 15 93 64 65 28 65 13 78 78 47 115 138 28 115 72 136 124 145 150 62 105 78 71 102 109 150 27 130 62 7 93 72 93 62 7 124 72 21 62 18 62 7 108 78 148 149\r\n", "output": "0 0 0 0 0 0 1863 0 0 0 0 0 604 0 97 0 0 1026 0 0 12 0 0 0 0 0 208 2982 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 67 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1400 0 158 371 0 0 0 0 0 92 296 0 0 0 0 1 199 0 0 0 0 0 0 0 0 0 0 0 0 0 0 48 0 0 0 0 0 0 0 0 151 0 0 1 0 0 230 37 0 6 0 0 0 44 0 0 0 0 0 0 0 0 174 0 0 0 0 0 2 0 0 0 0 0 4 0 729 0 0 0 0 0 0 4 0 0 4 486 29 \r\n"}, {"input": "1\r\n1\r\n", "output": "1 \r\n"}, {"input": "2\r\n2 1\r\n", "output": "2 1 \r\n"}]
false
stdio
null
true
19/A
19
A
Python 3
TESTS
3
92
0
217704800
class Solve: def __init__(self) -> None: self.n = 0; self.team = dict(); def vao(self) -> None: self.n = int(input()); for _ in range(self.n): x = input(); self.team[x] = 0; def tinhDiem(self, team1, team2, diem) -> None: diem1, diem2 = map(int,diem.split(":")); if(diem1 > diem2): self.team[team1] += 3; elif(diem1 < diem2): self.team[team2] += 3; else: self.team[team1] += 1; self.team[team2] += 1; def lam(self) -> None: size = (self.n * (self.n - 1)) // 2; for _ in range(size): s1, s2 = input().split(" "); team1, team2 = s1.split("-"); self.tinhDiem(team1, team2, s2); k = sorted(self.team.items(), key= lambda item: item[1], reverse= True); ketQua = [k[i][0] for i in range(self.n // 2)]; ketQua.sort(); for i in ketQua: print(i); def ra(self) -> None: pass; def main(): p = Solve(); p.vao(); p.lam(); p.ra(); main();
29
92
0
158272335
class Info: def __init__(self, newTeamName, newPoints, newGoalDiff, newScoredGoals): self.teamName = newTeamName self.points = newPoints self.goalDiff = newGoalDiff self.scoredGoals = newScoredGoals def __str__(self): return f"[teamName: '{self.teamName}',points: {self.points}," \ f"goalDiff: {self.goalDiff},socreadGoals: {self.scoredGoals}]" # obj = Info('Manchester', 10, 20, 30) # print(obj) def output(cont): for item in cont: print(item) def findIndexByName(cont, searchName): for i in range(len(cont)): if cont[i].teamName == searchName: return i return -1 N = int(input()) cont = [] for i in range(N): # newTeamName = input() # obj = Info(newTeamName, 0, 0, 0) # cont.append(obj) cont.append(Info(input(), 0, 0, 0)) # output(cont) ''' 01234567890123456789 line = "Barcelona-Real 15:10" ''' for i in range(N * (N - 1) // 2): Line = input() dashInd = Line.index('-') spaceInd = Line.index(' ') colonInd = Line.index(':') team1Name = Line[:dashInd] team2Name = Line[dashInd + 1:spaceInd] goal1 = int(Line[spaceInd + 1:colonInd]) goal2 = int(Line[colonInd + 1:]) team1Index = findIndexByName(cont, team1Name) team2Index = findIndexByName(cont, team2Name) # update points if goal1 > goal2: cont[team1Index].points += 3 elif goal2 > goal1: cont[team2Index].points += 3 else: cont[team1Index].points += 1 cont[team2Index].points += 1 # updateGoaldiff cont[team1Index].goalDiff += goal1 - goal2 cont[team2Index].goalDiff += goal2 - goal1 # updateScoredGoals cont[team1Index].scoredGoals += goal1 cont[team2Index].scoredGoals += goal2 cont.sort(key=lambda obj: (obj.points, obj.goalDiff, obj.scoredGoals), reverse=True) del cont[N // 2:] cont.sort(key=lambda obj: obj.teamName) for item in cont: print(item.teamName) #output(cont)
Codeforces Beta Round 19
ICPC
2,010
2
64
World Football Cup
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations: - the final tournament features n teams (n is always even) - the first n / 2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity. You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
The first input line contains the only integer n (1 ≤ n ≤ 50) — amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following n·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 — names of the teams; num1, num2 (0 ≤ num1, num2 ≤ 100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output n / 2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
null
null
[{"input": "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3", "output": "A\nD"}, {"input": "2\na\nA\na-A 2:1", "output": "a"}]
1,400
["implementation"]
29
[{"input": "4\r\nA\r\nB\r\nC\r\nD\r\nA-B 1:1\r\nA-C 2:2\r\nA-D 1:0\r\nB-C 1:0\r\nB-D 0:3\r\nC-D 0:3\r\n", "output": "A\r\nD\r\n"}, {"input": "2\r\na\r\nA\r\na-A 2:1\r\n", "output": "a\r\n"}, {"input": "2\r\nEULEUbCmfrmqxtzvg\r\nuHGRmKUhDcxcfqyruwzen\r\nuHGRmKUhDcxcfqyruwzen-EULEUbCmfrmqxtzvg 13:92\r\n", "output": "EULEUbCmfrmqxtzvg\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 2:2\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "A\r\nB\r\nF\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 7:7\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "B\r\nC\r\nF\r\n"}]
false
stdio
null
true
467/D
467
D
Python 3
TESTS
0
15
307,200
213192184
from sys import stdin from collections import defaultdict def main(): num = {} stdin.readline() essay = [] for word in stdin.readline().strip().lower().split(): count_r = word.count('r') length = len(word) num.setdefault(word, len(num)) essay.append((count_r, length, num[word])) n_synonyms = int(stdin.readline()) synonyms = defaultdict(list) for _ in range(n_synonyms): word, rep = stdin.readline().strip().lower().split() count_word_r = word.count('r') count_rep_r = rep.count('r') num.setdefault(word, len(num)) num.setdefault(rep, len(num)) synonyms[num[rep]].append(num[word]) essay.sort(reverse=True) best = {} for n_r, length, word in essay: if word not in best: best[word] = n_r, length for rep in synonyms[word]: if rep not in best: essay.append((n_r, length, rep)) sum_n_r, sum_len = 0, 0 for n_r, length, word in essay: n_r, length = best[word] sum_n_r += n_r sum_len += length print(sum_n_r, sum_len) if __name__ == '__main__': main()
68
311
36,454,400
229763337
if 1: import sys input = lambda: sys.stdin.readline().rstrip("\r\n") def I(): return input() def II(): return int(input()) def MII(): return map(int, input().split()) def LI(): return list(input().split()) def LII(): return list(map(int, input().split())) def LFI(): return list(map(float, input().split())) def GMI(): return map(lambda x: int(x) - 1, input().split()) def LGMI(): return list(map(lambda x: int(x) - 1, input().split())) inf = float('inf') if 1: from io import BytesIO, IOBase import math import random import os import bisect import typing from collections import Counter, defaultdict, deque from copy import deepcopy from functools import cmp_to_key, lru_cache, reduce from heapq import heapify, heappop, heappush, heappushpop, nlargest, nsmallest from itertools import accumulate, combinations, permutations, count from operator import add, iand, ior, itemgetter, mul, xor from string import ascii_lowercase, ascii_uppercase, ascii_letters from typing import * BUFSIZE = 4096 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin = IOWrapper(sys.stdin) sys.stdout = IOWrapper(sys.stdout) dfs, hashing = 0, 0 if dfs: from types import GeneratorType def bootstrap(f, stack=[]): def wrappedfunc(*args, **kwargs): if stack: return f(*args, **kwargs) else: to = f(*args, **kwargs) while True: if type(to) is GeneratorType: stack.append(to) to = next(to) else: stack.pop() if not stack: break to = stack[-1].send(to) return to return wrappedfunc if hashing: RANDOM = random.getrandbits(20) class Wrapper(int): def __init__(self, x): int.__init__(x) def __hash__(self): return super(Wrapper, self).__hash__() ^ RANDOM read_from_file = 0 if read_from_file: file = open("input.txt", "r").readline().strip()[1:-1] fin = open(file, 'r') input = lambda: fin.readline().strip() output_file = open("output.txt", "w") def fprint(*args, **kwargs): print(*args, **kwargs, file=output_file) class UnionFind: def __init__(self, n): self.parent = list(range(n)) def find(self, a): a = self.parent[a] acopy = a while a != self.parent[a]: a = self.parent[a] while acopy != a: self.parent[acopy], acopy = a, self.parent[acopy] return a def merge(self, a, b): pa, pb = self.find(a), self.find(b) if pa == pb: return False self.parent[pb] = pa return True tmp = set() n = II() essay = LI() for i in range(n): essay[i] = essay[i].lower() tmp.add(essay[i]) m = II() pairs = [] for _ in range(m): s1, s2 = LI() s1 = s1.lower() s2 = s2.lower() tmp.add(s1) tmp.add(s2) pairs.append((s2, s1)) tmp = list(tmp) tot = len(tmp) v1 = [inf] * tot v2 = [inf] * tot mapping = {s: i for i, s in enumerate(tmp)} path = [[] for _ in range(tot)] for a, b in pairs: path[mapping[a]].append(mapping[b]) for i in sorted(range(tot), key=lambda x: (tmp[x].count('r'), len(tmp[x]))): if v1[i] == inf: a, b = tmp[i].count('r'), len(tmp[i]) v1[i] = a v2[i] = b stack = [i] while stack: u = stack.pop() for v in path[u]: if v1[v] == inf: v1[v] = a v2[v] = b stack.append(v) ans1 = 0 ans2 = 0 for w in essay: ans1 += v1[mapping[w]] ans2 += v2[mapping[w]] print(ans1, ans2)
Codeforces Round 267 (Div. 2)
CF
2,014
2
256
Fedor and Essay
After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn't want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn't like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language. Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times. As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn't matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay. Please note that in this problem the case of letters doesn't matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.
The first line contains a single integer m (1 ≤ m ≤ 105) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won't exceed 105 characters. The next line contains a single integer n (0 ≤ n ≤ 105) — the number of pairs of words in synonym dictionary. The i-th of the next n lines contains two space-separated non-empty words xi and yi. They mean that word xi can be replaced with word yi (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn't exceed 5·105 characters. All the words at input can only consist of uppercase and lowercase letters of the English alphabet.
Print two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.
null
null
[{"input": "3\nAbRb r Zz\n4\nxR abRb\naA xr\nzz Z\nxr y", "output": "2 6"}, {"input": "2\nRuruRu fedya\n1\nruruRU fedor", "output": "1 10"}]
2,400
["dfs and similar", "dp", "graphs", "hashing", "strings"]
68
[{"input": "3\r\nAbRb r Zz\r\n4\r\nxR abRb\r\naA xr\r\nzz Z\r\nxr y\r\n", "output": "2 6\r\n"}, {"input": "2\r\nRuruRu fedya\r\n1\r\nruruRU fedor\r\n", "output": "1 10\r\n"}, {"input": "1\r\nffff\r\n1\r\nffff r\r\n", "output": "0 4\r\n"}, {"input": "2\r\nYURA YUrA\r\n1\r\nyura fedya\r\n", "output": "0 10\r\n"}, {"input": "5\r\nhello my name is fedya\r\n2\r\nhello hi\r\nis i\r\n", "output": "0 14\r\n"}, {"input": "5\r\nawiuegjsrkjshegkjshegseg g soeigjseg www s\r\n3\r\nwww s\r\nawiuegjsrkjshegkjshegseg wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww\r\nwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww www\r\n", "output": "0 13\r\n"}, {"input": "5\r\naa bb cc ee ff\r\n5\r\naa a\r\nbb aa\r\ncc bb\r\nee cc\r\nff bb\r\n", "output": "0 5\r\n"}, {"input": "7\r\nraki vezde est awjgkawkgjn ttttt raki raks\r\n4\r\nraks rks\r\nrks raks\r\nraki raks\r\nvezde pss\r\n", "output": "3 31\r\n"}, {"input": "5\r\nfedor fedya www awwwwwww a\r\n5\r\nr a\r\nfedor fedr\r\nwww a\r\nawwwwwww www\r\na r\r\n", "output": "1 12\r\n"}, {"input": "1\r\nYURA\r\n1\r\nyura lesha\r\n", "output": "0 5\r\n"}, {"input": "2\r\nABBABAABBAABABBABAABABBAABBABAABBAABABBAABBABAABABBABAABBAABABBA ABBABAABBAABABBABAABABBAABBABAABBAABABBAABBABAABABBABAABBAABABA\r\n2\r\nABBABAABBAABABBABAABABBAABBABAABBAABABBAABBABAABABBABAABBAABABA neuzaiheshi\r\nABBABAABBAABABBABAABABBAABBABAABBAABABBAABBABAABABBABAABBAABABBA ABBABAABBAABABBABAABABBAABBABAABBAABABBAABBABAABABBABAABBAABABA\r\n", "output": "0 22\r\n"}, {"input": "10\r\nlalka lolka yura lesha fedya bredor tourist www qqq gruihdrkgjp\r\n11\r\nlalka lolka\r\nlolka lalka\r\nyura lolka\r\nlalka poka\r\nfedya bredor\r\nbredor yura\r\ntourist bredor\r\nwww qqq\r\nqqq w\r\nw g\r\ngruihdrkgjp bredor\r\n", "output": "0 35\r\n"}, {"input": "1\r\nR\r\n0\r\n", "output": "1 1\r\n"}, {"input": "3\r\nreka greka rak\r\n11\r\nrek rak\r\nrak grek\r\nreka rak\r\ngreka reka\r\nrak reka\r\nrak greka\r\ngreka rak\r\nlol rek\r\nlol rak\r\nLO lol\r\nABA BA\r\n", "output": "3 9\r\n"}, {"input": "3\r\nreka greka rak\r\n13\r\nrek rak\r\nrak grek\r\nreka rak\r\ngreka reka\r\nrak reka\r\nrak greka\r\ngreka rak\r\nlol rek\r\nlol rak\r\nlol LO\r\nABA BA\r\nLOLKA rak\r\nrak lol\r\n", "output": "0 6\r\n"}, {"input": "1\r\nr\r\n0\r\n", "output": "1 1\r\n"}, {"input": "5\r\nfEdOR Is A bAd BoY\r\n2\r\nboy boYy\r\nFeDor fedyaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\r\n", "output": "0 70\r\n"}, {"input": "1\r\nyrwlqadsfw\r\n2\r\nmnqdxczpyo a\r\na mnqdxczpyo\r\n", "output": "1 10\r\n"}, {"input": "4\r\nr rr rrr rrrr\r\n9\r\nrr rrr\r\nrrrr rr\r\nr rr\r\nr rrrr\r\nrrr rr\r\nrrr rrr\r\nrr rrr\r\nrr r\r\nr r\r\n", "output": "4 4\r\n"}]
false
stdio
null
true
959/B
959
B
Python 3
TESTS
2
30
0
212073868
n, k, m = map(int, input().split()) d = dict() x = list(map(str, input().split())) a = list(map(int, input().split())) #for i in range(n): # d[x[i]] = a[i] #print(d) for i in range(k): y = list(map( int, input().split())) if y[0] > 1: y = y[1::] for i in range(len(y)-1): #print(a[y[i]-1]) #print(a[y[i+1]-1]) if a[y[i]-1] > a[y[i+1]-1]: a[y[i]-1] = a[y[i+1]-1] else: a[y[i+1]-1] = a[y[i]-1] #print(a) for i in range(n): d[x[i]] = a[i] #print(d) s = input().split() ans = 0 for i in s: ans += d[i] print(ans)
22
249
40,038,400
189251762
import sys input = sys.stdin.buffer.readline n, k, m = map(int, input().split()) s = [x.decode() for x in input().split()] a = [int(x) for x in input().split()] for _ in range(k) : x, *b = map(int, input().split()) mn = min(a[i - 1] for i in b) for i in b : a[i - 1] = mn c = {x : i for i, x in enumerate(s)} s = [x.decode() for x in input().split()] print(sum(a[c[x]] for x in s))
Codeforces Round 473 (Div. 2)
CF
2,018
2
256
Mahmoud and Ehab and the message
Mahmoud wants to send a message to his friend Ehab. Their language consists of n words numbered from 1 to n. Some words have the same meaning so there are k groups of words such that all the words in some group have the same meaning. Mahmoud knows that the i-th word can be sent with cost ai. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message? The cost of sending the message is the sum of the costs of sending every word in it.
The first line of input contains integers n, k and m (1 ≤ k ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively. The second line contains n strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct. The third line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) where ai is the cost of sending the i-th word. The next k lines describe the groups of words of same meaning. The next k lines each start with an integer x (1 ≤ x ≤ n) which means that there are x words in this group, followed by x integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group. The next line contains m space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.
The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.
null
In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107. In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.
[{"input": "5 4 4\ni loser am the second\n100 1 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second", "output": "107"}, {"input": "5 4 4\ni loser am the second\n100 20 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second", "output": "116"}]
1,200
["dsu", "greedy", "implementation"]
22
[{"input": "5 4 4\r\ni loser am the second\r\n100 1 1 5 10\r\n1 1\r\n1 3\r\n2 2 5\r\n1 4\r\ni am the second\r\n", "output": "107"}, {"input": "5 4 4\r\ni loser am the second\r\n100 20 1 5 10\r\n1 1\r\n1 3\r\n2 2 5\r\n1 4\r\ni am the second\r\n", "output": "116"}, {"input": "1 1 1\r\na\r\n1000000000\r\n1 1\r\na\r\n", "output": "1000000000"}, {"input": "1 1 10\r\na\r\n1000000000\r\n1 1\r\na a a a a a a a a a\r\n", "output": "10000000000"}]
false
stdio
null
true
732/D
732
D
PyPy 3-64
TESTS
0
30
0
231111067
N,K=map(int,input().split()) A=[int(x)-1 for x in input().split()] P=[int(x) for x in input().split()] def judge(X): last_day = [-1]*K for d in range(X): if A[d]!=-1: last_day[A[d]]=d if min(last_day)==-1: return False last_day_ls = [(a,i) for i,a in enumerate(last_day)] last_day_ls.sort() U=P[:] s = 0 t = 0 for d in range(X): if d == last_day_ls[t][0]: t += 1 if t>s: return False else: U[last_day_ls[s][1]]-=1 if U[last_day_ls[s][1]]==0: s+=1 if s==K: return True ng = 0 ok = N+1 while ok-ng>1: mid=(ok+ng)//2 if judge(mid): ok=mid else: ng=mid if ok>K: print(-1) else: print(ok)
53
218
13,926,400
208657531
import sys from heapq import heappush, heappop # https://codeforces.com/problemset/problem/732/D params = sys.stdin.readline().split(' ') n = int(params[0]) m = int(params[1]) d = [int(d) for d in sys.stdin.readline().split(' ')] exams = [int(e) for e in sys.stdin.readline().split(' ')] # Minimal time for doing exams and study minimal_time = sum(exams) + len(exams) def isPossible(d, exams, max, m): # Set priority exams keys = {} for day in range(max): if d[day] != 0: keys[d[day]] = day if len(keys) < m : return False priority_queue = [] for key in keys: heappush(priority_queue, (keys[key], key)) # Check possibility possible = True time = 0 while possible and len(priority_queue) > 0: # Take the exams exam = heappop(priority_queue) # Learning time learning_duration = exams[exam[1] - 1] # Add to the time + time for the exam time += learning_duration + 1 # Check if it's always possible if time > exam[0] + 1 : possible = False return possible bestValue = -1 if minimal_time > n: print(-1) else : min = minimal_time max = n while min <= max : mid = min + ((max - min) // 2) possible = isPossible(d, exams, mid, m) # Too small number if not possible : min = mid + 1 else : bestValue = mid max = mid - 1 print(bestValue)
Codeforces Round 377 (Div. 2)
CF
2,016
1
256
Exams
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m. About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day. On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest. About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way. Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects. The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i. The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
null
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day. In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day. In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
[{"input": "7 2\n0 1 0 2 1 0 2\n2 1", "output": "5"}, {"input": "10 3\n0 0 1 2 3 0 2 0 1 2\n1 1 4", "output": "9"}, {"input": "5 1\n1 1 1 1 1\n5", "output": "-1"}]
1,700
["binary search", "greedy", "sortings"]
53
[{"input": "7 2\r\n0 1 0 2 1 0 2\r\n2 1\r\n", "output": "5\r\n"}, {"input": "10 3\r\n0 0 1 2 3 0 2 0 1 2\r\n1 1 4\r\n", "output": "9\r\n"}, {"input": "5 1\r\n1 1 1 1 1\r\n5\r\n", "output": "-1\r\n"}, {"input": "100 10\r\n1 1 6 6 6 2 5 7 6 5 3 7 10 10 8 9 7 6 9 2 6 7 8 6 7 5 2 5 10 1 10 1 8 10 2 9 7 1 6 8 3 10 9 4 4 8 8 6 6 1 5 5 6 5 6 6 6 9 4 7 5 4 6 6 1 1 2 1 8 10 6 2 1 7 2 1 8 10 9 2 7 3 1 5 10 2 8 10 10 10 8 9 5 4 6 10 8 9 6 6\r\n2 4 10 11 5 2 6 7 2 15\r\n", "output": "74\r\n"}, {"input": "1 1\r\n1\r\n1\r\n", "output": "-1\r\n"}, {"input": "3 2\r\n0 0 0\r\n2 1\r\n", "output": "-1\r\n"}, {"input": "4 2\r\n0 1 0 2\r\n1 1\r\n", "output": "4\r\n"}, {"input": "10 1\r\n0 1 0 0 0 0 0 0 0 1\r\n1\r\n", "output": "2\r\n"}, {"input": "5 1\r\n0 0 0 0 1\r\n1\r\n", "output": "5\r\n"}, {"input": "7 2\r\n0 0 0 0 0 1 2\r\n1 1\r\n", "output": "7\r\n"}, {"input": "10 3\r\n0 0 1 2 2 0 2 0 1 3\r\n1 1 4\r\n", "output": "10\r\n"}, {"input": "6 2\r\n1 1 1 1 1 2\r\n1 1\r\n", "output": "6\r\n"}, {"input": "6 2\r\n1 0 0 0 0 2\r\n1 1\r\n", "output": "-1\r\n"}]
false
stdio
null
true
19/A
19
A
PyPy 3
TESTS
3
122
0
164372330
a = [] n = int(input()) for i in range(0, n): a.append(input()) points = [] for j in range(0, n*(n-1)//2): s = input() teams = [] goals = [] d = "" for k in range(0, len(s)): if s[k] == "-" or s[k] == " ": teams.append(d) d = "" if s[k] == " ": break else: d += s[k] e = "" for m in range(k+1, len(s)): if s[m] == ":": goals.append(int(e)) e = "" else: e += s[m] goals.append(int(e)) if goals[0] > goals[1]: points.append([teams[0], str(3), str(goals[0])]) points.append([teams[1], str(0), str(goals[1])]) elif goals[0] < goals[1]: points.append([teams[0], str(0), str(goals[0])]) points.append([teams[1], str(3), str(goals[1])]) else: points.append([teams[0], str(1), str(goals[0])]) points.append([teams[1], str(1), str(goals[1])]) totalpoints = [] b = [] for j in range(0, len(points)): if points[j][0] not in b: b.append(points[j][0]) totpoi = 0 goalie = 0 for k in range(0, len(points)): if points[k][0] == points[j][0]: totpoi += int(points[k][1]) goalie += int(points[k][2]) totalpoints.append([points[j][0], str(totpoi), str(goalie)]) totalpoints.sort(key=lambda x: int(x[1]), reverse=True) for j in range(0, len(totalpoints)): for k in range(0, len(totalpoints)-j-1): if int(totalpoints[k][1]) == int(totalpoints[k+1][1]): if int(totalpoints[k][2]) < int(totalpoints[k+1][2]): totalpoints[k], totalpoints[k+1] = totalpoints[k+1], totalpoints[k] for j in range((n//2)-1, -1, -1): print(totalpoints[j][0])
29
92
0
195020000
d = {} n = int(input()) for _ in range(n): # idx 0: pts # idx 1: diff # idx 2: goals scored d[input()] = [0, 0, 0] for _ in range(n*(n-1) // 2): pair, result = input().split() team1, team2 = pair.split('-') sc1, sc2 = map(int, result.split(':')) d[team1][1] += sc1 - sc2 d[team1][2] += sc1 d[team2][1] += sc2 - sc1 d[team2][2] += sc2 if sc1 == sc2: d[team1][0] += 1 d[team2][0] += 1 elif sc1 > sc2: d[team1][0] += 3 else: d[team2][0] += 3 res = sorted(d, key=d.get, reverse=True)[:n//2] print(*sorted(res), sep='\n') ######################################### ## ## ## Implemented by brownfox2k6 ## ## ## #########################################
Codeforces Beta Round 19
ICPC
2,010
2
64
World Football Cup
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations: - the final tournament features n teams (n is always even) - the first n / 2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity. You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
The first input line contains the only integer n (1 ≤ n ≤ 50) — amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following n·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 — names of the teams; num1, num2 (0 ≤ num1, num2 ≤ 100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output n / 2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
null
null
[{"input": "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3", "output": "A\nD"}, {"input": "2\na\nA\na-A 2:1", "output": "a"}]
1,400
["implementation"]
29
[{"input": "4\r\nA\r\nB\r\nC\r\nD\r\nA-B 1:1\r\nA-C 2:2\r\nA-D 1:0\r\nB-C 1:0\r\nB-D 0:3\r\nC-D 0:3\r\n", "output": "A\r\nD\r\n"}, {"input": "2\r\na\r\nA\r\na-A 2:1\r\n", "output": "a\r\n"}, {"input": "2\r\nEULEUbCmfrmqxtzvg\r\nuHGRmKUhDcxcfqyruwzen\r\nuHGRmKUhDcxcfqyruwzen-EULEUbCmfrmqxtzvg 13:92\r\n", "output": "EULEUbCmfrmqxtzvg\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 2:2\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "A\r\nB\r\nF\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 7:7\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "B\r\nC\r\nF\r\n"}]
false
stdio
null
true
320/B
320
B
Python 3
TESTS
3
92
0
193923340
def insert(a, b, num): #who are my neighbors? connections[num] = set() for v in vertices: if vertices[v][0]<a<vertices[v][1] or vertices[v][0]<b<vertices[v][1]: connections[num].add(v) connections[v].add(num) vertices[num] = (a, b) #who are my connections? visited = [0]*(num+1) queue = [] for v in connections[num]: queue.append(v) while len(queue)>0: v = queue.pop(0) if visited[v]: continue connections[num].update(connections[v]) connections[v].add(num) visited[v] = 1 for c in connections[v]: queue.append(c) for c in connections[num]: connections[c].update(connections[num]) n = int(input()) #1 = insert #2 = query vertices = {} connections = {} count = 1 for q in range(n): query, x, y = input().split(" ") if(int(query)==2): if int(y) in connections[int(x)]: print("YES") else: print("NO") else: insert(int(x), int(y), count) count+=1
24
92
0
192405239
import sys intervals = [] def depth_first_search(z, temp): if (temp[z] != -1): return temp[z] temp[z] = 0 for i in range(len(intervals)): if ((intervals[z][0] > intervals[i][0]) and (intervals[z][0] < intervals[i][1])) or \ ((intervals[z][1] > intervals[i][0]) and (intervals[z][1] < intervals[i][1])): temp[z] = temp[z] or depth_first_search(i, temp) return temp[z] def ping_pong(intervals): in_queries = int(input()) i = 0 while i < in_queries: current_line = list(map(int, input().split())) t = current_line[0] x = current_line[1] y = current_line[2] if t == 1: intervals += [[x,y]] else: result = 0 # Search temp = [-1] * in_queries temp[y-1] = 1 result = depth_first_search(x - 1, temp) if result == 0: print("NO") else: print("YES") i += 1 ping_pong(intervals)
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
320/B
320
B
PyPy 3-64
TESTS
3
92
0
222823967
from collections import defaultdict def insert(graph, intervals, count, a, b): if count not in graph: graph[count] = set() for node in graph: (x, y) = intervals[node] if x < a < y or x < b < y: graph[node].add(count) graph[count].add(node) def dfs(graph, src, dst, seen): if src in seen: return False if dst in graph[src]: return True seen.add(src) for nei in graph[src]: if dfs(graph, nei, dst, seen): return True seen.remove(src) return False graph = {} intervals = {1: (1, 5), 2: (5, 11), 3: (2, 9)} count = 0 for _ in range(int(input())): query, a, b = list(map(int, input().split())) if query == 1: # add (a, b) to the set count += 1 intervals[count] = (a, b) insert(graph, intervals, count, a, b) else: if dfs(graph, a, b, set()): print("YES") else: print("NO")
24
92
0
207542094
# /** # * author: brownfox2k6 # * created: 28/05/2023 14:35:05 Hanoi, Vietnam # **/ intervals = [[-1, -1]] size = 1 for _ in range(int(input())): a, x, y = map(int, input().split()) if a == 1: intervals.append([x, y]) size += 1 else: vis = [False for _ in range(size)] st = [x] while st: i = st.pop() if i == y: print("YES") break vis[i] = True for j in range(1, size): if i != j and not vis[j] and (intervals[j][0] < intervals[i][0] < intervals[j][1] or intervals[j][0] < intervals[i][1] < intervals[j][1]): st.append(j) else: print("NO")
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
320/B
320
B
PyPy 3-64
TESTS
3
92
0
223538769
nodes = [] parents = [] ranks = [] def findParent(spot): return spot if parents[spot] == spot else findParent(parents[spot]) def union(spot1, spot2): parent1 = findParent(spot1) parent2 = findParent(spot2) if parent1 != parent2: if ranks[parent1] >= ranks[parent2]: parents[parent2] = parent1 ranks[parent1] = max(ranks[parent1], ranks[parent2]+1) else: parents[parent1] = parent2 for _ in range(int(input())): a, b, c = list(map(int, input().split())) if a == 1: ranks.append(1) parents.append(len(nodes)) for index, node in enumerate(nodes): if node[0]<b<node[1] or node[0]<c<node[1]: union(index, len(nodes)) nodes.append((b,c)) else: print("YES" if findParent(b-1) == findParent(c-1) else "NO")
24
92
0
224539159
def add_interval(graph, intervals, x, y): interval_id = len(intervals) intervals.append((x, y)) graph.append([]) for i, interval in enumerate(intervals[:-1]): if (x < interval[0] < y) or (x < interval[1] < y): graph[i].append(interval_id) if (interval[0] < x < interval[1]) or (interval[0] < y < interval[1]): graph[interval_id].append(i) def does_path_exist(graph, a, b): visited = [False] * len(graph) def dfs(node): visited[node] = True if node == b: return True for neighbor in graph[node]: if not visited[neighbor]: if dfs(neighbor): return True return False return "YES" if dfs(a) else "NO" n = int(input()) graph = [] intervals = [] for _ in range(n): query = input().split() if query[0] == "1": x, y = map(int, query[1:]) add_interval(graph, intervals, x, y) elif query[0] == "2": a, b = map(int, query[1:]) print(does_path_exist(graph, a - 1, b - 1))
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
959/D
959
D
PyPy 3
TESTS
2
234
26,009,600
177903058
import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline def fast_prime_factorization(n): d = [(i + 1) % 2 * 2 for i in range(n + 1)] d[0], d[1] = 0, 1 for i in range(3, n + 1): if d[i]: continue for j in range(i, n + 1, 2 * i): if not d[j]: d[j] = i return d n = int(input()) a = list(map(int, input().split())) l = 3 * pow(10, 6) d = fast_prime_factorization(l) visit = [0] * l ans = [] for i in a: u = i s = set() while d[u] ^ u: s.add(u // d[u]) u //= d[u] if u ^ 1: s.add(u) ok = 1 for j in s: if visit[j]: ok = 0 if ok: ans.append(i) for j in s: visit[j] = 1 else: j = i + 1 while d[j] ^ j or visit[j]: j += 1 ans.append(j) visit[j] = 1 break i = 2 while len(ans) ^ n: if d[i] == i and not visit[i]: ans.append(i) i += 1 sys.stdout.write(" ".join(map(str, ans)))
55
1,091
127,283,200
201584351
import sys from array import array input = lambda: sys.stdin.buffer.readline().decode().rstrip() inp = lambda dtype: [dtype(x) for x in input().split()] debug = lambda *x: print(*x, file=sys.stderr) ceil_ = lambda a, b: (a + b - 1) // b sum_n = lambda n: (n * (n + 1)) // 2 get_bit = lambda x, i: (x >> i) & 1 Mint, Mlong, out = 2 ** 30 - 1, 2 ** 62 - 1, [] def count_prime(n): prim = array('i', [0] * (n + 1)) for i in range(2, n + 1): if not prim[i]: for j in range(i, n + 1, i): prim[j] = i return prim def prime_factor(x): fac = set() while x > 1: div = prim[x] fac.add(div) while x % div == 0: x //= div return fac def solve(i): global primes while True: cfacs = prime_factor(ans[i]) for j in cfacs: if j in primes: break else: primes |= cfacs break ans[i] += 1 def finish(ix): cur = 2 for i in range(ix, n): ans.append(cur) solve(i) cur = ans[-1] + 1 prim = count_prime(10 ** 7) for _ in range(1): n, a = int(input()), array('i', inp(int)) primes, ans = set(), array('i') for i in range(n): ans.append(a[i]) solve(i) if a[i] < ans[i]: finish(i + 1) break out.append(' '.join(map(str, ans))) print('\n'.join(map(str, out)))
Codeforces Round 473 (Div. 2)
CF
2,018
3
256
Mahmoud and Ehab and another array construction task
Mahmoud has an array a consisting of n integers. He asked Ehab to find another array b of the same length such that: - b is lexicographically greater than or equal to a. - bi ≥ 2. - b is pairwise coprime: for every 1 ≤ i < j ≤ n, bi and bj are coprime, i. e. GCD(bi, bj) = 1, where GCD(w, z) is the greatest common divisor of w and z. Ehab wants to choose a special array so he wants the lexicographically minimal array between all the variants. Can you find it? An array x is lexicographically greater than an array y if there exists an index i such than xi > yi and xj = yj for all 1 ≤ j < i. An array x is equal to an array y if xi = yi for all 1 ≤ i ≤ n.
The first line contains an integer n (1 ≤ n ≤ 105), the number of elements in a and b. The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 105), the elements of a.
Output n space-separated integers, the i-th of them representing bi.
null
Note that in the second sample, the array is already pairwise coprime so we printed it.
[{"input": "5\n2 3 5 4 13", "output": "2 3 5 7 11"}, {"input": "3\n10 3 7", "output": "10 3 7"}]
1,900
["constructive algorithms", "greedy", "math", "number theory"]
55
[{"input": "5\r\n2 3 5 4 13\r\n", "output": "2 3 5 7 11 "}, {"input": "3\r\n10 3 7\r\n", "output": "10 3 7 "}, {"input": "5\r\n7 10 2 5 5\r\n", "output": "7 10 3 11 13 "}, {"input": "7\r\n20 9 7 6 7 9 15\r\n", "output": "20 9 7 11 13 17 19 "}, {"input": "10\r\n5 3 2 2 3 3 3 4 2 5\r\n", "output": "5 3 2 7 11 13 17 19 23 29 "}, {"input": "3\r\n3 18 2\r\n", "output": "3 19 2 "}]
false
stdio
null
true
777/C
777
C
PyPy 3-64
TESTS
2
248
29,593,600
174562230
import sys input = sys.stdin.buffer.readline def process(A, Q): n = len(A) m = len(A[0]) k = len(Q) queries = [[] for i in range(n+1)] answer = ['No' for i in range(k)] for i in range(k): l, r = Q[i] if l==r: answer[i] = 'Yes' else: queries[r].append([l, i]) for i in range(n+1): queries[r] = sorted(queries[r]) for j in range(m): curr = None for i in range(1, n+1): if i > 1: if A[i-2][j] > A[i-1][j]: curr = i while len(queries[i]) > 0 and (curr is None or queries[i][-1][0] >= curr): l, i1 = queries[i].pop() answer[i1] = 'Yes' for x in answer: sys.stdout.write(f'{x}\n') n, m = [int(x) for x in input().split()] A = [] for i in range(n): row = [int(x) for x in input().split()] A.append(row) k = int(input()) Q = [] for i in range(k): l, r = [int(x) for x in input().split()] Q.append([l, r]) process(A, Q)
114
264
34,099,200
230765900
import sys input = sys.stdin.readline m, n = map(int, input().split()) grid = [list(map(int, input().split())) for _ in range(m)] dp = [[0] * n for _ in range(m)] mx = [0] * m for i in range(m - 2,-1,-1): for j in range(n): if grid[i][j] <= grid[i + 1][j]: dp[i][j] = dp[i + 1][j] + 1 mx[i] = max(dp[i]) q = int(input()) for _ in range(q): l,r = map(int, input().split()) l -= 1 r -= 1 if mx[l] >= r - l: print("Yes") else: print("No")
Codeforces Round 401 (Div. 2)
CF
2,017
1
256
Alyona and Spreadsheet
During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables. Now she has a table filled with integers. The table consists of n rows and m columns. By ai, j we will denote the integer located at the i-th row and the j-th column. We say that the table is sorted in non-decreasing order in the column j if ai, j ≤ ai + 1, j for all i from 1 to n - 1. Teacher gave Alyona k tasks. For each of the tasks two integers l and r are given and Alyona has to answer the following question: if one keeps the rows from l to r inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such j that ai, j ≤ ai + 1, j for all i from l to r - 1 inclusive. Alyona is too small to deal with this task and asks you to help!
The first line of the input contains two positive integers n and m (1 ≤ n·m ≤ 100 000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table. Each of the following n lines contains m integers. The j-th integers in the i of these lines stands for ai, j (1 ≤ ai, j ≤ 109). The next line of the input contains an integer k (1 ≤ k ≤ 100 000) — the number of task that teacher gave to Alyona. The i-th of the next k lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n).
Print "Yes" to the i-th line of the output if the table consisting of rows from li to ri inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".
null
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3.
[{"input": "5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5", "output": "Yes\nNo\nYes\nYes\nYes\nNo"}]
1,600
["binary search", "data structures", "dp", "greedy", "implementation", "two pointers"]
114
[{"input": "5 4\r\n1 2 3 5\r\n3 1 3 2\r\n4 5 2 3\r\n5 5 3 2\r\n4 4 3 4\r\n6\r\n1 1\r\n2 5\r\n4 5\r\n3 5\r\n1 3\r\n1 5\r\n", "output": "Yes\r\nNo\r\nYes\r\nYes\r\nYes\r\nNo\r\n"}, {"input": "1 1\r\n1\r\n1\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "10 1\r\n523130301\r\n127101624\r\n15573616\r\n703140639\r\n628818570\r\n957494759\r\n161270109\r\n386865653\r\n67832626\r\n53360557\r\n17\r\n4 5\r\n4 7\r\n8 8\r\n9 9\r\n3 9\r\n8 10\r\n8 9\r\n7 9\r\n4 5\r\n2 9\r\n4 6\r\n2 4\r\n2 6\r\n4 6\r\n7 9\r\n2 4\r\n8 10\r\n", "output": "No\r\nNo\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\n"}, {"input": "15 1\r\n556231456\r\n573340933\r\n626155933\r\n397229387\r\n10255952\r\n376567394\r\n906742013\r\n269437009\r\n31298788\r\n712285290\r\n620239975\r\n379221898\r\n229140718\r\n95080095\r\n997123854\r\n18\r\n5 15\r\n1 12\r\n4 10\r\n2 15\r\n12 15\r\n15 15\r\n2 2\r\n15 15\r\n15 15\r\n13 13\r\n10 14\r\n3 6\r\n14 15\r\n3 6\r\n4 4\r\n14 15\r\n12 14\r\n1 9\r\n", "output": "No\r\nNo\r\nNo\r\nNo\r\nNo\r\nYes\r\nYes\r\nYes\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\nNo\r\nYes\r\nYes\r\nNo\r\nNo\r\n"}, {"input": "11 1\r\n501465490\r\n366941771\r\n415080944\r\n385243536\r\n445132523\r\n697044413\r\n894369800\r\n812743722\r\n23684788\r\n466526046\r\n953916313\r\n45\r\n2 4\r\n8 9\r\n7 7\r\n4 9\r\n2 9\r\n2 11\r\n4 4\r\n5 7\r\n1 2\r\n5 10\r\n4 6\r\n1 7\r\n4 4\r\n1 6\r\n4 7\r\n10 11\r\n1 8\r\n6 11\r\n8 8\r\n8 10\r\n1 1\r\n5 10\r\n9 10\r\n6 9\r\n6 11\r\n1 1\r\n9 9\r\n5 11\r\n1 2\r\n9 11\r\n2 6\r\n3 7\r\n11 11\r\n6 7\r\n11 11\r\n7 8\r\n5 8\r\n11 11\r\n5 6\r\n4 5\r\n2 6\r\n5 10\r\n9 9\r\n1 1\r\n1 1\r\n", "output": "No\r\nNo\r\nYes\r\nNo\r\nNo\r\nNo\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\nNo\r\nYes\r\nNo\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\nNo\r\nYes\r\nNo\r\nYes\r\nNo\r\nNo\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\nNo\r\nNo\r\nYes\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\nYes\r\nYes\r\nNo\r\nNo\r\nYes\r\nYes\r\nYes\r\n"}]
false
stdio
null
true
652/F
652
F
PyPy 3-64
TESTS
3
62
0
176110242
n,l,t=map(int,input().split()) ants=[] last=[] xwi=[] for i in range(n): x,w=input().split() x=int(x) if w=='L': w=2 else: w=1 xwi.append((x,w,i)) xwi.sort(key=lambda x:x[0]) for i in range(n): x,w=xwi[i][:2] ants.append((x,w)) if w==1: last.append(((x+t)%l,1,i)) else: last.append(((x-t)%l,0,i)) ans=[] if ants[0][1]==1: last.sort() for i in range(n): if last[i][2]==0: idx0=i x0=ants[0][0] cross=0 for x,w in ants[1:]: if w==2: cross+=(l-(x-x0)+2*t)//l cross%=n for i in range(n): ans.append(last[(idx0-cross+i)%n][0]) else: last.sort() for i in range(n): if last[i][2]==0: idx0=i x0=ants[0][0] cross=0 for x,w in ants[1:]: if w==1: cross+=(x-x0+2*t)//l cross%=n for i in range(n): ans.append(last[(idx0+cross+i)%n][0]) print(*[ans[xwi[i][2]] for i in range(n)])
30
1,357
45,772,800
144095748
I = lambda: [int(i) for i in input().split()] import sys input = sys.stdin.readline n, L, T = I() a, b = [0]*n, [0]*n ans = [0] * n cnt = 0 for i in range(n): x, ch = input().split() a[i] = int(x) - 1 b[i] = (a[i], i) if ch == "R": w = a[i] + T a[i] = (a[i] + T) % L else: w = a[i] - T a[i] = (a[i] - T) % L cnt += w//L 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
626/D
626
D
Python 3
TESTS
4
249
0
20326015
def main(): n = int(input()) a = list(map(int, input().split())) max_element = max(a) + 1 diff_freq = [0 for i in range(max_element)] a.sort() for i in range(n): for j in range(i): diff_freq[a[i] - a[j]] += 1 largest = [0 for i in range(max_element)] for i in range(max_element): for j in range(i + 1, max_element): largest[i] += diff_freq[j] good_ones = 0 # print('diff_freq', diff_freq) # print('largest', largest) for i in range(max_element): for j in range(i + 1): if i + j < max_element: good_ones += diff_freq[i] * diff_freq[j] * largest[i + j] # print(good_ones) ans = good_ones / ((n*(n - 1) / 2) ** 3) print(ans) main()
34
186
3,481,600
178622675
n = int(input()) a = [int(x) for x in input().split()] diff = [0] * 5000 for i in range(n): for j in range(i + 1, n): (A, B) = (a[i], a[j]) if a[i] < a[j] else (a[j], a[i]) diff[B - A] += 1 suff = [0] * 5000 suff[-1] = diff[-1] for i in range(len(suff) - 2, -1, -1): suff[i] = diff[i] + suff[i+1] ans = 0 for i in range(5000): for j in range(5000): if i + j + 1 >= 5000: continue ans += diff[i] * diff[j] * suff[i + j + 1] total = (n*(n-1)//2)**3 print("{:.10f}".format(ans/total))
8VC Venture Cup 2016 - Elimination Round
CF
2,016
2
256
Jerry's Protest
Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds. In each round, Andrew and Jerry draw randomly without replacement from a jar containing n balls, each labeled with a distinct positive integer. Without looking, they hand their balls to Harry, who awards the point to the player with the larger number and returns the balls to the jar. The winner of the game is the one who wins at least two of the three rounds. Andrew wins rounds 1 and 2 while Jerry wins round 3, so Andrew wins the game. However, Jerry is unhappy with this system, claiming that he will often lose the match despite having the higher overall total. What is the probability that the sum of the three balls Jerry drew is strictly higher than the sum of the three balls Andrew drew?
The first line of input contains a single integer n (2 ≤ n ≤ 2000) — the number of balls in the jar. The second line contains n integers ai (1 ≤ ai ≤ 5000) — the number written on the ith ball. It is guaranteed that no two balls have the same number.
Print a single real value — the probability that Jerry has a higher total, given that Andrew wins the first two rounds and Jerry wins the third. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if $$\frac{|a-b|}{\max(1,b)} \leq 10^{-6}$$.
null
In the first case, there are only two balls. In the first two rounds, Andrew must have drawn the 2 and Jerry must have drawn the 1, and vice versa in the final round. Thus, Andrew's sum is 5 and Jerry's sum is 4, so Jerry never has a higher total. In the second case, each game could've had three outcomes — 10 - 2, 10 - 1, or 2 - 1. Jerry has a higher total if and only if Andrew won 2 - 1 in both of the first two rounds, and Jerry drew the 10 in the last round. This has probability $$\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{27}$$.
[{"input": "2\n1 2", "output": "0.0000000000"}, {"input": "3\n1 2 10", "output": "0.0740740741"}]
1,800
["brute force", "combinatorics", "dp", "probabilities"]
34
[{"input": "2\r\n1 2\r\n", "output": "0.0000000000\r\n"}, {"input": "3\r\n1 2 10\r\n", "output": "0.0740740741\r\n"}, {"input": "3\r\n1 2 3\r\n", "output": "0.0000000000\r\n"}, {"input": "4\r\n2 4 1 3\r\n", "output": "0.0416666667\r\n"}, {"input": "11\r\n1 2 4 8 16 32 64 128 256 512 1024\r\n", "output": "0.2773433509\r\n"}, {"input": "18\r\n69 88 49 91 42 56 58 13 38 93 77 99 18 32 82 81 92 46\r\n", "output": "0.1974898208\r\n"}, {"input": "16\r\n14 34 22 53 11 40 50 12 91 86 32 75 36 33 41 9\r\n", "output": "0.2145798611\r\n"}, {"input": "11\r\n54 91 4 6 88 23 38 71 77 26 84\r\n", "output": "0.1890428249\r\n"}, {"input": "20\r\n11 37 81 56 61 2 26 27 59 69 24 7 71 76 45 54 89 17 95 20\r\n", "output": "0.1931218837\r\n"}, {"input": "19\r\n55 91 63 21 30 82 14 38 9 67 50 44 90 93 68 6 73 65 29\r\n", "output": "0.1920376960\r\n"}, {"input": "13\r\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096\r\n", "output": "0.2881011986\r\n"}, {"input": "15\r\n1 2 3 4 5 10 20 40 80 160 320 640 1280 2560 5000\r\n", "output": "0.2950679192\r\n"}, {"input": "4\r\n1000 5000 1001 4999\r\n", "output": "0.0925925926\r\n"}, {"input": "10\r\n1000 1001 1002 1003 1004 1005 4001 4002 4003 4004\r\n", "output": "0.1796872428\r\n"}]
false
stdio
import sys def main(): if len(sys.argv) != 4: print("Usage: checker.py input_path output_path submission_output_path") sys.exit(1) input_path = sys.argv[1] output_path = sys.argv[2] submission_path = sys.argv[3] # Read reference output try: with open(output_path, 'r') as f: ref_lines = [line.strip() for line in f.readlines()] if len(ref_lines) != 1: print(0) return ref_str = ref_lines[0] except: print(0) return try: b = float(ref_str) except: print(0) return # Read submission output try: with open(submission_path, 'r') as f: sub_lines = [line.strip() for line in f.readlines()] if len(sub_lines) != 1: print(0) return sub_str = sub_lines[0] except: print(0) return try: a = float(sub_str) except: print(0) return # Compare a and b according to problem's criteria delta = abs(a - b) denominator = max(1.0, abs(b)) if delta / denominator <= 1e-6: print(1) else: print(0) if __name__ == "__main__": main()
true
959/D
959
D
Python 3
TESTS
2
30
0
170709356
import collections import heapq import sys import math import itertools import bisect from io import BytesIO, IOBase import os ###################################################################################### #--------------------------------------func-----------------------------------------# ###################################################################################### def valid(i,j,n,m): if i<n and i>=0 and j>=0 and j< m :return True #and l[i][j]==1 and visit[i][j] return False def sumn(i,n): return (n-i)*(i+n)/2 def sqfun(a,b,c): return (-b+math.sqrt(b*b-4*a*c))/2*a ###################################################################################### #--------------------------------------vars-----------------------------------------# ###################################################################################### # index=[[1,0],[0,1],[-1,0],[0,-1]] ###################################################################################### #--------------------------------------Input-----------------------------------------# ###################################################################################### def value(): return tuple(map(int, input().split())) def values(): return tuple(map(int, sys.stdin.readline().split())) def inlst(): return [int(i) for i in input().split()] def inlsts(): return [int(i) for i in sys.stdin.readline().split()] def inp(): return int(input()) def inps(): return int(sys.stdin.readline()) def instr(): return input() def stlst(): return [i for i in input().split()] ###################################################################################### #--------------------------------------code here-------------------------------------# ###################################################################################### def get(num): while True: if num not in s and all(num%i!=0 for i in range(2,int(math.sqrt(num))+1)): return (num) num+=1 def solve(): global s n = inp() l=inlsts() b=[l[0]] s=set() s.add(b[0]) bb=False for i in range(1,n): if not bb:b.append(get((l[i]))) else:b.append(get(2)) if b[-1]>l[i]:bb=True s.add(b[-1]) print (*b) if __name__ == "__main__": # for i in range(inp()): solve()
55
1,107
13,004,800
161949830
def get_primes(n): sieve = [True] * n p = [] for i in range(2, n): if sieve[i]: p.append(i) for j in range(i * i, n, i): sieve[j] = False return p def prime_factors(a, p): f = [] for x in p: if x * x > a: break if a % x == 0: f.append(x) while a % x == 0: a //= x if a > 1: f.append(a) return f p = get_primes(1310000) n = int(input()) factors = set() b = [] for i, x in enumerate(map(int, input().split())): f = prime_factors(x, p) if factors.isdisjoint(f): b.append(x) factors.update(f) continue x += 1 while not factors.isdisjoint(f := prime_factors(x, p)): x += 1 b.append(x) factors.update(f) for f in factors: p.remove(f) b += p[: n - len(b)] break print(" ".join(str(x) for x in b))
Codeforces Round 473 (Div. 2)
CF
2,018
3
256
Mahmoud and Ehab and another array construction task
Mahmoud has an array a consisting of n integers. He asked Ehab to find another array b of the same length such that: - b is lexicographically greater than or equal to a. - bi ≥ 2. - b is pairwise coprime: for every 1 ≤ i < j ≤ n, bi and bj are coprime, i. e. GCD(bi, bj) = 1, where GCD(w, z) is the greatest common divisor of w and z. Ehab wants to choose a special array so he wants the lexicographically minimal array between all the variants. Can you find it? An array x is lexicographically greater than an array y if there exists an index i such than xi > yi and xj = yj for all 1 ≤ j < i. An array x is equal to an array y if xi = yi for all 1 ≤ i ≤ n.
The first line contains an integer n (1 ≤ n ≤ 105), the number of elements in a and b. The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 105), the elements of a.
Output n space-separated integers, the i-th of them representing bi.
null
Note that in the second sample, the array is already pairwise coprime so we printed it.
[{"input": "5\n2 3 5 4 13", "output": "2 3 5 7 11"}, {"input": "3\n10 3 7", "output": "10 3 7"}]
1,900
["constructive algorithms", "greedy", "math", "number theory"]
55
[{"input": "5\r\n2 3 5 4 13\r\n", "output": "2 3 5 7 11 "}, {"input": "3\r\n10 3 7\r\n", "output": "10 3 7 "}, {"input": "5\r\n7 10 2 5 5\r\n", "output": "7 10 3 11 13 "}, {"input": "7\r\n20 9 7 6 7 9 15\r\n", "output": "20 9 7 11 13 17 19 "}, {"input": "10\r\n5 3 2 2 3 3 3 4 2 5\r\n", "output": "5 3 2 7 11 13 17 19 23 29 "}, {"input": "3\r\n3 18 2\r\n", "output": "3 19 2 "}]
false
stdio
null
true
320/B
320
B
PyPy 3-64
TESTS
3
124
0
202099216
import sys input = lambda: sys.stdin.readline().rstrip() class UnionFind: def __init__(self, n): self.parent = list(range(n)) def find(self, a): acopy = a while a != self.parent[a]: a = self.parent[a] while acopy != a: self.parent[acopy], acopy = a, self.parent[acopy] return a def merge(self, a, b): self.parent[self.find(b)] = self.find(a) N = int(input()) uni = UnionFind(N) A = [] for _ in range(N): t,x,y = map(int, input().split()) if t==1: m = len(A) for i,(a,b) in enumerate(A): if a<x<b or a<y<b: uni.merge(i,m) A.append((x,y)) else: if uni.find(x-1)==uni.find(y-1): print('YES') else: print('NO')
24
92
0
224539518
def add_interval(graph, intervals, x, y): interval_id = len(intervals) intervals.append((x, y)) graph.append([]) for i, interval in enumerate(intervals[:-1]): if (x < interval[0] < y) or (x < interval[1] < y): graph[i].append(interval_id) if (interval[0] < x < interval[1]) or (interval[0] < y < interval[1]): graph[interval_id].append(i) def dfs(graph, visited, a, b): visited[a] = True if a == b: return True for neighbor in graph[a]: if not visited[neighbor]: if dfs(graph, visited, neighbor, b): return True return False def does_path_exists(graph, a, b): visited = [False] * len(graph) return "YES" if dfs(graph, visited, a, b) else "NO" n = int(input()) graph = [] intervals = [] for _ in range(n): query = input().split() if query[0] == "1": x, y = map(int, query[1:]) add_interval(graph, intervals, x, y) elif query[0] == "2": a, b = map(int, query[1:]) print(does_path_exists(graph, a - 1, b - 1))
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
320/B
320
B
PyPy 3-64
TESTS
3
124
0
210664775
from collections import deque def bfs(g, start, finish): q = deque() visited = [] q.append(start) while q: node = q.popleft() if node == finish: return True if node not in visited: visited.append(node) for v in g[node]: q.append(v) return False def main(): n = int(input()) g = {i:[] for i in range(n)} pairs = [] for i in range(n): t = list(map(int, input().split())) if t[0] == 1: pairs.append([t[1], t[2]]) for j in range(len(pairs)-1): if pairs[j][0]<pairs[-1][0]<pairs[j][1] or pairs[j][0]<pairs[-1][1]<pairs[j][1]: g[j].append(len(pairs)-1) g[len(pairs)-1].append(j) else: res = bfs(g, t[1]-1, t[2]-1) if res == False: print('NO') else: print('YES') if __name__ == "__main__": main()
24
92
0
225183589
graph = [] def build_graph(interval_list): graph = [[] for _ in range(len(interval_list))] for i in range(len(interval_list)): for j in range(len(interval_list)): if i != j: if (interval_list[j][0] < interval_list[i][0] and interval_list[i][0] < interval_list[j][1]) or (interval_list[j][0] < interval_list[i][1] and interval_list[i][1] < interval_list[j][1]): graph[i].append(j) return graph def dfs(graph, start, end): visited = [False] * len(graph) stack = [start] while stack: cur = stack.pop() if cur == end: return True if not visited[cur]: visited[cur] = True for neighbor in graph[cur]: stack.append(neighbor) return False n = int(input()) interval_list = [] graph = [] for _ in range(n): op, x, y = input().split() x = int(x); y = int(y) if op == '1': interval_list.append([x, y]) graph = build_graph(interval_list) #print(graph) else: if dfs(graph, x - 1, y - 1): print("YES") else: print("NO")
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
320/B
320
B
PyPy 3
TESTS
3
122
0
146674497
def DFS(a, b) -> bool: # int_a = intervals[a - 1] # int_b = intervals[b - 1] visited = [False] * len(intervals) return DFS_helper(a, b, visited) def DFS_helper(a, b, visited) -> bool: if visited[a - 1]: return False interval_a = intervals[a - 1] interval_b = intervals[b - 1] a_start, a_end = interval_a[0], interval_a[1] b_start, b_end = interval_b[0], interval_b[1] if a_start == b_start and a_end == b_end: return True visited[a - 1] = True # do all possible moves for i in range(len(intervals)): interval = intervals[i] n = i + 1 n_start = interval[0] n_end = interval[1] if n_start > a_start and n_start < a_end: if (DFS_helper(n, b, visited)): return True elif n_end > a_start and n_end < a_end: if (DFS_helper(n, b, visited)): return True return False num_q = int(input()) intervals = list() for i in range(num_q): q = input().strip().split(" ") q_type = int(q[0]) a = int(q[1]) b = int(q[2]) if q_type == 1: intervals.append([a, b]) else: if (DFS(a, b)): print("YES") else: print("NO")
24
92
102,400
171021087
s, t, i = {}, list(), 1 for n in range(int(input())): c, a, b = map(int, input().split()) if c > 1: if b in s[a]: print('YES') else: print('NO') else: s[i] = {i} for j, (x,y) in enumerate(t, 1): if (x < a and a < y) or (x < b and b < y): s[i].update(s[j]) r = set(j for j, (x, y) in enumerate(t, 1) if a < x < b or a < y < b) for j in range(1, len(t) + 1): if r & s[j]: s[j].update(s[i]) t.append((a, b)) i += 1
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
320/B
320
B
PyPy 3-64
TESTS
3
92
1,638,400
224532290
adjlist = {} intervals = [] n = 0 def DFS(a, b, visited): if a == b: return True if visited[a] or a not in adjlist: return False visited[a] = True for u in adjlist[a]: if u == b: return True if DFS(u,b,visited): visited[u] = False return True visited[a] = False return False for _ in range(int(input())): a, b, c = list(map(int,input().split())) if a == 1: for index, interval in enumerate(intervals): if interval[0] < b < interval[1] or interval[0] < c < interval[1]: if index+1 not in adjlist: adjlist[index+1] = [] adjlist[index+1].append(n+1) if b < interval[0] < c or b < interval[1] < c: if n+1 not in adjlist: adjlist[n+1] = [] adjlist[n+1].append(index+1) intervals.append((b,c)) n+=1 else: print("YES" if DFS(b,c, [False]*(n+1)) else "NO")
24
92
102,400
172295999
import sys from collections import deque input = sys.stdin.readline def inlt(): return(list(map(int,input().split()))) n = int(input()) edges = [] for i in range(n): query = inlt() # print("QUERY", query, i) if query[0] == 1: edges.append((query[1], query[2])) else: start = query[1] dest = query[2] # print(start, dest) Q = deque() Q.append(start) visited = set() found = False while Q: node = Q.popleft() if node == dest: found = True break if node in visited: continue visited.add(node) a, b = edges[node - 1] for i in range(1, len(edges) + 1): if i != node: # print(edges[i - 1]) # print(a, b) c, d = edges[i - 1] if c < a < d or c < b < d: Q.append(i) if found: print('YES') else: print('NO')
Codeforces Round 189 (Div. 2)
CF
2,013
2
256
Ping-Pong (Easy Version)
In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2. Your program should handle the queries of the following two types: 1. "1 x y" (x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals. 2. "2 a b" (a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval? Answer all the queries. Note, that initially you have an empty set of intervals.
The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value. It's guaranteed that all queries are correct.
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
null
null
[{"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2", "output": "NO\nYES"}]
1,500
["dfs and similar", "graphs"]
24
[{"input": "5\r\n1 1 5\r\n1 5 11\r\n2 1 2\r\n1 2 9\r\n2 1 2\r\n", "output": "NO\r\nYES\r\n"}, {"input": "10\r\n1 -311 -186\r\n1 -1070 -341\r\n1 -1506 -634\r\n1 688 1698\r\n2 2 4\r\n1 70 1908\r\n2 1 2\r\n2 2 4\r\n1 -1053 1327\r\n2 5 4\r\n", "output": "NO\r\nNO\r\nNO\r\nYES\r\n"}, {"input": "10\r\n1 -1365 -865\r\n1 1244 1834\r\n2 1 2\r\n1 -1508 -752\r\n2 3 2\r\n2 2 1\r\n1 -779 595\r\n1 -1316 877\r\n2 2 1\r\n1 -698 1700\r\n", "output": "NO\r\nNO\r\nNO\r\nNO\r\n"}, {"input": "20\r\n1 1208 1583\r\n1 -258 729\r\n1 -409 1201\r\n1 194 1938\r\n1 -958 1575\r\n1 -1466 1752\r\n2 1 2\r\n2 1 2\r\n2 6 5\r\n1 -1870 1881\r\n1 -2002 2749\r\n1 -2002 2984\r\n1 -2002 3293\r\n2 2 4\r\n2 8 10\r\n2 9 6\r\n1 -2002 3572\r\n1 -2002 4175\r\n1 -2002 4452\r\n1 -2002 4605\r\n", "output": "YES\r\nYES\r\nYES\r\nYES\r\nYES\r\nNO\r\n"}, {"input": "9\r\n1 1 4\r\n1 5 20\r\n1 11 30\r\n1 29 60\r\n1 59 100\r\n1 100 200\r\n2 1 5\r\n2 1 6\r\n2 2 5\r\n", "output": "NO\r\nNO\r\nYES\r\n"}]
false
stdio
null
true
989/D
989
D
Python 3
TESTS
3
77
0
39679237
inp=input('') [n,l,wm]=list(map(float,inp.split())) li=[] for i in range(int(n)): lpos,v=list(map(float,input().split())) li.append((lpos,v)) sl=set(li) counter=0 for e1 in sl: for e2 in sl: r1=e1[0]+l r2=e2[0]+l v1=e1[1] v2=e2[1] if(v1==v2): continue a=r1+(r1-r2)/(v2-v1)*(v1+wm) b=r1+(r1-r2)/(v2-v1)*(v1-wm) if a*b<0 or (b>0 and b<l): counter+=1 print(int(counter/2))
30
841
6,860,800
39193022
# Codeforces Round #487 (Div. 2)import collections from functools import cmp_to_key #key=cmp_to_key(lambda x,y: 1 if x not in y else -1 ) import sys def getIntList(): return list(map(int, input().split())) import bisect N,L,WM = getIntList() z = {} z[-1] = {1:[], -1:[]} z[0] = {1:[], -1:[]} z[1] = {1:[], -1:[]} for i in range(N): x0,v = getIntList() t = (x0,v) if x0+L <=0: z[-1][v].append(t) elif x0>=0: z[1][v].append(t) else: z[0][v].append(t) res = 0 res += len(z[-1][1] ) * len(z[ 1][-1] ) res += len(z[0][1] ) * len(z[ 1][-1] ) res += len(z[-1][1] ) * len(z[ 0][-1] ) if WM==1: print(res) sys.exit() z[1][-1].sort() z[-1][1].sort() #print(z[-1][1]) tn = len(z[1][-1]) for t in z[1][1]: g = (-WM-1) * t[0] / (-WM+1) - L g = max(g, t[0]+ 0.5) p = bisect.bisect_right(z[1][-1], (g,2) ) res += tn-p tn = len(z[-1][1]) for t in z[-1][-1]: g = (WM+1) * (t[0] + L)/ (WM-1) g = min(g, t[0] - 0.1) p = bisect.bisect_left(z[-1][1], (g,-2) ) res += p print(res)
Codeforces Round 487 (Div. 2)
CF
2,018
2
256
A Shade of Moonlight
The sky can be seen as a one-dimensional axis. The moon is at the origin whose coordinate is $$$0$$$. There are $$$n$$$ clouds floating in the sky. Each cloud has the same length $$$l$$$. The $$$i$$$-th initially covers the range of $$$(x_i, x_i + l)$$$ (endpoints excluded). Initially, it moves at a velocity of $$$v_i$$$, which equals either $$$1$$$ or $$$-1$$$. Furthermore, no pair of clouds intersect initially, that is, for all $$$1 \leq i \lt j \leq n$$$, $$$\lvert x_i - x_j \rvert \geq l$$$. With a wind velocity of $$$w$$$, the velocity of the $$$i$$$-th cloud becomes $$$v_i + w$$$. That is, its coordinate increases by $$$v_i + w$$$ during each unit of time. Note that the wind can be strong and clouds can change their direction. You are to help Mino count the number of pairs $$$(i, j)$$$ ($$$i < j$$$), such that with a proper choice of wind velocity $$$w$$$ not exceeding $$$w_\mathrm{max}$$$ in absolute value (possibly negative and/or fractional), the $$$i$$$-th and $$$j$$$-th clouds both cover the moon at the same future moment. This $$$w$$$ doesn't need to be the same across different pairs.
The first line contains three space-separated integers $$$n$$$, $$$l$$$, and $$$w_\mathrm{max}$$$ ($$$1 \leq n \leq 10^5$$$, $$$1 \leq l, w_\mathrm{max} \leq 10^8$$$) — the number of clouds, the length of each cloud and the maximum wind speed, respectively. The $$$i$$$-th of the following $$$n$$$ lines contains two space-separated integers $$$x_i$$$ and $$$v_i$$$ ($$$-10^8 \leq x_i \leq 10^8$$$, $$$v_i \in \{-1, 1\}$$$) — the initial position and the velocity of the $$$i$$$-th cloud, respectively. The input guarantees that for all $$$1 \leq i \lt j \leq n$$$, $$$\lvert x_i - x_j \rvert \geq l$$$.
Output one integer — the number of unordered pairs of clouds such that it's possible that clouds from each pair cover the moon at the same future moment with a proper choice of wind velocity $$$w$$$.
null
In the first example, the initial positions and velocities of clouds are illustrated below. The pairs are: - $$$(1, 3)$$$, covering the moon at time $$$2.5$$$ with $$$w = -0.4$$$; - $$$(1, 4)$$$, covering the moon at time $$$3.5$$$ with $$$w = -0.6$$$; - $$$(1, 5)$$$, covering the moon at time $$$4.5$$$ with $$$w = -0.7$$$; - $$$(2, 5)$$$, covering the moon at time $$$2.5$$$ with $$$w = -2$$$. Below is the positions of clouds at time $$$2.5$$$ with $$$w = -0.4$$$. At this moment, the $$$1$$$-st and $$$3$$$-rd clouds both cover the moon. In the second example, the only pair is $$$(1, 4)$$$, covering the moon at time $$$15$$$ with $$$w = 0$$$. Note that all the times and wind velocities given above are just examples among infinitely many choices.
[{"input": "5 1 2\n-2 1\n2 1\n3 -1\n5 -1\n7 -1", "output": "4"}, {"input": "4 10 1\n-20 1\n-10 -1\n0 1\n10 -1", "output": "1"}]
2,500
["binary search", "geometry", "math", "sortings", "two pointers"]
30
[{"input": "5 1 2\r\n-2 1\r\n2 1\r\n3 -1\r\n5 -1\r\n7 -1\r\n", "output": "4\r\n"}, {"input": "4 10 1\r\n-20 1\r\n-10 -1\r\n0 1\r\n10 -1\r\n", "output": "1\r\n"}, {"input": "1 100000000 98765432\r\n73740702 1\r\n", "output": "0\r\n"}, {"input": "10 2 3\r\n-1 -1\r\n-4 1\r\n-6 -1\r\n1 1\r\n10 -1\r\n-8 -1\r\n6 1\r\n8 1\r\n4 -1\r\n-10 -1\r\n", "output": "5\r\n"}, {"input": "3 100000000 100000000\r\n-100000000 1\r\n100000000 1\r\n0 -1\r\n", "output": "1\r\n"}, {"input": "9 25000000 989\r\n-100000000 -1\r\n-75000000 1\r\n75000000 1\r\n50000000 -1\r\n-50000000 1\r\n0 1\r\n25000000 1\r\n-25000000 -1\r\n100000000 -1\r\n", "output": "11\r\n"}, {"input": "2 5 1\r\n-2 1\r\n5 -1\r\n", "output": "1\r\n"}, {"input": "2 5 1\r\n-9 -1\r\n-2 1\r\n", "output": "0\r\n"}, {"input": "3 4 5\r\n9 1\r\n-4 1\r\n-8 -1\r\n", "output": "0\r\n"}, {"input": "5 1 1\r\n-6 1\r\n15 1\r\n-7 1\r\n-13 -1\r\n12 -1\r\n", "output": "2\r\n"}, {"input": "50 1 19\r\n-5213 -1\r\n2021 -1\r\n-4479 1\r\n1569 -1\r\n1618 1\r\n-8318 1\r\n3854 1\r\n8190 -1\r\n9162 1\r\n8849 1\r\n-5545 -1\r\n-7898 -1\r\n728 1\r\n-2175 -1\r\n6453 -1\r\n2999 1\r\n4716 1\r\n-2192 -1\r\n7938 -1\r\n1910 -1\r\n-6863 -1\r\n5230 -1\r\n-2782 -1\r\n-2587 -1\r\n-3389 1\r\n-332 -1\r\n5915 1\r\n-2604 1\r\n-8907 1\r\n-2019 1\r\n2992 1\r\n-3279 -1\r\n6720 1\r\n4332 1\r\n8789 -1\r\n2003 1\r\n-8046 -1\r\n-594 -1\r\n-4133 -1\r\n-7954 -1\r\n-6270 -1\r\n4042 -1\r\n3650 1\r\n-8569 1\r\n2529 -1\r\n266 -1\r\n3405 -1\r\n-9753 1\r\n1205 -1\r\n6437 -1\r\n", "output": "262\r\n"}, {"input": "50 100 40\r\n4843 -1\r\n7653 1\r\n5391 1\r\n-1651 1\r\n-8530 -1\r\n9770 1\r\n2721 1\r\n7321 1\r\n-3636 -1\r\n-1525 -1\r\n-3060 1\r\n1877 -1\r\n3771 -1\r\n-7651 1\r\n581 -1\r\n1127 -1\r\n6966 -1\r\n-6089 1\r\n1465 -1\r\n3147 -1\r\n-6927 -1\r\n4477 1\r\n-6535 1\r\n5991 -1\r\n-2740 1\r\n5021 1\r\n-7761 -1\r\n4626 1\r\n9958 1\r\n4275 1\r\n5695 1\r\n8835 -1\r\n7791 -1\r\n189 -1\r\n-170 1\r\n-4468 -1\r\n-708 1\r\n34 -1\r\n-9068 1\r\n6424 -1\r\n-2066 -1\r\n-7367 1\r\n6224 1\r\n3329 1\r\n-1809 -1\r\n7105 1\r\n-4607 -1\r\n-3174 -1\r\n-9782 -1\r\n1350 -1\r\n", "output": "253\r\n"}]
false
stdio
null
true
835/D
835
D
PyPy 3
TESTS
4
638
104,140,800
116519645
st=input() n=len(st) dp=[[0]*(n) for i in range(n)] for i in range(n): for j in range(i+1): dp[i][j]=1 pt=n-2 var=0 while pt>=0: var=pt+1 while var<n: if st[pt]==st[var]: dp[pt][var]=dp[pt+1][var-1] var+=1 pt-=1 ans=[0]*(n+1) for i in range(n): for j in range(i,n): flag=0 if dp[i][j]>0: if 2*j-i+1 < n and dp[j+1][2*j-i+1]==1 and dp[i][2*j-i+1]==1: dp[i][2*j-i+1]+=1 if 2*j-i+2 < n and dp[j+2][2*j-i+2]==1 and dp[i][2*j-i+2]==1: dp[i][2*j-i+2]=dp[i][j]+1 ans[dp[i][j]]+=1 for i in range(n-2,-1,-1): ans[i]=ans[i]+ans[i+1] print(*ans[1:])
65
764
205,926,400
199901359
s = input() n = len(s) res = [0] * n res[0] = n dp = [[0] * n for _ in range(n)] for i in range(n): dp[i][i] = 1 for i in range(n - 1): if s[i] == s[i + 1]: dp[i][i + 1] = 2 res[1] += 1 for length in range(2, n): for l in range(0, n - length): r = l + length if s[l] != s[r] or dp[l + 1][r - 1] == 0: continue m = (l + r - 1) // 2 dp[l][r] = dp[l][m] + 1 res[dp[l][r] - 1] += 1 for i in range(n - 2, -1, -1): res[i] += res[i + 1] print(*res)
Codeforces Round 427 (Div. 2)
CF
2,017
3
256
Palindromic characteristics
Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.
The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.
Print |s| integers — palindromic characteristics of string s.
null
In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.
[{"input": "abba", "output": "6 1 0 0"}, {"input": "abacaba", "output": "12 4 1 0 0 0 0"}]
1,900
["brute force", "dp", "hashing", "strings"]
65
[{"input": "abba\r\n", "output": "6 1 0 0 \r\n"}, {"input": "abacaba\r\n", "output": "12 4 1 0 0 0 0 \r\n"}, {"input": "qqqpvmgd\r\n", "output": "11 3 0 0 0 0 0 0 \r\n"}, {"input": "wyemcafatp\r\n", "output": "11 1 0 0 0 0 0 0 0 0 \r\n"}]
false
stdio
null
true
835/D
835
D
PyPy 3-64
TESTS
5
2,698
206,028,800
186789017
import random import sys, os, io input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline def is_Prime(n): if n == 1: return False for i in range(2, min(int(n ** (1 / 2)) + 2, n)): if n % i == 0: return False return True def random_mod(): mod = random.randint(7 * pow(10, 11), pow(10, 12)) if not mod % 2: mod += 1 while not is_Prime(mod): mod += 2 while not min(mod % 3, mod % 5): mod += 2 return mod def f(u, v): return u * n + v def same(l1, r1, l2, r2): u = (h[r1 + 1] - h[l1]) % mod * p[l2 - l1] % mod v = (h[r2 + 1] - h[l2]) % mod return True if u == v else False s = list(input().rstrip()) n = len(s) b = 29 mod = random_mod() p = [1] h = [0] for i in s: p.append(b * p[-1] % mod) h.append(h[-1] + (i - 96) * p[-1]) m = n * n dp = [-1] * m for i in range(n): dp[f(i, i)] = 0 l, r = i - 1, i + 1 while 0 <= l and r < n and s[l] == s[r]: dp[f(l, r)] = 0 l -= 1 r += 1 for i in range(n - 1): l, r = i, i + 1 while 0 <= l and r < n and s[l] == s[r]: dp[f(l, r)] = 0 l -= 1 r += 1 for i in range(1, n): d = (i - 1) // 2 for l in range(n - i): r = l + i u = f(l, r) if same(l, l + d, r - d, r): dp[u] = max(dp[u], dp[f(l, l + d)] + 1) ans = [0] * n for i in dp: if i ^ -1: ans[i] += 1 for i in range(n - 2, -1, -1): ans[i] += ans[i + 1] sys.stdout.write(" ".join(map(str, ans)))
65
889
114,995,200
97046958
def PanlidromicCharacteristics(string): n = len(string) res = [[0 for i in range (n)] for j in range (n)] count = [0 for i in range (n + 1)] # for i in range (n): # res[i][i] = 1 # count[1] += 1 for length in range (1, n + 1): for i in range (n-length + 1): j = i + length - 1 if length == 1: res[i][j] = 1 elif length == 2 and string[i] == string[j]: res[i][j] = 2 elif string[i] == string[j] and res[i + 1][j - 1] > 0: res[i][j] = res[i][i + length//2 - 1] + 1 count[res[i][j]] += 1 # k-palindrome is also a (k - 1)-palindrome for i in range (len(count) - 1, 0, -1): count[i - 1] += count[i] for i in range (1, len(count)): print(count[i], end = " ") return string = input() PanlidromicCharacteristics(string)
Codeforces Round 427 (Div. 2)
CF
2,017
3
256
Palindromic characteristics
Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.
The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.
Print |s| integers — palindromic characteristics of string s.
null
In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.
[{"input": "abba", "output": "6 1 0 0"}, {"input": "abacaba", "output": "12 4 1 0 0 0 0"}]
1,900
["brute force", "dp", "hashing", "strings"]
65
[{"input": "abba\r\n", "output": "6 1 0 0 \r\n"}, {"input": "abacaba\r\n", "output": "12 4 1 0 0 0 0 \r\n"}, {"input": "qqqpvmgd\r\n", "output": "11 3 0 0 0 0 0 0 \r\n"}, {"input": "wyemcafatp\r\n", "output": "11 1 0 0 0 0 0 0 0 0 \r\n"}]
false
stdio
null
true
379/B
379
B
PyPy 3-64
TESTS
0
46
0
212362250
def get_run(start,right): s = '' current = start if right: for i in put_indeces: s += 'R' * (i - current) s += 'P' current = i else: l = len(put_indeces) for i in range(l): s += 'L' * (current - put_indeces[l-i-1]) s += 'P' current = put_indeces[l-i-1] return s, current n = int(input()) a = list(map(int,input().split())) s = '' a_max = max(a) going_right = True index = 0 for run in range(a_max): put_indeces = [] for i in range(n): if a[i] > 0: put_indeces.append(i) t_s, index = get_run(index,going_right) s += t_s if run < a_max-1: if index < n-1: s += 'RL' else: s += 'LR' going_right = not going_right print(s)
15
46
409,600
144292617
n=int(input()) a=[*map(int,input().split())] r='PRL'*a[0] for i in range(1,n):r+='R'+'PLR'*a[i] print(r)
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
106/D
106
D
PyPy 3
TESTS
2
248
0
91182293
def main(): n, m = map(int, input().split()) island = []; features = [] for _ in range(n): s = input() for i in range(m): if s[i] not in "#.": features.append((i, _)) island.append(s) k = int(input()) #maxes = [0, 0, 0, 0] vector = [0, 0] for __ in range(k): d, l = input().split() l = int(l) if d == "N": #maxes[0] = max(maxes[0], l) vector[1] -= l elif d == "S": #maxes[1] = max(maxes[1], l) vector[1] += l elif d == "E": #maxes[2] = max(maxes[2], l) vector[0] += l else: #maxes[3] = max(maxes[3], l) vector[0] -= l good = [] for feature in features: endpoint = (feature[0] + vector[0], feature[1] + vector[1]) if endpoint[0] < 1 or endpoint[0] > m - 2 or endpoint[1] < 1 or endpoint[1] > n - 2 or island[endpoint[1]][endpoint[0]] == "#": continue good.append(feature) if good == []: print("no solution") else: print("".join(sorted([island[po[1]][po[0]] for po in good]))) main()
70
1,090
28,160,000
153861321
import sys input = sys.stdin.readline n, m = map(int, input().split()) a, pos = [], [] for i in range(n): a.append(input().rstrip()) for j in range(m): if a[i][j] != '.' and a[i][j] != '#': pos.append((i, j)) W = [[0]*(m+3) for i in range(n+3)] for i in range(n): for j in range(m): if a[i][j] != '#': W[i][j] = W[i][j-1] + 1 E = [[0]*(m+3) for i in range(n+3)] for i in range(n): for j in range(m-1, -1, -1): if a[i][j] != '#': E[i][j] = E[i][j+1] + 1 N = [[0]*(m+3) for i in range(n+3)] for j in range(m): for i in range(n): if a[i][j] != '#': N[i][j] = N[i-1][j] + 1 S = [[0]*(m+3) for i in range(n+3)] for j in range(m): for i in range(n-1, -1, -1): if a[i][j] != '#': S[i][j] = S[i+1][j] + 1 k = int(input()) d, l = [0]*k, [0]*k for i in range(k): di, li = input().split() d[i], l[i] = di, int(li) kq = [] for x, y in pos: ok, dut = 1, a[x][y] for i in range(k): if d[i] == 'E': if E[x][y] - 1 < l[i]: ok = 0 break else: y += l[i] if d[i] == 'W': if W[x][y] - 1 < l[i]: ok = 0 break else: y -= l[i] if d[i] == 'N': if N[x][y] - 1 < l[i]: ok = 0 break else: x -= l[i] continue if d[i] == 'S': if S[x][y] - 1 < l[i]: ok = 0 break else: x += l[i] if ok == 1: kq.append(dut) if len(kq) == 0: print('no solution') else: kq.sort() print(''.join(kq))
Codeforces Beta Round 82 (Div. 2)
CF
2,011
2
256
Treasure Island
Our brave travelers reached an island where pirates had buried treasure. However as the ship was about to moor, the captain found out that some rat ate a piece of the treasure map. The treasure map can be represented as a rectangle n × m in size. Each cell stands for an islands' square (the square's side length equals to a mile). Some cells stand for the sea and they are impenetrable. All other cells are penetrable (i.e. available) and some of them contain local sights. For example, the large tree on the hills or the cave in the rocks. Besides, the map also has a set of k instructions. Each instruction is in the following form: "Walk n miles in the y direction" The possible directions are: north, south, east, and west. If you follow these instructions carefully (you should fulfill all of them, one by one) then you should reach exactly the place where treasures are buried. Unfortunately the captain doesn't know the place where to start fulfilling the instructions — as that very piece of the map was lost. But the captain very well remembers that the place contained some local sight. Besides, the captain knows that the whole way goes through the island's penetrable squares. The captain wants to know which sights are worth checking. He asks you to help him with that.
The first line contains two integers n and m (3 ≤ n, m ≤ 1000). Then follow n lines containing m integers each — the island map's description. "#" stands for the sea. It is guaranteed that all cells along the rectangle's perimeter are the sea. "." stands for a penetrable square without any sights and the sights are marked with uppercase Latin letters from "A" to "Z". Not all alphabet letters can be used. However, it is guaranteed that at least one of them is present on the map. All local sights are marked by different letters. The next line contains number k (1 ≤ k ≤ 105), after which k lines follow. Each line describes an instruction. Each instruction possesses the form "dir len", where dir stands for the direction and len stands for the length of the way to walk. dir can take values "N", "S", "W" and "E" for North, South, West and East correspondingly. At that, north is to the top, South is to the bottom, west is to the left and east is to the right. len is an integer from 1 to 1000.
Print all local sights that satisfy to the instructions as a string without any separators in the alphabetical order. If no sight fits, print "no solution" without the quotes.
null
null
[{"input": "6 10\n##########\n#K#..#####\n#.#..##.##\n#..L.#...#\n###D###A.#\n##########\n4\nN 2\nS 1\nE 1\nW 2", "output": "AD"}, {"input": "3 4\n####\n#.A#\n####\n2\nW 1\nN 2", "output": "no solution"}]
1,700
["brute force", "implementation"]
70
[{"input": "6 10\r\n##########\r\n#K#..#####\r\n#.#..##.##\r\n#..L.#...#\r\n###D###A.#\r\n##########\r\n4\r\nN 2\r\nS 1\r\nE 1\r\nW 2\r\n", "output": "AD"}, {"input": "3 4\r\n####\r\n#.A#\r\n####\r\n2\r\nW 1\r\nN 2\r\n", "output": "no solution"}, {"input": "10 10\r\n##########\r\n#K#..##..#\r\n##...ZB..#\r\n##.......#\r\n#D..#....#\r\n##...Y..##\r\n#...N...J#\r\n#.G...#.##\r\n#.S.I....#\r\n##########\r\n4\r\nE 2\r\nW 4\r\nS 3\r\nN 4\r\n", "output": "YZ"}, {"input": "6 6\r\n######\r\n#UPWT#\r\n#KJSL#\r\n#VCMA#\r\n#QOGB#\r\n######\r\n5\r\nN 1\r\nW 1\r\nS 1\r\nE 1\r\nN 1\r\n", "output": "ABCGJLMOS"}, {"input": "3 3\r\n###\r\n#A#\r\n###\r\n1\r\nN 1\r\n", "output": "no solution"}, {"input": "7 8\r\n########\r\n#..#O.##\r\n##U#YTI#\r\n##.#####\r\n##R.E.P#\r\n###W#Q##\r\n########\r\n2\r\nN 1\r\nW 1\r\n", "output": "QTUW"}, {"input": "4 9\r\n#########\r\n#AB#CDEF#\r\n#.......#\r\n#########\r\n1\r\nE 3\r\n", "output": "C"}, {"input": "4 9\r\n#########\r\n#AB#CDEF#\r\n#.......#\r\n#########\r\n1\r\nW 3\r\n", "output": "F"}, {"input": "4 9\r\n#########\r\n#AB#CDEF#\r\n#.......#\r\n#########\r\n1\r\nS 1\r\n", "output": "ABCDEF"}, {"input": "9 4\r\n####\r\n#AB#\r\n#C.#\r\n##.#\r\n#D.#\r\n#E.#\r\n#F.#\r\n#G.#\r\n####\r\n1\r\nN 3\r\n", "output": "G"}, {"input": "9 4\r\n####\r\n#AB#\r\n#C.#\r\n##.#\r\n#D.#\r\n#E.#\r\n#F.#\r\n#G.#\r\n####\r\n1\r\nS 3\r\n", "output": "BD"}, {"input": "10 11\r\n###########\r\n#.F.#.P.###\r\n#..#.#....#\r\n#Y#.U....##\r\n##....#...#\r\n##.OV.NK#.#\r\n#H####.#EA#\r\n#R.#....###\r\n##..#MS.D##\r\n###########\r\n3\r\nW 2\r\nE 1\r\nN 1\r\n", "output": "DNV"}]
false
stdio
null
true
383/E
383
E
PyPy 3-64
TESTS
1
46
0
169196973
l = [] for _ in range(int(input())): a = input() for i in ["a", "e", "i", "o", "u"]: if a.__contains__(i): break else: l.append(False) continue l.append(True) for _ in range(len(l)-1): l[0]=(l[0]==l[1]) l.pop(1) print(1 if l[0] else 0)
68
3,026
69,324,800
155280550
import sys readline=sys.stdin.readline N=int(readline()) dp=[0]*(1<<24) dp[-1]=N for _ in range(N): bit=(1<<24)-1 for a in readline().rstrip(): a=ord(a)-97 if bit&1<<a: bit^=1<<a dp[bit]-=1 for i in range(24): for bit in range((1<<24)-1,-1,-1): if not bit&1<<i: dp[bit]+=dp[bit^1<<i] ans=0 for x in dp: ans^=x**2 print(ans)
Codeforces Round 225 (Div. 1)
CF
2,014
4
256
Vowels
Iahubina is tired of so many complicated languages, so she decided to invent a new, simple language. She already made a dictionary consisting of n 3-words. A 3-word is a sequence of exactly 3 lowercase letters of the first 24 letters of the English alphabet (a to x). She decided that some of the letters are vowels, and all the others are consonants. The whole language is based on a simple rule: any word that contains at least one vowel is correct. Iahubina forgot which letters are the vowels, and wants to find some possible correct sets of vowels. She asks Iahub questions. In each question, she will give Iahub a set of letters considered vowels (in this question). For each question she wants to know how many words of the dictionary are correct, considering the given set of vowels. Iahubina wants to know the xor of the squared answers to all the possible questions. There are 224 different questions, they are all subsets of the set of the first 24 letters of the English alphabet. Help Iahub find that number.
The first line contains one integer, n (1 ≤ n ≤ 104). Each of the next n lines contains a 3-word consisting of 3 lowercase letters. There will be no two identical 3-words.
Print one number, the xor of the squared answers to the queries.
null
null
[{"input": "5\nabc\naaa\nada\nbcd\ndef", "output": "0"}]
2,700
["combinatorics", "divide and conquer", "dp"]
68
[{"input": "5\r\nabc\r\naaa\r\nada\r\nbcd\r\ndef\r\n", "output": "0\r\n"}, {"input": "100\r\namd\r\namj\r\natr\r\nbcp\r\nbjm\r\ncna\r\ncpj\r\ncse\r\ndij\r\ndjp\r\ndlv\r\nebk\r\nedf\r\nelw\r\nfbr\r\nfcl\r\nfhs\r\nflo\r\nfmj\r\ngcg\r\ngen\r\nghg\r\ngvb\r\ngxx\r\nhbe\r\nhbf\r\nhgu\r\nhlv\r\nhqa\r\nibg\r\nifp\r\nima\r\nitt\r\nivl\r\nixu\r\njle\r\njli\r\nket\r\nkit\r\nkws\r\nlep\r\nles\r\nleu\r\nmbp\r\nmci\r\nmdv\r\nmhf\r\nmih\r\nmll\r\nmop\r\nndp\r\nnfs\r\nngl\r\nnng\r\noic\r\nomo\r\nooj\r\noti\r\npax\r\npfo\r\npjd\r\npup\r\nqer\r\nrad\r\nrdg\r\nrfq\r\nrvt\r\nrwa\r\nrxj\r\nshc\r\nsjv\r\nswx\r\ntcu\r\ntlm\r\ntmb\r\ntml\r\ntmw\r\ntvr\r\ntvx\r\nuid\r\nuir\r\nukf\r\nulg\r\nvce\r\nves\r\nvfb\r\nvok\r\nvut\r\nvvi\r\nvwb\r\nwab\r\nwba\r\nwdf\r\nweq\r\nwog\r\nwsl\r\nxbk\r\nxiq\r\nxop\r\nxpp\r\n", "output": "13888\r\n"}, {"input": "100\r\naip\r\nbfl\r\nbld\r\nblh\r\nbpk\r\nbqd\r\nbtk\r\ncfu\r\nciv\r\nckf\r\ncog\r\ncro\r\nctt\r\ncve\r\ncvn\r\ndlj\r\neer\r\negw\r\negx\r\nffi\r\nfld\r\nggk\r\ngis\r\ngkv\r\ngnq\r\ngvj\r\nhdo\r\nhgf\r\nhgu\r\nhjt\r\nhla\r\nhni\r\nhnk\r\nifa\r\niir\r\niml\r\njfa\r\njgl\r\nkbf\r\nliv\r\nlqo\r\nmlw\r\nmot\r\nmpx\r\nnas\r\nnlo\r\nobt\r\nodo\r\nodx\r\nolr\r\nolw\r\nonc\r\npac\r\npdp\r\nphn\r\npku\r\npng\r\npsd\r\nptl\r\npuq\r\npvk\r\npvx\r\nqjj\r\nqju\r\nqpf\r\nqqv\r\nqsf\r\nrac\r\nrgj\r\nrrg\r\nsbm\r\nsdf\r\nsif\r\nsil\r\nsnv\r\nspt\r\nsxt\r\ntou\r\nttj\r\nufi\r\nuht\r\nujm\r\null\r\nupm\r\nuqf\r\nvof\r\nvpq\r\nwae\r\nwck\r\nwed\r\nwhd\r\nwjn\r\nwpp\r\nwvd\r\nxbx\r\nxdv\r\nxeh\r\nxmq\r\nxsm\r\nxsp\r\n", "output": "8624\r\n"}, {"input": "10\r\nhjk\r\nkkw\r\nmsw\r\nnht\r\noqu\r\npcx\r\npet\r\nshd\r\nutb\r\nwbw\r\n", "output": "0\r\n"}, {"input": "20\r\netf\r\nffq\r\ngqe\r\nhpj\r\nido\r\niep\r\nkbv\r\nlgs\r\nlkl\r\nlvg\r\nmhs\r\nocr\r\nonc\r\nonv\r\npmv\r\nqhk\r\nrck\r\nrgj\r\nsib\r\nuox\r\n", "output": "0\r\n"}, {"input": "30\r\nagf\r\naov\r\ncac\r\ncdq\r\nclc\r\ncue\r\ndmh\r\ndrr\r\ndxv\r\nfrv\r\njmg\r\nkih\r\nkii\r\nkqm\r\nkwc\r\nnri\r\nohw\r\nrfk\r\nrrd\r\nrrk\r\ntmp\r\ntsc\r\nuhg\r\nuhx\r\nujw\r\nvms\r\nvrg\r\nwer\r\nxml\r\nxuv\r\n", "output": "0\r\n"}, {"input": "40\r\nbhw\r\nblh\r\ncal\r\nccg\r\ncdd\r\ncsm\r\ndir\r\ndux\r\nefp\r\nfnw\r\ngcr\r\nhuc\r\niaf\r\nipv\r\niva\r\niwl\r\njeb\r\njwk\r\nlot\r\nmcf\r\nmnk\r\nnak\r\nopl\r\norb\r\noxj\r\nqws\r\nrbl\r\nsmo\r\nsuw\r\nsws\r\ntgt\r\numg\r\nvhn\r\nvud\r\nwml\r\nwqg\r\nxbv\r\nxgj\r\nxlm\r\nxxv\r\n", "output": "944\r\n"}, {"input": 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false
stdio
null
true
840/B
840
B
PyPy 3
TESTS
3
93
0
203369461
import sys input = sys.stdin.buffer.readline def bfs(n, g, root, A): parent = [None for i in range(n+1)] parent[root] = [-1, None] depth = [[root]] odds = [root] depth_count = [0 for i in range(n+1)] while True: next_s = [] for x in depth[-1]: for y, i in g[x]: if parent[y] is None: parent[y] = [x, i] depth_count[y] = depth_count[x]+1 next_s.append(y) if len(next_s)==0: break depth.append(next_s) tree_sum = [0 for i in range(n+1)] while len(depth) > 0: row = depth.pop() for x in row: if A[x-1]==1: tree_sum[x]+=1 a, i = parent[x] if i is not None: tree_sum[i]+=tree_sum[x] return tree_sum, depth_count, parent def process(n, A, G): if -1 not in A and sum(A) % 2==1: sys.stdout.write('-1\n') return one_count = [] neg_count = [] for i in range(n): if A[i]==1: one_count.append(i) elif A[i]==-1: neg_count.append(i) if len(one_count) % 2==1: i = neg_count.pop() A[i] = 1 for i2 in neg_count: A[i2] = 0 else: for i2 in neg_count: A[i2] = 0 if 1 not in A: sys.stdout.write('0\n') return g = [[] for i in range(n+1)] m = len(G) for i in range(m): u, v = G[i] g[u].append([v, i]) g[v].append([u, i]) degree = [len(g[i]) % 2 for i in range(n+1)] answer = [0 for i in range(m)] tree_sum, depth_count, parent = bfs(n, g, 1, A) for i in range(1, n+1): x, I = parent[i] if x != -1: answer[I] = 1 S = sum(A) for i in range(1, n+1): if A[i-1]==0 and degree[i]==1: for u, I in g[i]: if depth_count[u]==depth_count[i]+1: tree = tree_sum[u] else: tree = S-tree_sum[i] if tree % 2==0: answer[I] = 0 degree[u] = (degree[u]+1) % 2 degree[i] = (degree[i]+1) % 2 break elif A[i-1]==1 and degree[i]==0: for u, I in g[i]: if depth_count[u]==depth_count[i]+1: tree = tree_sum[u] else: tree = S-tree_sum[i] if tree % 2==0: answer[I] = 0 degree[u] = (degree[u]+1) % 2 degree[i] = (degree[i]+1) % 2 break final_answer = [] for i in range(m): if answer[i]==1: final_answer.append(i+1) print(len(final_answer)) final_answer = ' '.join(map(str, final_answer)) sys.stdout.write(f'{final_answer}\n') return n, m = [int(x) for x in input().split()] A = [int(x) for x in input().split()] G = [] for i in range(m): u, v = [int(x) for x in input().split()] G.append([u, v]) process(n, A, G)
73
2,979
87,756,800
167362143
import bisect import copy import decimal import fractions import heapq import itertools import math import random import sys import time from collections import Counter,deque,defaultdict from functools import lru_cache,reduce from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max def _heappush_max(heap,item): heap.append(item) heapq._siftdown_max(heap, 0, len(heap)-1) def _heappushpop_max(heap, item): if heap and item < heap[0]: item, heap[0] = heap[0], item heapq._siftup_max(heap, 0) return item from math import gcd as GCD read=sys.stdin.read readline=sys.stdin.readline readlines=sys.stdin.readlines class Graph: def __init__(self,V,edges=False,graph=False,directed=False,weighted=False,inf=float("inf")): self.V=V self.directed=directed self.weighted=weighted self.inf=inf if graph: self.graph=graph self.edges=[] for i in range(self.V): if self.weighted: for j,d in self.graph[i]: if self.directed or not self.directed and i<=j: self.edges.append((i,j,d)) else: for j in self.graph[i]: if self.directed or not self.directed and i<=j: self.edges.append((i,j)) else: self.edges=edges self.graph=[[] for i in range(self.V)] if weighted: for i,j,d in self.edges: self.graph[i].append((j,d)) if not self.directed: self.graph[j].append((i,d)) else: for i,j in self.edges: self.graph[i].append(j) if not self.directed: self.graph[j].append(i) def SIV_DFS(self,s,bipartite_graph=False,cycle_detection=False,directed_acyclic=False,euler_tour=False,linked_components=False,lowlink=False,parents=False,postorder=False,preorder=False,subtree_size=False,topological_sort=False,unweighted_dist=False,weighted_dist=False): seen=[False]*self.V finished=[False]*self.V if directed_acyclic or cycle_detection or topological_sort: dag=True if euler_tour: et=[] if linked_components: lc=[] if lowlink: order=[None]*self.V ll=[None]*self.V idx=0 if parents or cycle_detection or lowlink or subtree_size: ps=[None]*self.V if postorder or topological_sort: post=[] if preorder: pre=[] if subtree_size: ss=[1]*self.V if unweighted_dist or bipartite_graph: uwd=[self.inf]*self.V uwd[s]=0 if weighted_dist: wd=[self.inf]*self.V wd[s]=0 stack=[(s,0)] if self.weighted else [s] while stack: if self.weighted: x,d=stack.pop() else: x=stack.pop() if not seen[x]: seen[x]=True stack.append((x,d) if self.weighted else x) if euler_tour: et.append(x) if linked_components: lc.append(x) if lowlink: order[x]=idx ll[x]=idx idx+=1 if preorder: pre.append(x) for y in self.graph[x]: if self.weighted: y,d=y if not seen[y]: stack.append((y,d) if self.weighted else y) if parents or cycle_detection or lowlink or subtree_size: ps[y]=x if unweighted_dist or bipartite_graph: uwd[y]=uwd[x]+1 if weighted_dist: wd[y]=wd[x]+d elif not finished[y]: if (directed_acyclic or cycle_detection or topological_sort) and dag: dag=False if cycle_detection: cd=(y,x) elif not finished[x]: finished[x]=True if euler_tour: et.append(~x) if lowlink: bl=True for y in self.graph[x]: if self.weighted: y,d=y if ps[x]==y and bl: bl=False continue ll[x]=min(ll[x],order[y]) if x!=s: ll[ps[x]]=min(ll[ps[x]],ll[x]) if postorder or topological_sort: post.append(x) if subtree_size: for y in self.graph[x]: if self.weighted: y,d=y if y==ps[x]: continue ss[x]+=ss[y] if bipartite_graph: bg=[[],[]] for tpl in self.edges: x,y=tpl[:2] if self.weighted else tpl if uwd[x]==self.inf or uwd[y]==self.inf: continue if not uwd[x]%2^uwd[y]%2: bg=False break else: for x in range(self.V): if uwd[x]==self.inf: continue bg[uwd[x]%2].append(x) retu=() if bipartite_graph: retu+=(bg,) if cycle_detection: if dag: cd=[] else: y,x=cd cd=self.Route_Restoration(y,x,ps) retu+=(cd,) if directed_acyclic: retu+=(dag,) if euler_tour: retu+=(et,) if linked_components: retu+=(lc,) if lowlink: retu=(ll,) if parents: retu+=(ps,) if postorder: retu+=(post,) if preorder: retu+=(pre,) if subtree_size: retu+=(ss,) if topological_sort: if dag: tp_sort=post[::-1] else: tp_sort=[] retu+=(tp_sort,) if unweighted_dist: retu+=(uwd,) if weighted_dist: retu+=(wd,) if len(retu)==1: retu=retu[0] return retu N,M=map(int,readline().split()) D=list(map(int,readline().split())) edges=[] idx={} for i in range(M): u,v=map(int,readline().split()) u-=1;v-=1 edges.append((u,v)) idx[(u,v)]=i idx[(v,u)]=i G=Graph(N,edges=edges) parents,tour=G.SIV_DFS(0,parents=True,postorder=True) if D.count(1)%2: for i in range(N): if D[i]==-1: D[i]=1 break for i in range(N): if D[i]==-1: D[i]=0 if D.count(1)%2: print(-1) exit() ans_lst=[] for x in tour[:-1]: if D[x]: ans_lst.append(idx[(x,parents[x])]+1) D[x]^=1 D[parents[x]]^=1 print(len(ans_lst)) if ans_lst: print(*ans_lst,sep="\n")
Codeforces Round 429 (Div. 1)
CF
2,017
3
256
Leha and another game about graph
Leha plays a computer game, where is on each level is given a connected graph with n vertices and m edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer di, which can be equal to 0, 1 or - 1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, di = - 1 or it's degree modulo 2 is equal to di. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them.
The first line contains two integers n, m (1 ≤ n ≤ 3·105, n - 1 ≤ m ≤ 3·105) — number of vertices and edges. The second line contains n integers d1, d2, ..., dn ( - 1 ≤ di ≤ 1) — numbers on the vertices. Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n) — edges. It's guaranteed, that graph in the input is connected.
Print - 1 in a single line, if solution doesn't exist. Otherwise in the first line k — number of edges in a subset. In the next k lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1.
null
In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1.
[{"input": "1 0\n1", "output": "-1"}, {"input": "4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4", "output": "0"}, {"input": "2 1\n1 1\n1 2", "output": "1\n1"}, {"input": "3 3\n0 -1 1\n1 2\n2 3\n1 3", "output": "1\n2"}]
2,100
["constructive algorithms", "data structures", "dfs and similar", "dp", "graphs"]
73
[{"input": "1 0\r\n1\r\n", "output": "-1\r\n"}, {"input": "4 5\r\n0 0 0 -1\r\n1 2\r\n2 3\r\n3 4\r\n1 4\r\n2 4\r\n", "output": "0\r\n"}, {"input": "2 1\r\n1 1\r\n1 2\r\n", "output": "1\r\n1\r\n"}, {"input": "3 3\r\n0 -1 1\r\n1 2\r\n2 3\r\n1 3\r\n", "output": "1\r\n2\r\n"}, {"input": "10 10\r\n-1 -1 -1 -1 -1 -1 -1 -1 -1 -1\r\n6 7\r\n8 3\r\n6 4\r\n4 2\r\n9 2\r\n5 10\r\n9 8\r\n10 7\r\n5 1\r\n6 2\r\n", "output": "0\r\n"}, {"input": "3 2\r\n1 0 1\r\n1 2\r\n2 3\r\n", "output": "2\r\n2\r\n1\r\n"}]
false
stdio
import sys def main(input_path, output_path, submission_output_path): # Read input with open(input_path, 'r') as f: n, m = map(int, f.readline().split()) d = list(map(int, f.readline().split())) edges = [tuple(map(int, f.readline().split())) for _ in range(m)] # Read submission output with open(submission_output_path, 'r') as f: lines = f.read().strip().split('\n') if not lines: print(0) return first_line = lines[0].strip() if first_line == '-1': sum_d = sum(x for x in d if x != -1) has_neg1 = any(x == -1 for x in d) exists = (sum_d % 2 == 0) or (has_neg1 and (sum_d % 2 == 1)) print(0 if exists else 1) return else: try: k = int(first_line) if len(lines) < k + 1: print(0) return selected = list(map(int, lines[1:1 + k])) except: print(0) return # Check edge indices if any(e < 1 or e > m for e in selected): print(0) return if len(set(selected)) != len(selected): print(0) return # Compute degrees degree = [0] * (n + 1) # 1-based for e in selected: u, v = edges[e - 1] degree[u] += 1 degree[v] += 1 # Check each vertex valid = True for i in range(n): di = d[i] if di == -1: continue if degree[i + 1] % 2 != di: valid = False break print(1 if valid else 0) if __name__ == '__main__': input_path = sys.argv[1] output_path = sys.argv[2] submission_output_path = sys.argv[3] main(input_path, output_path, submission_output_path)
true
261/A
261
A
Python 3
TESTS
3
186
8,089,600
127903476
try: import sys from functools import lru_cache, cmp_to_key from heapq import merge, heapify, heappop, heappush from math import * from collections import defaultdict as dd, deque, Counter as C from itertools import combinations as comb, permutations as perm from bisect import bisect_left as bl, bisect_right as br, bisect, insort from time import perf_counter from fractions import Fraction import copy from copy import deepcopy import time starttime = time.time() mod = int(pow(10, 9) + 7) mod2 = 998244353 def data(): return sys.stdin.readline().strip() def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end) def L(): return list(sp()) def sl(): return list(ssp()) def sp(): return map(int, data().split()) def ssp(): return map(str, data().split()) def l1d(n, val=0): return [val for i in range(n)] def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)] # sys.setrecursionlimit(int(pow(10,6))) # from sys import stdin # input = stdin.buffer.readline # I = lambda : list(map(int,input().split())) # import sys # input=sys.stdin.readline sys.stdin = open("input.txt", "r") sys.stdout = open("output.txt", "w") except: pass m=L()[0] Q=L() n=L()[0] A=sorted(L()) for q in Q: for i in range(n//(q+2),0,-1): x=n-i*(q+2) if (x%q<=1 and (x//q)>0) or x==0: # print(x,q,i) A=A[::-1] cost=0 cnt=1 for j in range(n): if cnt<=i: if j%(q+2) in [q+1,q]: if j%(q+2)==q+1: cnt+=1 continue cost+=A[j] else: cost+=A[j] if x%q==1: cost-=A[-1] print(cost) exit()
45
404
9,216,000
118885816
m=int(input()) lstm = list(map(int, input().strip().split(' '))) n=int(input()) lstn = list(map(int, input().strip().split(' '))) m1=min(lstm) lstn.sort(reverse=True) i=0 s=0 while(i<n): if i+m1<=n: s+=sum(lstn[i:i+m1]) i+=m1 if i+2<=n: i+=2 elif i+1<=n: i+=1 else: s+=sum(lstn[i:n]) i=n print(s)
Codeforces Round 160 (Div. 1)
CF
2,013
2
256
Maxim and Discounts
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems. There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart. Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well. Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105). The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices. The numbers in the lines are separated by single spaces.
In a single line print a single integer — the answer to the problem.
null
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200. In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
[{"input": "1\n2\n4\n50 50 100 100", "output": "200"}, {"input": "2\n2 3\n5\n50 50 50 50 50", "output": "150"}, {"input": "1\n1\n7\n1 1 1 1 1 1 1", "output": "3"}]
1,400
["greedy", "sortings"]
45
[{"input": "1\r\n2\r\n4\r\n50 50 100 100\r\n", "output": "200\r\n"}, {"input": "2\r\n2 3\r\n5\r\n50 50 50 50 50\r\n", "output": "150\r\n"}, {"input": "1\r\n1\r\n7\r\n1 1 1 1 1 1 1\r\n", "output": "3\r\n"}, {"input": "18\r\n16 16 20 12 13 10 14 15 4 5 6 8 4 11 12 11 16 7\r\n15\r\n371 2453 905 1366 6471 4331 4106 2570 4647 1648 7911 2147 1273 6437 3393\r\n", "output": "38578\r\n"}, {"input": "2\r\n12 4\r\n28\r\n5366 5346 1951 3303 1613 5826 8035 7079 7633 6155 9811 9761 3207 4293 3551 5245 7891 4463 3981 2216 3881 1751 4495 96 671 1393 1339 4241\r\n", "output": "89345\r\n"}, {"input": "57\r\n3 13 20 17 18 18 17 2 17 8 20 2 11 12 11 14 4 20 9 20 14 19 20 4 4 8 8 18 17 16 18 10 4 7 9 8 10 8 20 4 11 8 12 16 16 4 11 12 16 1 6 14 11 12 19 8 20\r\n7\r\n5267 7981 1697 826 6889 1949 2413\r\n", "output": "11220\r\n"}, {"input": "1\r\n1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n2\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n1\r\n3\r\n3 1 1\r\n", "output": "3\r\n"}]
false
stdio
null
true
1006/A
1006
A
Python 3
TESTS
1
31
4,505,600
157613253
num = input() input = input() testcase_string = input.split(' ') testcase = [] numbers = [] for i in testcase_string: testcase.append(int(i)) numbers.append(int(i)) count = 0 i = 0 while i < len(testcase) + count: while i in testcase: index = testcase.index(i) if i % 2 == 0: testcase[index] = i - 1 count += 1 else: testcase[index] = i + 1 count += 1 i += 1 for i in testcase: print(i)
18
31
102,400
208969372
def main(): size = input() for i in input().split(): i = int(i) if i & 1: print(i, end=' ') else: print(i - 1, end=' ') if __name__ == '__main__': main()
Codeforces Round 498 (Div. 3)
ICPC
2,018
1
256
Adjacent Replacements
Mishka got an integer array $$$a$$$ of length $$$n$$$ as a birthday present (what a surprise!). Mishka doesn't like this present and wants to change it somehow. He has invented an algorithm and called it "Mishka's Adjacent Replacements Algorithm". This algorithm can be represented as a sequence of steps: - Replace each occurrence of $$$1$$$ in the array $$$a$$$ with $$$2$$$; - Replace each occurrence of $$$2$$$ in the array $$$a$$$ with $$$1$$$; - Replace each occurrence of $$$3$$$ in the array $$$a$$$ with $$$4$$$; - Replace each occurrence of $$$4$$$ in the array $$$a$$$ with $$$3$$$; - Replace each occurrence of $$$5$$$ in the array $$$a$$$ with $$$6$$$; - Replace each occurrence of $$$6$$$ in the array $$$a$$$ with $$$5$$$; - $$$\dots$$$ - Replace each occurrence of $$$10^9 - 1$$$ in the array $$$a$$$ with $$$10^9$$$; - Replace each occurrence of $$$10^9$$$ in the array $$$a$$$ with $$$10^9 - 1$$$. Note that the dots in the middle of this algorithm mean that Mishka applies these replacements for each pair of adjacent integers ($$$2i - 1, 2i$$$) for each $$$i \in\{1, 2, \ldots, 5 \cdot 10^8\}$$$ as described above. For example, for the array $$$a = [1, 2, 4, 5, 10]$$$, the following sequence of arrays represents the algorithm: $$$[1, 2, 4, 5, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$1$$$ with $$$2$$$) $$$\rightarrow$$$ $$$[2, 2, 4, 5, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$2$$$ with $$$1$$$) $$$\rightarrow$$$ $$$[1, 1, 4, 5, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$3$$$ with $$$4$$$) $$$\rightarrow$$$ $$$[1, 1, 4, 5, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$4$$$ with $$$3$$$) $$$\rightarrow$$$ $$$[1, 1, 3, 5, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$5$$$ with $$$6$$$) $$$\rightarrow$$$ $$$[1, 1, 3, 6, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$6$$$ with $$$5$$$) $$$\rightarrow$$$ $$$[1, 1, 3, 5, 10]$$$ $$$\rightarrow$$$ $$$\dots$$$ $$$\rightarrow$$$ $$$[1, 1, 3, 5, 10]$$$ $$$\rightarrow$$$ (replace all occurrences of $$$10$$$ with $$$9$$$) $$$\rightarrow$$$ $$$[1, 1, 3, 5, 9]$$$. The later steps of the algorithm do not change the array. Mishka is very lazy and he doesn't want to apply these changes by himself. But he is very interested in their result. Help him find it.
The first line of the input contains one integer number $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of elements in Mishka's birthday present (surprisingly, an array). The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.
Print $$$n$$$ integers — $$$b_1, b_2, \dots, b_n$$$, where $$$b_i$$$ is the final value of the $$$i$$$-th element of the array after applying "Mishka's Adjacent Replacements Algorithm" to the array $$$a$$$. Note that you cannot change the order of elements in the array.
null
The first example is described in the problem statement.
[{"input": "5\n1 2 4 5 10", "output": "1 1 3 5 9"}, {"input": "10\n10000 10 50605065 1 5 89 5 999999999 60506056 1000000000", "output": "9999 9 50605065 1 5 89 5 999999999 60506055 999999999"}]
800
["implementation"]
18
[{"input": "5\r\n1 2 4 5 10\r\n", "output": "1 1 3 5 9\r\n"}, {"input": "10\r\n10000 10 50605065 1 5 89 5 999999999 60506056 1000000000\r\n", "output": "9999 9 50605065 1 5 89 5 999999999 60506055 999999999\r\n"}, {"input": "1\r\n999999999\r\n", "output": "999999999\r\n"}, {"input": "1\r\n1000000000\r\n", "output": "999999999\r\n"}, {"input": "1\r\n210400\r\n", "output": "210399\r\n"}, {"input": "5\r\n100000000 100000000 100000000 100000000 100000000\r\n", "output": "99999999 99999999 99999999 99999999 99999999\r\n"}, {"input": "1\r\n2441139\r\n", "output": "2441139\r\n"}, {"input": "2\r\n2 2\r\n", "output": "1 1\r\n"}, {"input": "3\r\n2 2 2\r\n", "output": "1 1 1\r\n"}, {"input": "2\r\n4 4\r\n", "output": "3 3\r\n"}]
false
stdio
null
true
43/C
43
C
PyPy 3
TESTS
2
154
20,275,200
129770228
import sys from math import sqrt,ceil,floor,gcd from collections import Counter input = lambda:sys.stdin.readline() def int_arr(): return list(map(int,input().split())) def str_arr(): return list(map(str,input().split())) def get_str(): return map(str,input().split()) def get_int(): return map(int,input().split()) def get_flo(): return map(float,input().split()) def lcm(a,b): return (a*b) // gcd(a,b) mod = 1000000007 def solve(arr): d = {} for i in arr: if i in d: d[i] += 1 else: d[i] = 1 c = 0 for i in d.keys(): c += d[i]//2 print(c) # for _ in range(int(input())): n = int(input()) arr = int_arr() solve(arr)
21
92
409,600
192525785
a = [0] * 3 _ = input() for n in map(int, input().split()): a[n % 3] += 1 print(a[0] // 2 + min(a[1], a[2]))
Codeforces Beta Round 42 (Div. 2)
CF
2,010
2
256
Lucky Tickets
Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid also threw part of the pieces away. Having seen this, Vasya got terrified but still tried to restore the collection. He chose several piece pairs and glued each pair together so that each pair formed a lucky ticket. The rest of the pieces Vasya threw away reluctantly. Thus, after the gluing of the 2t pieces he ended up with t tickets, each of which was lucky. When Leonid tore the tickets in two pieces, one piece contained the first several letters of his number and the second piece contained the rest. Vasya can glue every pair of pieces in any way he likes, but it is important that he gets a lucky ticket in the end. For example, pieces 123 and 99 can be glued in two ways: 12399 and 99123. What maximum number of tickets could Vasya get after that?
The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to count as lucky. Vasya does not have to use all the pieces. The numbers on the pieces an on the resulting tickets may coincide.
Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together.
null
null
[{"input": "3\n123 123 99", "output": "1"}, {"input": "6\n1 1 1 23 10 3", "output": "1"}]
1,300
["greedy"]
21
[{"input": "3\r\n123 123 99\r\n", "output": "1\r\n"}, {"input": "6\r\n1 1 1 23 10 3\r\n", "output": "1\r\n"}, {"input": "3\r\n43440907 58238452 82582355\r\n", "output": "1\r\n"}, {"input": "4\r\n31450303 81222872 67526764 17516401\r\n", "output": "1\r\n"}, {"input": "5\r\n83280 20492640 21552119 7655071 47966344\r\n", "output": "2\r\n"}, {"input": "6\r\n94861402 89285133 30745405 41537407 90189008 83594323\r\n", "output": "1\r\n"}, {"input": "7\r\n95136773 99982752 97528336 79027944 96847471 96928960 89423004\r\n", "output": "2\r\n"}, {"input": "1\r\n19938466\r\n", "output": "0\r\n"}, {"input": "2\r\n55431511 35254032\r\n", "output": "0\r\n"}, {"input": "2\r\n28732939 23941418\r\n", "output": "1\r\n"}, {"input": "10\r\n77241684 71795210 50866429 35232438 22664883 56785812 91050433 75677099 84393937 43832346\r\n", "output": "4\r\n"}]
false
stdio
null
true
627/D
627
D
PyPy 3
TESTS
2
139
0
80967206
import sys input = sys.stdin.readline n, k = map(int, input().split()) a = [int(i) for i in input().split()] g = [[] for _ in range(n)] for i in range(n - 1): u, v = map(int, input().split()) g[u-1].append(v-1) g[v-1].append(u-1) stack = [0] done = [False] * n par = [0] * n order = [] while len(stack) > 0: x = stack.pop() done[x] = True order.append(x) for i in g[x]: if done[i] == False: par[i] = x stack.append(i) order = order[::-1] sub = [0] * n for i in order: sub[i] = 1 for j in g[i]: if par[j] == i: sub[i] += sub[j] def good(guess): cnt = [0] * n for i in order: if a[i] < guess: continue cnt[i] = 1 opt = 0 for j in g[i]: if par[j] == i: if cnt[j] == sub[j]: cnt[i] += cnt[j] else: opt = max(opt, cnt[j]) cnt[i] += opt return cnt[0] >= k l, r = 0, max(a) while l < r: mid = (l + r + 1) // 2 if good(mid): l = mid else: r = mid - 1 print(l)
69
5,116
50,380,800
80969556
import sys input = sys.stdin.readline n, k = map(int, input().split()) a = [int(i) for i in input().split()] g = [[] for _ in range(n)] for i in range(n - 1): u, v = map(int, input().split()) g[u-1].append(v-1) g[v-1].append(u-1) stack = [0] done = [False] * n par = [0] * n order = [] while len(stack) > 0: x = stack.pop() done[x] = True order.append(x) for i in g[x]: if done[i] == False: par[i] = x stack.append(i) order = order[::-1] sub = [0] * n for i in order: sub[i] = 1 for j in g[i]: if par[j] == i: sub[i] += sub[j] def good(guess): cnt = [0] * n for i in order: if a[i] < guess: continue cnt[i] = 1 opt = 0 for j in g[i]: if par[j] == i: if cnt[j] == sub[j]: cnt[i] += cnt[j] else: opt = max(opt, cnt[j]) cnt[i] += opt if cnt[0] >= k: return True up = [0] * n for i in order[::-1]: if a[i] < guess: continue opt, secondOpt = 0, 0 total = 1 for j in g[i]: val, size = 0, 0 if par[j] == i: val = cnt[j] size = sub[j] else: val = up[i] size = n - sub[i] if val == size: total += val else: if opt < val: opt, secondOpt = val, opt elif secondOpt < val: secondOpt = val for j in g[i]: if par[j] == i: up[j] = total add = opt if sub[j] == cnt[j]: up[j] -= cnt[j] elif cnt[j] == opt: add = secondOpt up[j] += add for i in range(n): if a[i] < guess: continue total, opt = 1, 0 for j in g[i]: val, size = 0, 0 if par[j] == i: val = cnt[j] size = sub[j] else: val = up[i] size = n - sub[i] if val == size: total += val else: opt = max(opt, val) if total + opt >= k: return True return False l, r = 0, max(a) while l < r: mid = (l + r + 1) // 2 if good(mid): l = mid else: r = mid - 1 print(l)
8VC Venture Cup 2016 - Final Round
CF
2,016
7
256
Preorder Test
For his computer science class, Jacob builds a model tree with sticks and balls containing n nodes in the shape of a tree. Jacob has spent ai minutes building the i-th ball in the tree. Jacob's teacher will evaluate his model and grade Jacob based on the effort he has put in. However, she does not have enough time to search his whole tree to determine this; Jacob knows that she will examine the first k nodes in a DFS-order traversal of the tree. She will then assign Jacob a grade equal to the minimum ai she finds among those k nodes. Though Jacob does not have enough time to rebuild his model, he can choose the root node that his teacher starts from. Furthermore, he can rearrange the list of neighbors of each node in any order he likes. Help Jacob find the best grade he can get on this assignment. A DFS-order traversal is an ordering of the nodes of a rooted tree, built by a recursive DFS-procedure initially called on the root of the tree. When called on a given node v, the procedure does the following: 1. Print v. 2. Traverse the list of neighbors of the node v in order and iteratively call DFS-procedure on each one. Do not call DFS-procedure on node u if you came to node v directly from u.
The first line of the input contains two positive integers, n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ n) — the number of balls in Jacob's tree and the number of balls the teacher will inspect. The second line contains n integers, ai (1 ≤ ai ≤ 1 000 000), the time Jacob used to build the i-th ball. Each of the next n - 1 lines contains two integers ui, vi (1 ≤ ui, vi ≤ n, ui ≠ vi) representing a connection in Jacob's tree between balls ui and vi.
Print a single integer — the maximum grade Jacob can get by picking the right root of the tree and rearranging the list of neighbors.
null
In the first sample, Jacob can root the tree at node 2 and order 2's neighbors in the order 4, 1, 5 (all other nodes have at most two neighbors). The resulting preorder traversal is 2, 4, 1, 3, 5, and the minimum ai of the first 3 nodes is 3. In the second sample, it is clear that any preorder traversal will contain node 1 as either its first or second node, so Jacob cannot do better than a grade of 1.
[{"input": "5 3\n3 6 1 4 2\n1 2\n2 4\n2 5\n1 3", "output": "3"}, {"input": "4 2\n1 5 5 5\n1 2\n1 3\n1 4", "output": "1"}]
2,600
["binary search", "dfs and similar", "dp", "graphs", "greedy", "trees"]
69
[{"input": "5 3\r\n3 6 1 4 2\r\n1 2\r\n2 4\r\n2 5\r\n1 3\r\n", "output": "3\r\n"}, {"input": "4 2\r\n1 5 5 5\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "1\r\n"}, {"input": "2 1\r\n1 100000\r\n2 1\r\n", "output": "100000\r\n"}, {"input": "2 2\r\n1 1000000\r\n1 2\r\n", "output": "1\r\n"}, {"input": "10 4\r\n104325 153357 265088 777795 337716 557321 702646 734430 464449 744072\r\n9 4\r\n8 1\r\n10 7\r\n8 6\r\n7 9\r\n8 2\r\n3 5\r\n8 3\r\n10 8\r\n", "output": "557321\r\n"}, {"input": "10 3\r\n703660 186846 317819 628672 74457 58472 247014 480113 252764 860936\r\n10 6\r\n7 4\r\n10 9\r\n9 5\r\n6 3\r\n6 2\r\n7 1\r\n10 7\r\n10 8\r\n", "output": "252764\r\n"}, {"input": "10 10\r\n794273 814140 758469 932911 607860 683826 987442 652494 952171 698608\r\n1 3\r\n3 8\r\n2 7\r\n2 1\r\n2 9\r\n3 10\r\n6 4\r\n9 6\r\n3 5\r\n", "output": "607860\r\n"}]
false
stdio
null
true
545/B
545
B
Python 3
TESTS
2
61
307,200
198691333
def girl(s, t): distance = 0 ls = len(s) mas_dif = [] for i in range(ls): if s[i] != t[i]: mas_dif.append(i) distance += 1 if distance%2!=0: print("impossible") return 0 else: gg = "" for i in range(len(s)-1): if mas_dif[0]==i: if distance > 0: distance-=1 gg += "1" else: gg += "0" gg += s[i] print(gg) return 0 s = input() t = input() girl(s, t)
54
93
1,126,400
11153223
import operator def gen(s, t, c): for a, b, in zip(s, t): if c and a != b: c -= 1 yield a else: yield b s = input() t = input() count = len(s) - sum(map(operator.eq, s, t)) if count & 1 == 0: print(str.join("", gen(s, t, count // 2))) 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
314/C
314
C
Python 3
TESTS
3
436
5,017,600
16949569
class FenwickTree(object): """Interface for Binary-Indexed Tree, which stores cumulative frequency tables. Note that original data is not preserved in this data structure. Construction takes O(n lg n) time; Item access takes O(lg n) time; Item update takes O(lg n) time.""" def __init__(self): self.tree = [0] def __len__(self): return len(self.tree) - 1 def __getitem__(self, key): """ Example: In list [1,2,3,4], FT[3] returns 6. :param key: 1-based index :return: cumulative sum up to key-th item """ psum = 0 while key > 0: psum += self.tree[key] key -= (key & -key) return psum def add(self, key, value): """ Example: FT.add(3,5) modifies list [1,2,3,4] to [1,2,8,4] :param key: 1-based index :param value: value to be modified to :return: None """ while key <= len(self): self.tree[key] += value key += (key & -key) def append(self, value): """ Example: FT.append(5) turns list [1,2,3,4] to [1,2,3,4,5] :param value: value to be appended :return: None """ ln = len(self) large = self[ln] small = self[(ln + 1) - ((ln + 1) & -(ln + 1))] self.tree.append(large - small + value) # Above imported from Common Algorithms library. n = int(input()) a = sorted(map(int, input().split())) q = [0] * (a[-1] + 1) FT = FenwickTree() for x in q: FT.append(x) for x in a: v = FT[x] * x + x FT.add(x, v - q[x]) q[x] = v print(FT[a[-1]])
24
1,402
41,267,200
169478092
import os import sys from io import BytesIO, IOBase # region fastio BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") MOD_NUM = 10**9+7 class SegTree: def __init__(self, n): sz = 1 while sz<n: sz *= 2 tree = [0]*(2*sz) self.tree = tree self.sz = sz self.n = n def update(self, i, x): sz = self.sz n = self.n tree = self.tree node = i + sz tree[node] = x tree[node] %= MOD_NUM node //= 2 while node>0: tree[node] = tree[2*node] + tree[2*node+1] tree[node] %= MOD_NUM node //= 2 def query(self, ql, qr, u=1, l=0, r=None): sz = self.sz n = self.n tree = self.tree if r is None: r=sz-1 if ql<=l and r<=qr: return tree[u]%MOD_NUM if r<ql or qr<l: return 0 mid = (l+r)//2 left = self.query(ql, qr, 2*u, l, mid) right = self.query(ql, qr, 2*u+1, mid+1, r) return (left + right)%MOD_NUM n = int(input()) a = list(map(int,input().split())) tree = SegTree(10**6+10) for x in a: pref_sum = tree.query(1, x) tree.update(x, pref_sum*x + x) ans = tree.query(0, 10**6) print(ans)
Codeforces Round 187 (Div. 1)
CF
2,013
2
256
Sereja and Subsequences
Sereja has a sequence that consists of n positive integers, a1, a2, ..., an. First Sereja took a piece of squared paper and wrote all distinct non-empty non-decreasing subsequences of sequence a. Then for each sequence written on the squared paper, Sereja wrote on a piece of lines paper all sequences that do not exceed it. A sequence of positive integers x = x1, x2, ..., xr doesn't exceed a sequence of positive integers y = y1, y2, ..., yr, if the following inequation holds: x1 ≤ y1, x2 ≤ y2, ..., xr ≤ yr. Now Sereja wonders, how many sequences are written on the lines piece of paper. Help Sereja, find the required quantity modulo 1000000007 (109 + 7).
The first line contains integer n (1 ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106).
In the single line print the answer to the problem modulo 1000000007 (109 + 7).
null
null
[{"input": "1\n42", "output": "42"}, {"input": "3\n1 2 2", "output": "13"}, {"input": "5\n1 2 3 4 5", "output": "719"}]
2,000
["data structures", "dp"]
24
[{"input": "1\r\n42\r\n", "output": "42\r\n"}, {"input": "3\r\n1 2 2\r\n", "output": "13\r\n"}, {"input": "5\r\n1 2 3 4 5\r\n", "output": "719\r\n"}, {"input": "4\r\n11479 29359 26963 24465\r\n", "output": "927446239\r\n"}, {"input": "5\r\n5706 28146 23282 16828 9962\r\n", "output": "446395832\r\n"}, {"input": "6\r\n492 2996 11943 4828 5437 32392\r\n", "output": "405108414\r\n"}, {"input": "7\r\n14605 3903 154 293 12383 17422 18717\r\n", "output": "975867090\r\n"}, {"input": "8\r\n19719 19896 5448 21727 14772 11539 1870 19913\r\n", "output": "937908482\r\n"}, {"input": "9\r\n3 3 2 2 3 2 1 1 2\r\n", "output": "93\r\n"}, {"input": "10\r\n3 3 3 2 1 2 3 2 3 1\r\n", "output": "529\r\n"}, {"input": "42\r\n8 5 1 10 5 9 9 3 5 6 6 2 8 2 2 6 3 8 7 2 5 3 4 3 3 2 7 9 6 8 7 2 9 10 3 8 10 6 5 4 2 3\r\n", "output": "608660833\r\n"}, {"input": "42\r\n68 35 1 70 25 79 59 63 65 6 46 82 28 62 92 96 43 28 37 92 5 3 54 93 83 22 17 19 96 48 27 72 39 70 13 68 100 36 95 4 12 23\r\n", "output": "56277550\r\n"}, {"input": "42\r\n18468 6335 26501 19170 15725 11479 29359 26963 24465 5706 28146 23282 16828 9962 492 2996 11943 4828 5437 32392 14605 3903 154 293 12383 17422 18717 19719 19896 5448 21727 14772 11539 1870 19913 25668 26300 17036 9895 28704 23812 31323\r\n", "output": "955898058\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n1000000\r\n", "output": "1000000\r\n"}]
false
stdio
null
true
28/A
28
A
PyPy 3
TESTS
0
122
0
218407688
n, m = map(int, input().split()) nails = [list(map(int, input().split())) for _ in range(n)] rod_lengths = list(map(int, input().split())) result = [-1] * n for i in range(n): x1, y1 = nails[i] x2, y2 = nails[(i + 1) % n] dx = abs(x2 - x1) dy = abs(y2 - y1) if dx > dy: direction = 0 # Fold the rod horizontally else: direction = 1 # Fold the rod vertically for j in range(m): if rod_lengths[j] > max(dx, dy) and result[i] == -1: result[i] = j + 1 rod_lengths[j] = -1 # Mark the rod as used # Check if all nails have been used if -1 in result: print("NO") else: print("YES") print(*result)
51
778
9,625,600
93132115
#Code by Sounak, IIESTS #------------------------------warmup---------------------------- import os import sys import math from io import BytesIO, IOBase from fractions import Fraction import collections from itertools import permutations from collections import defaultdict from collections import deque import threading BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") #-------------------game starts now----------------------------------------------------- class Factorial: def __init__(self, MOD): self.MOD = MOD self.factorials = [1, 1] self.invModulos = [0, 1] self.invFactorial_ = [1, 1] def calc(self, n): if n <= -1: print("Invalid argument to calculate n!") print("n must be non-negative value. But the argument was " + str(n)) exit() if n < len(self.factorials): return self.factorials[n] nextArr = [0] * (n + 1 - len(self.factorials)) initialI = len(self.factorials) prev = self.factorials[-1] m = self.MOD for i in range(initialI, n + 1): prev = nextArr[i - initialI] = prev * i % m self.factorials += nextArr return self.factorials[n] def inv(self, n): if n <= -1: print("Invalid argument to calculate n^(-1)") print("n must be non-negative value. But the argument was " + str(n)) exit() p = self.MOD pi = n % p if pi < len(self.invModulos): return self.invModulos[pi] nextArr = [0] * (n + 1 - len(self.invModulos)) initialI = len(self.invModulos) for i in range(initialI, min(p, n + 1)): next = -self.invModulos[p % i] * (p // i) % p self.invModulos.append(next) return self.invModulos[pi] def invFactorial(self, n): if n <= -1: print("Invalid argument to calculate (n^(-1))!") print("n must be non-negative value. But the argument was " + str(n)) exit() if n < len(self.invFactorial_): return self.invFactorial_[n] self.inv(n) # To make sure already calculated n^-1 nextArr = [0] * (n + 1 - len(self.invFactorial_)) initialI = len(self.invFactorial_) prev = self.invFactorial_[-1] p = self.MOD for i in range(initialI, n + 1): prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p self.invFactorial_ += nextArr return self.invFactorial_[n] class Combination: def __init__(self, MOD): self.MOD = MOD self.factorial = Factorial(MOD) def ncr(self, n, k): if k < 0 or n < k: return 0 k = min(k, n - k) f = self.factorial return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD #------------------------------------------------------------------------- def dist(a, b): return abs(a[0] - b[0]) + abs(a[1] - b[1]) def get_sorted_required_pruts(dists): res = [dists[i * 2] + dists[i * 2 + 1] for i in range(len(dists) // 2)] res = [(i, x) for i, x in enumerate(res)] return sorted(res, key=lambda x: x[1]) def get_answer(pruts, required_pruts): i = 0 j = 0 answer = "YES" seq = [] while i < len(required_pruts): if j == len(pruts): answer = "NO" return answer, None if pruts[j][1] > required_pruts[i][1]: answer = "NO" return answer, None if pruts[j][1] < required_pruts[i][1]: j += 1 else: seq.append((required_pruts[i][0], pruts[j][0] + 1)) i += 1 j += 1 return answer, [x[1] for x in sorted(seq)] n, m = map(int,input().split()) gvozdi = [None] * n for i in range(n): gvozdi[i] = list(map(int,input().split())) pruts = list(map(int,input().split())) pruts = [(i, p) for i, p in enumerate(pruts)] pruts = sorted(pruts, key=lambda x: x[1]) dists = [dist(gvozdi[i], gvozdi[i + 1]) for i in range(len(gvozdi) - 1)] dists.append(dist(gvozdi[0], gvozdi[-1])) even_required_pruts = get_sorted_required_pruts(dists) # print(dists[-1:] + dists[:-1]) odd_required_pruts = get_sorted_required_pruts(dists[-1:] + dists[:-1]) even_answer, even_seq = get_answer(pruts, even_required_pruts) odd_answer, odd_seq = get_answer(pruts, odd_required_pruts) if even_answer == "NO" and odd_answer == "NO": print("NO") elif even_answer == "YES": print("YES") even_seq = [even_seq[i // 2] if i % 2 == 1 else -1 for i in range(n)] print(" ".join(map(str, even_seq))) else: print("YES") # print(odd_seq) odd_seq = [odd_seq[i // 2] if i % 2 == 0 else -1 for i in range(n)] print(" ".join(map(str, odd_seq)))
Codeforces Beta Round 28 (Codeforces format)
CF
2,010
2
256
Bender Problem
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods. Help Bender to solve this difficult task.
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod. If there are multiple solutions, print any of them.
null
null
[{"input": "4 2\n0 0\n0 2\n2 2\n2 0\n4 4", "output": "YES\n1 -1 2 -1"}, {"input": "6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n3 2 3", "output": "YES\n1 -1 2 -1 3 -1"}, {"input": "6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n2 2 3", "output": "NO"}]
1,600
["implementation"]
51
[{"input": "4 2\r\n0 0\r\n0 2\r\n2 2\r\n2 0\r\n4 4\r\n", "output": "YES\r\n1 -1 2 -1 \r\n"}, {"input": "6 3\r\n0 0\r\n1 0\r\n1 1\r\n2 1\r\n2 2\r\n0 2\r\n3 2 3\r\n", "output": "YES\r\n1 -1 2 -1 3 -1 \r\n"}, {"input": "6 3\r\n0 0\r\n1 0\r\n1 1\r\n2 1\r\n2 2\r\n0 2\r\n2 2 3\r\n", "output": "NO\r\n"}, {"input": "4 4\r\n0 0\r\n0 1\r\n1 1\r\n1 0\r\n1 1 1 1\r\n", "output": "NO\r\n"}, {"input": "6 2\r\n0 0\r\n1 0\r\n1 1\r\n2 1\r\n2 2\r\n0 2\r\n2 2\r\n", "output": "NO\r\n"}, {"input": "6 3\r\n0 0\r\n2 0\r\n2 2\r\n1 2\r\n1 1\r\n0 1\r\n4 2 2\r\n", "output": "YES\r\n-1 1 -1 2 -1 3 "}, {"input": "4 4\r\n-8423 7689\r\n6902 7689\r\n6902 2402\r\n-8423 2402\r\n20612 20612 91529 35617\r\n", "output": "YES\r\n1 -1 2 -1 \r\n"}, {"input": "4 4\r\n1679 -198\r\n9204 -198\r\n9204 -5824\r\n1679 -5824\r\n18297 92466 187436 175992\r\n", "output": "NO\r\n"}, {"input": "4 2\r\n0 0\r\n0 2\r\n2 2\r\n2 0\r\n200000 200000\r\n", "output": "NO\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, m = map(int, f.readline().split()) nails = [tuple(map(int, f.readline().split())) for _ in range(n)] rods = list(map(int, f.readline().split())) with open(output_path) as f: ref_lines = f.read().splitlines() with open(submission_path) as f: sub_lines = f.read().splitlines() if not sub_lines: print(0) return ref_is_no = ref_lines and ref_lines[0].strip() == "NO" sub_is_no = sub_lines and sub_lines[0].strip() == "NO" if ref_is_no: print(1 if sub_is_no else 0) return else: if not sub_lines or sub_lines[0].strip() != "YES": print(0) return if len(sub_lines) < 2: print(0) return try: sub_list = list(map(int, sub_lines[1].split())) except: print(0) return if len(sub_list) != n: print(0) return used_rods = set() for num in sub_list: if num == -1: continue if num < 1 or num > m or num in used_rods: print(0) return used_rods.add(num) for i in range(n): rod_num = sub_list[i] if rod_num == -1: continue prev_i = i - 1 if i > 0 else n - 1 next_i = (i + 1) % n x_prev, y_prev = nails[prev_i] x_curr, y_curr = nails[i] x_next, y_next = nails[next_i] len_prev = abs(y_curr - y_prev) if x_prev == x_curr else abs(x_curr - x_prev) len_next = abs(y_next - y_curr) if x_curr == x_next else abs(x_next - x_curr) sum_len = len_prev + len_next rod_index = rod_num - 1 if rod_index >= len(rods) or rods[rod_index] != sum_len: print(0) return print(1) if __name__ == "__main__": main()
true
797/E
797
E
PyPy 3
TESTS
7
639
8,499,200
214253125
n = int(input()) a = [int(i) for i in input().split(' ')] q = int(input()) for test in range(n): p, k = map(int, input().split()) ans = 0 while p <= n: p += a[p-1] + k ans += 1 print(ans)
53
685
137,523,200
158654991
import sys import io, os input = sys.stdin.buffer.readline #input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline import math n = int(input()) A = list(map(int, input().split())) limit = int(math.sqrt(n)) dp = [[-1]*n for i in range(limit+1)] for k in range(1, limit+1): for i in range(n-1, -1, -1): if i+A[i]+k >= n: dp[k][i] = 1 else: dp[k][i] = dp[k][i+A[i]+k]+1 q = int(input()) for i in range(q): p, k = map(int, input().split()) if k >= limit: ans = 0 while p <= n: p = p+A[p-1]+k ans += 1 print(ans) else: print(dp[k][p-1])
Educational Codeforces Round 19
ICPC
2,017
2
256
Array Queries
a is an array of n positive integers, all of which are not greater than n. You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.
The first line contains one integer n (1 ≤ n ≤ 100000). The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n). The third line containts one integer q (1 ≤ q ≤ 100000). Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).
Print q integers, ith integer must be equal to the answer to ith query.
null
Consider first example: In first query after first operation p = 3, after second operation p = 5. In next two queries p is greater than n after the first operation.
[{"input": "3\n1 1 1\n3\n1 1\n2 1\n3 1", "output": "2\n1\n1"}]
2,000
["brute force", "data structures", "dp"]
53
[{"input": "3\r\n1 1 1\r\n3\r\n1 1\r\n2 1\r\n3 1\r\n", "output": "2\r\n1\r\n1\r\n"}, {"input": "10\r\n3 5 4 3 7 10 6 7 2 3\r\n10\r\n4 5\r\n2 10\r\n4 6\r\n9 9\r\n9 2\r\n5 1\r\n6 4\r\n1 1\r\n5 6\r\n6 4\r\n", "output": "1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n1\r\n"}, {"input": "50\r\n6 2 5 6 10 2 5 8 9 2 9 5 10 4 3 6 10 6 1 1 3 7 2 1 7 8 5 9 6 2 7 6 1 7 2 10 10 2 4 2 8 4 3 10 7 1 7 8 6 3\r\n50\r\n23 8\r\n12 8\r\n3 3\r\n46 3\r\n21 6\r\n7 4\r\n26 4\r\n12 1\r\n25 4\r\n18 7\r\n29 8\r\n27 5\r\n47 1\r\n50 2\r\n46 7\r\n13 6\r\n44 8\r\n12 2\r\n18 3\r\n35 10\r\n38 1\r\n7 10\r\n4 2\r\n22 8\r\n36 3\r\n25 2\r\n47 3\r\n33 5\r\n10 5\r\n12 9\r\n7 4\r\n26 4\r\n19 4\r\n3 8\r\n12 3\r\n35 8\r\n31 4\r\n25 5\r\n3 5\r\n46 10\r\n37 6\r\n8 9\r\n20 5\r\n36 1\r\n41 9\r\n6 7\r\n40 5\r\n24 4\r\n41 10\r\n14 8\r\n", "output": "3\r\n4\r\n6\r\n2\r\n4\r\n5\r\n3\r\n8\r\n3\r\n3\r\n2\r\n2\r\n1\r\n1\r\n1\r\n3\r\n1\r\n6\r\n5\r\n2\r\n3\r\n3\r\n7\r\n2\r\n2\r\n4\r\n1\r\n3\r\n4\r\n3\r\n5\r\n3\r\n5\r\n5\r\n6\r\n2\r\n3\r\n2\r\n4\r\n1\r\n1\r\n3\r\n4\r\n2\r\n1\r\n5\r\n2\r\n4\r\n1\r\n3\r\n"}, {"input": "50\r\n49 21 50 31 3 19 10 40 20 5 47 12 12 30 5 4 5 2 11 23 5 39 3 30 19 3 23 40 4 14 39 50 45 14 33 37 1 15 43 30 45 23 32 3 9 17 9 5 14 12\r\n50\r\n36 29\r\n44 24\r\n38 22\r\n30 43\r\n15 19\r\n39 2\r\n13 8\r\n29 50\r\n37 18\r\n32 19\r\n42 39\r\n20 41\r\n14 11\r\n4 25\r\n14 35\r\n17 23\r\n1 29\r\n3 19\r\n8 6\r\n26 31\r\n9 46\r\n36 31\r\n21 49\r\n17 38\r\n47 27\r\n5 21\r\n42 22\r\n36 34\r\n50 27\r\n11 45\r\n26 41\r\n1 16\r\n21 39\r\n18 43\r\n21 37\r\n12 16\r\n22 32\r\n7 18\r\n10 14\r\n43 37\r\n18 50\r\n21 32\r\n27 32\r\n48 16\r\n5 6\r\n5 11\r\n5 6\r\n39 29\r\n40 13\r\n14 6\r\n", "output": "1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n2\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n2\r\n1\r\n2\r\n3\r\n1\r\n1\r\n1\r\n1\r\n1\r\n3\r\n3\r\n3\r\n1\r\n1\r\n2\r\n"}]
false
stdio
null
true
545/B
545
B
Python 3
TESTS
2
124
8,192,000
194939999
# def GetHammingDistance(firstString: str, secondString: str, length: int) -> int: # result = 0 # for index in range(0, length): # if firstString[index] != secondString[index]: # result += 1 # return result def GetInfo(firstString: str, secondString: str, length : int) -> list: indexs = [] for index in range(0, length): if firstString[index] != secondString[index]: indexs.append(index) if firstString.count("1") > secondString.count("1"): indexs.append(firstString) else: indexs.append(firstString) return indexs def Solution(): firstString = input() secondString = input() stringsLength = len(firstString) if stringsLength % 2 != 0: return "impossible" else: data = GetInfo(firstString, secondString, stringsLength) dataLength = len(data) if (dataLength - 1) % 2 != 0: return "impossible" smallString = data[len(data) - 1] resultString = "" for element in range(0, dataLength - 2): index = data[element] resultString = smallString[:index] + "1" + smallString[index + 1:] return resultString print(Solution())
54
93
2,355,200
208637051
s1 = input() s2 = input() nd = 0 for i in range(len(s1)): if s1[i] != s2[i]: nd+=1 if nd%2 != 0: print('impossible') else: sm = [] n2 = 0 for i in range(len(s1)): if s1[i] != s2[i]: if n2 < nd/2: sm.append(s1[i]) n2 = n2+1 else: sm.append(s2[i]) else: sm.append(s1[i]) print(''.join(sm))
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
19/A
19
A
PyPy 3-64
TESTS
3
124
0
227188826
n = int(input()) teams_names = [] t_points ={} for i in range(n): teams_names.append(input()) t_points[teams_names[i]] = 0 for i in range((n*(n-1))//2): t,s = input().split() t1, t2 = t.split('-') s1, s2 = s.split(':') if s1>s2: t_points[t1] += 3 elif s2>s1: t_points[t2] += 3 else: t_points[t1] += 1 t_points[t2] += 1 sorted_t_points = sorted(sorted(t_points.items(), key=lambda x:x[1], reverse=True)[:n//2]) for i in range(n//2): print(sorted_t_points[i][0])
29
92
0
226368602
# LUOGU_RID: 127228014 a=int(input());b={} for i in range(a):b[input()]=[0,0,0] for i in range(a*(a-1)//2): c=input().split();c[0]=c[0].split('-');c[1]=list(map(int,c[1].split(':')));b[c[0][0]][1]+=c[1][0]-c[1][1];b[c[0][0]][2]+=c[1][0];b[c[0][1]][1]+=c[1][1]-c[1][0];b[c[0][1]][2]+=c[1][1] if c[1][0]>c[1][1]:b[c[0][0]][0]+=3 elif c[1][0]<c[1][1]:b[c[0][1]][0]+=3 else:b[c[0][0]][0]+=1;b[c[0][1]][0]+=1 def r(i):return b[i] for i in sorted(sorted(b,key=r,reverse=1)[:a//2]):print(i)
Codeforces Beta Round 19
ICPC
2,010
2
64
World Football Cup
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations: - the final tournament features n teams (n is always even) - the first n / 2 teams (according to the standings) come through to the knockout stage - the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals - it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity. You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
The first input line contains the only integer n (1 ≤ n ≤ 50) — amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following n·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 — names of the teams; num1, num2 (0 ≤ num1, num2 ≤ 100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output n / 2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
null
null
[{"input": "4\nA\nB\nC\nD\nA-B 1:1\nA-C 2:2\nA-D 1:0\nB-C 1:0\nB-D 0:3\nC-D 0:3", "output": "A\nD"}, {"input": "2\na\nA\na-A 2:1", "output": "a"}]
1,400
["implementation"]
29
[{"input": "4\r\nA\r\nB\r\nC\r\nD\r\nA-B 1:1\r\nA-C 2:2\r\nA-D 1:0\r\nB-C 1:0\r\nB-D 0:3\r\nC-D 0:3\r\n", "output": "A\r\nD\r\n"}, {"input": "2\r\na\r\nA\r\na-A 2:1\r\n", "output": "a\r\n"}, {"input": "2\r\nEULEUbCmfrmqxtzvg\r\nuHGRmKUhDcxcfqyruwzen\r\nuHGRmKUhDcxcfqyruwzen-EULEUbCmfrmqxtzvg 13:92\r\n", "output": "EULEUbCmfrmqxtzvg\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 2:2\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "A\r\nB\r\nF\r\n"}, {"input": "6\r\nA\r\nB\r\nC\r\nD\r\nE\r\nF\r\nA-B 1:0\r\nA-C 0:0\r\nA-D 1:0\r\nA-E 5:5\r\nA-F 0:1\r\nB-C 1:0\r\nB-D 1:0\r\nB-E 1:0\r\nB-F 0:2\r\nC-D 7:7\r\nC-E 1:0\r\nC-F 1:0\r\nD-E 1:0\r\nD-F 1:0\r\nE-F 0:1\r\n", "output": "B\r\nC\r\nF\r\n"}]
false
stdio
null
true
28/A
28
A
Python 3
TESTS
3
216
307,200
67613492
n,m = map(int,input().split()) x = 0 y = 0 sx = [] sy = [] for i in range(n): A = map(int,input().split()) A = list(A) if A[0] in sx: sx = sx else: sx.append(A[0]) if A[1] in sy: sy = sy else: sy.append(A[1]) q = 2 * sx[-1] + 2 * sy[-1] ss = map(int,input().split()) ss = list(ss) if sum(ss) >= q: print('YES') ss = [] for i in range(1,m+1): ss.append(i) ss.append(-1) print(" ".join(map(str,ss))) else: print('NO')
51
92
204,800
212540864
def main(): n, m = map(int, input().split()) nails = [] for _ in range(n): nails.append(tuple(map(int, input().split()))) rods = list(map(int, input().split())) rods_count1, rods_count2 = {}, {} for i, rod in enumerate(rods): if rod not in rods_count1: rods_count1[rod] = [i] else: rods_count1[rod].append(i) if rod not in rods_count2: rods_count2[rod] = [i] else: rods_count2[rod].append(i) ans = [] for i in range(0, n, 2): if i < n - 2: rod_len = abs(nails[i][0] - nails[i + 2][0]) + abs(nails[i][1] - nails[i + 2][1]) else: rod_len = abs(nails[i][0] - nails[0][0]) + abs(nails[i][1] - nails[0][1]) if rod_len not in rods_count1: break else: ans.append(-1) ans.append(rods_count1[rod_len].pop() + 1) if not rods_count1[rod_len]: del rods_count1[rod_len] else: print("YES") print(' '.join(map(str, ans))) return last_nail = nails.pop() nails.insert(0, last_nail) ans = [] for i in range(0, n, 2): if i < n - 2: rod_len = abs(nails[i][0] - nails[i + 2][0]) + abs(nails[i][1] - nails[i + 2][1]) else: rod_len = abs(nails[i][0] - nails[0][0]) + abs(nails[i][1] - nails[0][1]) if rod_len not in rods_count2: break else: ans.append(rods_count2[rod_len].pop() + 1) ans.append(-1) if not rods_count2[rod_len]: del rods_count2[rod_len] else: print("YES") print(' '.join(map(str, ans))) return print("NO") if __name__ == "__main__": main()
Codeforces Beta Round 28 (Codeforces format)
CF
2,010
2
256
Bender Problem
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods. Help Bender to solve this difficult task.
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod. If there are multiple solutions, print any of them.
null
null
[{"input": "4 2\n0 0\n0 2\n2 2\n2 0\n4 4", "output": "YES\n1 -1 2 -1"}, {"input": "6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n3 2 3", "output": "YES\n1 -1 2 -1 3 -1"}, {"input": "6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n2 2 3", "output": "NO"}]
1,600
["implementation"]
51
[{"input": "4 2\r\n0 0\r\n0 2\r\n2 2\r\n2 0\r\n4 4\r\n", "output": "YES\r\n1 -1 2 -1 \r\n"}, {"input": "6 3\r\n0 0\r\n1 0\r\n1 1\r\n2 1\r\n2 2\r\n0 2\r\n3 2 3\r\n", "output": "YES\r\n1 -1 2 -1 3 -1 \r\n"}, {"input": "6 3\r\n0 0\r\n1 0\r\n1 1\r\n2 1\r\n2 2\r\n0 2\r\n2 2 3\r\n", "output": "NO\r\n"}, {"input": "4 4\r\n0 0\r\n0 1\r\n1 1\r\n1 0\r\n1 1 1 1\r\n", "output": "NO\r\n"}, {"input": "6 2\r\n0 0\r\n1 0\r\n1 1\r\n2 1\r\n2 2\r\n0 2\r\n2 2\r\n", "output": "NO\r\n"}, {"input": "6 3\r\n0 0\r\n2 0\r\n2 2\r\n1 2\r\n1 1\r\n0 1\r\n4 2 2\r\n", "output": "YES\r\n-1 1 -1 2 -1 3 "}, {"input": "4 4\r\n-8423 7689\r\n6902 7689\r\n6902 2402\r\n-8423 2402\r\n20612 20612 91529 35617\r\n", "output": "YES\r\n1 -1 2 -1 \r\n"}, {"input": "4 4\r\n1679 -198\r\n9204 -198\r\n9204 -5824\r\n1679 -5824\r\n18297 92466 187436 175992\r\n", "output": "NO\r\n"}, {"input": "4 2\r\n0 0\r\n0 2\r\n2 2\r\n2 0\r\n200000 200000\r\n", "output": "NO\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, m = map(int, f.readline().split()) nails = [tuple(map(int, f.readline().split())) for _ in range(n)] rods = list(map(int, f.readline().split())) with open(output_path) as f: ref_lines = f.read().splitlines() with open(submission_path) as f: sub_lines = f.read().splitlines() if not sub_lines: print(0) return ref_is_no = ref_lines and ref_lines[0].strip() == "NO" sub_is_no = sub_lines and sub_lines[0].strip() == "NO" if ref_is_no: print(1 if sub_is_no else 0) return else: if not sub_lines or sub_lines[0].strip() != "YES": print(0) return if len(sub_lines) < 2: print(0) return try: sub_list = list(map(int, sub_lines[1].split())) except: print(0) return if len(sub_list) != n: print(0) return used_rods = set() for num in sub_list: if num == -1: continue if num < 1 or num > m or num in used_rods: print(0) return used_rods.add(num) for i in range(n): rod_num = sub_list[i] if rod_num == -1: continue prev_i = i - 1 if i > 0 else n - 1 next_i = (i + 1) % n x_prev, y_prev = nails[prev_i] x_curr, y_curr = nails[i] x_next, y_next = nails[next_i] len_prev = abs(y_curr - y_prev) if x_prev == x_curr else abs(x_curr - x_prev) len_next = abs(y_next - y_curr) if x_curr == x_next else abs(x_next - x_curr) sum_len = len_prev + len_next rod_index = rod_num - 1 if rod_index >= len(rods) or rods[rod_index] != sum_len: print(0) return print(1) if __name__ == "__main__": main()
true