AtCoder abc148

C题
求GCD

D题
顺序遍历

E题
trailing zero只与5的个数有关，因此算一下5/25/125…的倍数

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100050)def main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn = int(fp.readline())if n & 1:print(0)else:t = 5 * 2ans = 0while t <= n:ans += n // tt *= 5print(ans)if __name__ == "__main__":main()

F题
首先对每个点分别求到A和到T的最短路径。记为da与dt数组，接下来稍微要动一下脑子。
需要找到T能“躲藏”的最长路径，也就是da[i] > dt[i]的i。A只需要等待在前一个格子让T遇上A即可。

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100050)def bfs(g, u):n = len(g)dist = [-1] * nqu = deque()dist[u] = 0qu.append(u)while qu:cur = qu.popleft()for v in g[cur]:if dist[v] == -1:dist[v] = dist[cur] + 1qu.append(v)return distdef main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn, u, v = map(int, fp.readline().split())u, v = u - 1, v - 1g = [[] for _ in range(n)]for i in range(n - 1):x, y = map(int, fp.readline().split())x, y = x - 1, y - 1g[x].append(y)g[y].append(x)du, dv = bfs(g, u), bfs(g, v)ans = 0for i in range(n):if du[i] < dv[i]:ans = max(ans, dv[i] - 1)print(ans)if __name__ == "__main__":main()