http://codeforces.com/problemset/problem/182/D 题意:如果把字符串a重复m次可以得到字符串b,那么我们称字符串a为字符串b的一个因子,现在给定两个字符串S1和S2,求它们的公共因子个数. 思路: 先求最小循环节,如果最小循环节不同,那么肯定是没有公共因子的.如果相同的话,那就看循环节长度为1,2,3...是否可行. #include<iostream> #include<cstdio> #include<cstring> u…

Codeforces Round #117 (Div. 2) 代码 Codeforces Round #117 (Div. 2) A. Battlefield any trench in meters numerically does not exceed b. 这个条件意味着每次都是在蓄能开始时走向下一条线段,也就是说每条线段相当于花费了\(a+b\)的时间. bfs,用\(d_i\)表示到达线段i需要经过最少的线段条数,到达\(B\)的时候直接计算欧几里得距离. B. Vasya's Cal…

链接: https://codeforces.com/contest/1271/problem/E 题意: At first, let's define function f(x) as follows: f(x)={x2x−1if x is evenotherwise We can see that if we choose some value v and will apply function f to it, then apply f to f(v), and so on, we'll…

题目链接:https://codeforces.com/contest/1384/problem/A 题意 构造 $n+1$ 个字符串,使得 $n$ 对相邻字符串的相同前缀长度对应于数组 $a$ . 题解 构造一个足够长的字符串,每次反转前缀不同处的字符即可. 代码 #include <bits/stdc++.h> using namespace std; void solve() { int n; cin >> n; string s(200, 'a'); cout <&l…

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题意:对于一个数\(x\),有函数\(f(x)\),如果它是偶数,则\(x/=2\),否则\(x-=1\),不断重复这个过程,直到\(x-1\),我们记\(x\)到\(1\)的这个过程为\(path(x)\),它表示这个过程中所有\(x\)的值,现在给你一个数\(n\)和\(k\),要你找出最大的数\(x\),并且\(x\)在\(path[1,n]\)中出现的次数不小于\(k\). 题解:我们对于\(x\)可以选择分奇偶来讨论. 1.假如\(x\)为奇数,那么根据题意,我们知道\(x,2x,2x…

题目链接: http://codeforces.com/contest/731/problem/A A. Night at the Museum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Grigoriy, like the hero of one famous comedy film, found a job as a…

B. Obsessive String   Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following proble…

D. Messenger time limit per test2 seconds memory limit per test512 megabytes inputstandard input outputstandard output Each employee of the "Blake Techologies" company uses a special messaging app "Blake Messenger". All the stuff likes…

Educational Codeforces Round 117 (Rated for Div. 2) A. Distance https://codeforces.com/contest/1612/problem/A 题目给出的条件是 距离为曼哈顿距离,而曼哈顿距离等价于步长. 由题目的一半条件,可以得到步长和为AB步长,各自步长为AB步长的一半. 所以显然可以推出: \[1.如果和为奇数则,不存在\\ 2.如果都为偶数,显然只需要取步长一半即可\\ \] //原始代码 #include<bi…