C. Vasya and String 题目连接: http://www.codeforces.com/contest/676/problem/C Description High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as th…

C. Vasya and String time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' onl…

C. Vasya and String time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output ->  Link  <- 题目就不贴出来了,链接在上面: 题意:给定n,k,然后是一个长度为n的只含字母a.b的字符串,问至多改变k次使得一个子串字母都相同且长度最大,如所给样例: input 10 1 bbabbabbba output…

题目链接: http://codeforces.com/contest/676/problem/C 题解: 把连续的一段压缩成一个数,对新的数组求前缀和,用两个指针从左到右线性扫一遍. 一段值改变一部分的情况考虑的不够周到,wa了两次. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vect…

C. Vasya and Basketball time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each succe…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

题目传送门 /* 构造水题:对于0的多个位数的NO,对于位数太大的在后面补0,在9×k的范围内的平均的原则 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN]; int main(void) //Codeforces Round #…

题目大意 两个人轮流在一个字符串上删掉一个字符,没有字符可删的人输掉游戏 删字符的规则如下: 1. 每次从一个字符串中选取一个字符,它是一个长度至少为 3 的奇回文串的中心 2. 删掉该字符,同时,他选择的那个字符串分成了两个独立的字符串 现在问,先手是否必胜,如果先手必胜,输出第一步应该删掉第几个字符,有多解的话,输出序号最小的那个 字符串的长度不超过5000,只包含小写英文字母 做法分析 可以这样考虑:将所有的长度大于等于 3(其实只需要找长度为 3 的就行)的奇回文串的中心标记出来 我们将…

// Codeforces Round #365 (Div. 2) // C - Chris and Road 二分找切点 // 题意:给你一个凸边行,凸边行有个初始的速度往左走,人有最大速度,可以停下来,竖直走. // 问走到终点的最短时间 // 思路: // 1.贪心来做 // 2.我觉的二分更直观 // 可以抽象成:一条射线与凸边行相交,判断交点.二分找切点 #include <bits/stdc++.h> using namespace std; #define LL long lon…

Codeforces Round #297 (Div. 2)B. Pasha and String Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/525/problem/B Description Pasha got a very beautiful string s for his birthday, the string consists o…