思路: 尺取,写挫了,debug了半天. #include <bits/stdc++.h> using namespace std; typedef long long LL; const int N=1e5+10; char ch[N]; int vis[1001]; int main() { int s,t; scanf("%s",ch); memset(vis,0,sizeof(vis)); int len=strlen(ch); int num=0; int ans…

题目链接 http://codeforces.com/problemset/gymProblem/100703/I Description standard input/outputStatements As a matter of fact, Dragon knows what Prince is interested in now. Prince uses to spend his rare off days learning different courses and trainings.…

Problem 1072: The longest same color grid Time Limits:  1000 MS   Memory Limits:  65536 KB 64-bit interger IO format:  %lld   Java class name:  Main Description There are n grid, m kind of color. Grid number 1 to N, color number 1 to M.  The color of…

Bob’s Race Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads…

题目链接:http://codeforces.com/problemset/problem/180/E 给你n个数,每个数代表一种颜色,给你1到m的m种颜色.最多可以删k个数,问你最长连续相同颜色的序列的长度是多少. 将相同颜色的下标存到对应颜色的容器中,比如ans[a[i]].push_back(i)就是将下标为i颜色为a[i]的数存到ans[a[i]]容器中. 对于每种颜色序列,尺取一下 在差距小于k的情况下取能取到的最大长度. //#pragma comment(linker, "/STA…

Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to come in two…

King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves. After a con…

题目链接 Problem Description Alice are given an array A[1..N] with N numbers. Now Alice want to build an array B by a parameter K as following rules: Initially, the array B is empty. Consider each interval in array A. If the length of this interval is le…

<题目链接> 题目大意:给定一段序列,每次进行两次操作,输入1 x代表插入x元素(x元素一定大于等于之前的所有元素),或者输入2,表示输出这个序列的任意子集$s$,使得$max(s)-mean(s)$表示这个集合的最大值与平均值的最大差值. 解题分析:首先,因为输入的$x$是非递减的,所以要使$max(s)-mean(s)$最大,肯定$max(s)$就是最后一个输入元素的大小.$x$已经确定了,现在就是尽可能的使$mean(s)$尽可能的小.如何使得平均值最小呢?肯定是从最前面的最小的元素开始…

尺取,写起来有点麻烦 枚举左端点,然后找到右端点,,使得区间[l,r]里各种颜色花朵的数量满足b数组中各种花朵的数量,然后再judge区间[l,r]截取出后能否可以供剩下的n-1个人做花环 /* 给定序列A,分成n段,每段k个, 然后删掉A中的一些,但任要能组成n段,使得A中的一段包含序列B中的所有元素 如果可以输出删掉的元素,否则输出-1 */ #include<bits/stdc++.h> using namespace std; #define maxn 500005 int m,k,n…