Kingdom of Black and White Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5583 Description In the Kingdom of Black and White (KBW), there are two kinds of frogs: black frog and white frog. Now N frogs are stand…

Problem Description In the Kingdom of Black and White (KBW), there are two kinds of frogs: black frog and white frog. Now N frogs are standing in a line, some of them are black, the others are white. The total strength of those frogs are calculated b…

Kingdom of Black and White Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 585    Accepted Submission(s): 193 Problem Description In the Kingdom of Black and White (KBW), there are two kinds of…

http://acm.hdu.edu.cn/showproblem.php?pid=5583 题意: 给出一个01串,现在对这串进行分组,连续相同的就分为一组,如果该组内有x个数,那么就对答案贡献x*x,现在最多可以修改原串中的一个字符,问答案最大可以为多少. 思路:暴力求解. 一开始只需要预处理分块,计算出每一分块的个数,然后暴力处理一下即可,需要注意分块数为1的情况,此时左右两边都要合并起来. #include<iostream> #include<cstdio> #inclu…

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Higher Math Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2219    Accepted Submission(s): 1219 Problem Description You are building a house. You’d prefer if all the walls have a precise right…

这里是杭电hdu上的链接:http://acm.hdu.edu.cn/showproblem.php?pid=3999  Problem Description: As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely: 1.  insert a key k to a empty tree, then the tree becom…

Problem Description 谁画8画的好,画的快,今后就发的快,学业发达,事业发达,祝大家发,发,发. Input 输入的第一行为一个整数N,表示后面有N组数据. 每组数据中有一个字符和一个整数,字符表示画笔,整数(>=5)表示高度. Output 画横线总是一个字符粗,竖线随着总高度每增长6而增加1个字符宽.当总高度从5增加到6时,其竖线宽度从1增长到2.下圈高度不小于上圈高度,但应尽量接近上圈高度,且下圈的内径呈正方形. 每画一个"8"应空一行,但最前和最后都无空…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5793    Accepted Submission(s): 3459 Problem Description Eddy's interest is very extensive, recently he is interested in prime n…

分析:水题. #include<iostream> using namespace std; #define N 5050 char a[N],b[N],tmp[N]; void Read(char p[]) { getchar(); gets(tmp); while(gets(tmp)) { if(strcmp(tmp,"END")==0) break; if(strlen(tmp)!=0) strcat(p,tmp); strcat(p,"\n");…