#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <string> #include <map> #include <cmath> #include <vector> #define Faster ios::sync_with_stdio(false),cin.tie(0) #define…

String painter Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2520    Accepted Submission(s): 1134 Problem Description There are two strings A and B with equal length. Both strings are made up…

String painter Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2792 Accepted Submission(s): 1272 Problem Description There are two strings A and B with equal length. Both strings are made up of lo…

题目 String painter 给出两个字符串s1,s2.对于每次操作可以将 s1 串中的任意一个子段变成另一个字符.问最少需要多少步操作能将s1串变为s2串. 解析 太妙了这个题,mark一下. 这个题先考虑怎么由空串转化s2, \(f[i][j]\)表示从空串到s2最少的次数, 则有\(f[i][j]=s[i+1][j]+1\), 若\([i+1,j]\)存在一个\(k\),使\(s2[i]==s2[k]\),则\(f[i][j]=min\{f[i+1][k]+f[k+1][j]\}\)…

http://acm.hdu.edu.cn/showproblem.php?pid=2476 String painter Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6863    Accepted Submission(s): 3330 Problem Description There are two strings A and…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=2476 Problem Description There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you c…

// uva live 4394 String painter // // 问题是,在培训指导dp运动主题,乍一看,我以为只是一点点复杂 // A A磕磕磕,两个半小时后,.发现超过例子.然而,鉴于他们跪在数据 // 还要wa.才发现,自己的方法是有问题的,假设是将两个串同一时候考虑 // 的话.比方: dp[i][j] 表示从i到j,s串刷成目标b串所须要的最小的花费 // 然后依据区间的端点的字符特点,进行状态转移.然而可能是我太搓了. // 发现这种状态转移是不正确的,比方edc和cde.…

题意 String painter Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6669    Accepted Submission(s): 3230 Problem Description There are two strings A and B with equal length. Both strings are made…

pack("H*", $string) 转化成这样: //16进制字符串转[]byte func HexToByte(hex string) []byte { length := len(hex) / 2 slice := make([]byte, length) rs := []rune(hex) for i := 0; i < length; i++ { s := string(rs[i*2 : i*2+2]) value, _ := strconv.ParseInt(s,…

题目: Problem Description There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any o…