Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 2232    Accepted Submission(s): 889 Problem Description In mathematics, a subsequence is a sequence that can be derived fr…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 558    Accepted Submission(s): 203 Problem Description In mathematics, a subsequence is a sequence that can be derived fro…

Palindrome subsequence Problem Description In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others) Total Submission(s): 2836 Accepted Submission(s): 1160 Problem Description In mathematics, a subsequence is a sequence that can be derived from a…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 88    Accepted Submission(s): 26 Problem Description In mathematics, a subsequence is a sequence that can be derived from…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others) Total Submission(s): 2595    Accepted Submission(s): 1039 Problem Description In mathematics, a subsequence is a sequence that can be derived…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 4513    Accepted Submission(s): 1935 Problem Description In mathematics, a subsequence is a sequence that can be derived f…

标签(空格分隔): 区间qp Palindrome subsequence \[求一个string的 回文子序列 的个数 \] 少废话,上代码. #include<bits/stdc++.h> //头文件没有什么问题 using namespace std; //名字空间没有什么问题 const int maxn = 1000 +5; // 字符串的最大长度 const int mod = 10007;//题目规定的模数 int n,cnt;//n:有多少个字符串,cnt:记录当前有多少个ca…

Problem Description In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A,…

A palindrome is a nonempty string over some alphabet that reads the same forwardand backward. Examples of palindromes are all strings of length 1, civic,racecar, and aibohphobia (fear of palindromes).Give an efficient algorithm to find the longest pa…

In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A, B, C, D, E, F>. (…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4632 题意:求回文串子串的的个数. 思路:看转移方程就能理解了. dp[i][j] 表示区间i j 之间的回文子串的个数. 状态转移方程:dp[i][j]=dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1];         if(str[i]==str[j]) dp[i][j]=dp[i][j]+dp[i+1][j-1]+1; 代码: #include<iostrea…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 题意:给你一个序列,问你该序列中有多少个回文串子序列,可以不连续. 思路:dp[i][j]表示序列i到j中具有的回文串序列数. 当s[i]==s[j], dp[i][j]=dp[i+1][j]+dp[i][j-1]+1 否则 dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]: #include <iostream> #include <cstdio…

题目链接 做的我很无奈,当时思路很乱,慌乱之中,起了一个想法,可以做,但是需要优化.尼玛,思路跑偏了,自己挖个坑,封榜之后,才从坑里出来,过的队那么多,开始的时候过的那么快,应该就不是用这种扯淡方法做的. 表示很无奈,没有想到简单的递推式,搞了一个MLE+TLE的方法. 最初版本,多了一个for的复杂度,只要标记一下就好,可是在递归了不好处理,让我折腾了老一会,才弄好. 复制代码 int dfs(int l,int r) { ; if(l == r) ; else if(l > r) ; if(…

http://acm.hdu.edu.cn/showproblem.php?pid=4632 简单DP 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<string> #include<cstring> #include<cmath> #include<set> #include<vector> #include<l…

link:http://acm.hdu.edu.cn/showproblem.php?pid=4632 refer to: o(╯□╰)o……明明百度找的题解,然后后来就找不到我看的那份了,这位哥们对不住了…… #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <cctype> #i…

题目 参考自博客:http://blog.csdn.net/u011498819/article/details/38356675 题意:查找这样的子回文字符串(未必连续,但是有从左向右的顺序)个数. 简单的区间dp,哎,以为很神奇的东西,其实也是dp,只是参数改为区间,没做过此类型的题,想不到用dp,以后就 知道了,若已经知道[0,i],推[0,i+1], 显然还要从i+1 处往回找,dp方程也简单: dp[j][i]=(dp[j+1][i]+dp[j][i-1]+10007-dp[j+1][…

题意 给定一个字符串,问有多少个回文子串(两个子串可以一样). 思路 注意到任意一个回文子序列收尾两个字符一定是相同的,于是可以区间dp,用dp[i][j]表示原字符串中[i,j]位置中出现的回文子序列的个数,有递推关系: dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]  (*) 如果i和j位置出现的字符相同,那么dp[i][j]可以由dp[i+1][j-1]中的子序列加上这两个字符构成回文子序列,也就是 dp[i][j]+=dp[i+1][j-1],注意…

题意1:问你一个串有几个不连续子序列(相同字母不同位置视为两个) 题意2:问你一个串有几种不连续子序列(相同字母不同位置视为一个,空串视为一个子序列) 思路1:由容斥可知当两个边界字母相同时 dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + dp[i + 1][j - 1] + 1;当两个字母不同时 dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1].然后区间DP…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 问题要求回答一串字符串中回文子序列的数量,例如acbca就有 a,c,b,c,a,cc,aa,aca,aca(注意这两个aca的c是不同位置的c,都要累计),aba,cbc,acca,acbca.共13种. 我们如果构造dp[i][j]为区间从i-j的回文子序列个数,当i==j时dp[i][j]=1,当i!=j时,如果字符串i,j位相等,他们便可以从dp[i+1,j-1]转移而来,即dp[i]…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 题目大意:给你若干个字符串,回答每个字符串有多少个回文子序列(可以不连续的子串).解题思路: 设dp[i][j]为[i,j]的回文子序列数,那么得到状态转移方程: dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+MOD)%MOD if(str[i]==str[j]) dp[i][j]+=dp[i-1][j+1]+1 代码: #include<cstdi…

思路:设dp[i][j] 为i到j内回文子串的个数.先枚举所有字符串区间.再依据容斥原理. 那么状态转移方程为   dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1] 如果 a[i] = a[j] , dp[i][j] += (dp[i+1][j-1] + 1). 或者 dp[i][j] =dp[i][j-1] + dp[i+1][j] 如果 a[i] != a[j] , dp[i][j] -=  dp[i+1][j-1]. 需要注意的是 假如区间…

题目链接 题目大意 给你几个字符串 (1<len(s)<1000) ,要你求每个字符串的回文序列个数.对于10008取模. Solution 区间DP. 比较典型的例题. 状态定义: 令 \(f[i][j]\) 表示 \(i\) 到 \(j\) 的回文序列个数,\(s\) 为给出的字符串. 状态转移: \(s[i]\neq s[j]\) 那么此时 \(f[i][j]\) 即为\(f[i][j-1]\),\(f[i+1][j]\)之和. 但由于 \(i+1->j-1\)的我们明显重复统计了…

HDU 4632 Palindrome subsequence dp[x][y]表示区间[x,y]构成回文串的方案数. 若str[x]==str[y],dp[x][y]=dp[x+1][y]+dp[x][y-1]-dp[x+1][y-1]+(dp[x+1][y-1]+1)=dp[x+1][y]+dp[x][y-1]+1. 若str[x]!=str[y],dp[x][y]=dp[x+1][y]+dp[x][y-1]-dp[x+1][y-1]. #include<cstdio> #include&…

HDU-4632 Palindrome subsequence 题意:给定一个字符串,长度最长为1000,问该串有多少个回文子串. 分析:设dp[i][j]表示从 i 到 j 有多少个回文子串,则有动态规划方程: str[i] != str[j]:dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1];str[i]  = str[j]:dp[i][j] = dp[i+1][j] + dp[i][j-1] + 1. #include <cstdlib>…

http://acm.hdu.edu.cn/showproblem.php?pid=4632 题意: 一个字符串,有多少个subsequence是回文串. 别人的题解: 用dp[i][j]表示这一段里有多少个回文串,那首先dp[i][j]=dp[i+1][j]+dp[i][j-1],但是dp[i+1][j]和dp[i][j-1]可能有公共部分,所以要减去dp[i+1][j-1]. 如果str[i]==str[j]的话,还要加上dp[i+1][j-1]+1. 但是自己却是这样想的,把每个区间都要看…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 1977    Accepted Submission(s): 822 Problem Description In mathematics, a subsequence is a sequence that can be derived fr…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others) Total Submission(s): 1538    Accepted Submission(s): 635 Problem Description In mathematics, a subsequence is a sequence that can be derived f…

也写了好几天的区间DP了,这里稍微总结一下(感觉还是不怎么会啊!). 但是多多少少也有了点感悟: 一.在有了一点思路之后,一定要先确定好dp数组的含义,不要模糊不清地就去写状态转移方程. 二.还么想好...想到了再加上去.... 之前也看了别人的总结,也给出了不少区间DP的模板.这里我也贴一下基本的模板: 区间DP模板 ; len < n; len++) { //操作区间的长度 ; i+len <= n; i++) { //始末 int j=i+len; //检查是否匹配(非必须) for (…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)Total Submission(s): 2858    Accepted Submission(s): 1168 Problem Description In mathematics, a subsequence is a sequence that can be derived f…