HDU4632：Palindrome subsequence(区间DP)

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, B, C, D, E, F>.
(http://en.wikipedia.org/wiki/Subsequence)

Given a string S, your task is to find out how many different subsequence of S is palindrome. Note that for any two subsequence X = <S _x1, S _x2, ..., S _xk> and Y = <S _y1, S _y2, ..., S _yk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if S _xi  = S _yi. Also two subsequences with different length should be considered different.

 

 

Input

The first line contains only one integer T (T<=50), which is the number of test cases. Each test case contains a string S, the length of S is not greater than 1000 and only contains lowercase letters.

 

 

Output

For each test case, output the case number first, then output the number of different subsequence of the given string, the answer should be module 10007.

 

 

Sample Input

4 a aaaaa goodafternooneveryone welcometoooxxourproblems

 

 

Sample Output

Case 1: 1 Case 2: 31 Case 3: 421 Case 4: 960

 

 

题意：找到一个字符串里又多少个回文串子序列，要注意子序列的定义哦！

思路：这几天做了几道区间DP,这算是比较简单的一道区间DP了，dp数组仍然是储存区间内的回文串数，然后推导出状态即可

 

 

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;



const int mod = 10007;

char str[1005];

int dp[1005][1005];



int main()

{

    int t,i,j,k,len,cas = 1;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%s",str);

        len = strlen(str);

        for(i = 0; i<len; i++)

            dp[i][i] = 1;//单个字符肯定是一个回文子串

        for(i = 1; i<len; i++)

        {

            for(j = i-1; j>=0; j--)

            {

                dp[j][i] = (dp[j+1][i]+dp[j][i-1]-dp[j+1][i-1]+mod)%mod;//之前没有加mod，wa了，j~i的区间的回文数是j+1~i与j~i-1区间回文数的和，但是要注意这里会有重复的

                if(str[i] == str[j])

                    dp[j][i] = (dp[j][i]+dp[j+1][i-1]+1+mod)%mod;//如果区间两头是相等的，则要加上dp[j+1][i-1]+1，因为首尾是可以组成一个回文子串的，而且首尾可以与中间任何一个回文子串组成新的回文子串

            }

        }

        printf("Case %d: %d\n",cas++,dp[0][len-1]);

    }



    return 0;

}