Parlindrome subsequence_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】给出一个字符串，求其中含有多少个回文串。

【思路】dp[i][j]表示i到j区间内的回文串数。

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int mod=10007;
const int N=1010;
char str[N];
int dp[N][N];

int main()
{
    int t,cas=1;
    scanf("%d",&t);
    
    while(t--)
    {
        scanf("%s",str);
        int len=strlen(str);
        
        memset(dp,0,sizeof(dp));
        
        for(int i=0;i<len;i++)
        {
            dp[i][i]=1;
        }
        for(int i=0;i<len;i++)
            for(int j=i-1;j>=0;j--)
            {
                dp[j][i]=(dp[j+1][i]+dp[j][i-1]-dp[j+1][i-1]+mod)%mod;//加上mod是以免被取模的数为负
//中间j+1到i-1的部分重复了
                if(str[i]==str[j])
                    dp[j][i]=(dp[j][i]+dp[j+1][i-1]+1+mod)%mod;//两端相等的话，要加上dp[j+1][i-1]+1;
//两端可以组成一个回文串，两端可以与其间任意一个回文串组成新的回文串，数量将增加dp[j+1][i-1]+1;
            }
        printf("Case %d: %d
",cas++,dp[0][len-1]);
    }
    return 0;
}