Palindrome subsequence_区间DP

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65535K (Java/Other)

Total Submission(s) : 53   Accepted Submission(s) : 22

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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

用dp[i][j]表示i到j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]。（画图可知）

如果str[i]==str[j]的话，还要加上dp[i+1][j-1]+1。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=1005;
const int mod=10007;
char str[N];
int dp[N][N];
int main()
{
    int t,cas=0;
    cin>>t;
    while(t--)
    {
       scanf("%s",str);
        memset(dp,0,sizeof(dp));
        int len=strlen(str);
        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][i-1]+dp[j+1][i]-dp[j+1][i-1]+mod)%mod;
            if(str[i]==str[j]) dp[j][i]=(dp[j][i]+dp[j+1][i-1]+1+mod)%mod;
        }
        
        printf("Case %d: %d
",++cas,dp[0][len-1]);
    }
    return 0;

}