区间dp

http://acm.hdu.edu.cn/showproblem.php?pid=4632

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
   
   
   
   
   
   
   
   


   
   
   
   

令dp[i][j] 表示[i,j]区间内含有的回文子序列，dp[i][j]=d[i+1][j]+dp[i][j-1]-dp[i+1][j-1];如果dp[i][j]在加上dp[i+1][j]+dp[i][j-1]+1;

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
const int MOD=10007;
char a[1005];
int dp[1005][1005];
int main()
{
    int n,T;
    int tt=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",a+1);
        n=strlen(a+1);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            dp[i][i]=1;
        for(int i=n-1;i>0;i--)
        {
            for(int j=i+1;j<=n;j++)
            {
                dp[i][j]=max(dp[i][j],dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+MOD)%MOD;
                if(a[i]==a[j])
                    dp[i][j]=(dp[i][j]+1+dp[i+1][j-1])%MOD;
            }
        }
        printf("Case %d: %d\n",tt++,dp[1][n]);
    }
    return 0;
}