hdu4632

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 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

一个DP，比赛时DP的状态搞错了.....

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <vector>
#include <stack>
using namespace std;
#define ll long long int
char a[1005];
ll b[1005][1005];
int main()
{
    int n,i,m,j,t;
    cin>>n;
    for(i=0;i<n;i++)
    {
        scanf("%s",a);
        memset(b,0,sizeof(b));
        m=strlen(a);
        for(j=0;j<m;j++)
         b[j][j]=1;
         for(j=1;j<m;j++)
         for(t=j-1;t>=0;t--)
         {
             if(a[t]!=a[j])
              b[t][j]=(b[t+1][j]+b[t][j-1]-b[t+1][j-1]+10007)%10007;
              else 
              b[t][j]=(b[t+1][j]+b[t][j-1]+1)%10007;
         }
         printf("Case %d: %d
",i+1,b[0][m-1]);
    }
}