Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2280    Accepted Submission(s): 913

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

Source

2013 Multi-University Training Contest 4

dp[i][j] 表示 包含 a[i]的，到 j 的回文个数

转移：

dp[i][j] = 1 ;

dp[i][j] = dp[i][j-1] ; i < j

if(a[i]==a[j]) dp[i][j] += dp[k][j-1]  ； i < k < j

这样是n^3的，会T

所以用 sum[i][j] 表示到回文个数，优化转移

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define maxn 100010
#define LL long long
#define mod 10007
using namespace std;

int dp[1010][1010],sum[1010][1010];
char a[1010] ;
int main()
{
    int n,i,j,m;
    int T,case1=0;
    cin >> T ;
    while( T-- )
    {
        scanf("%s",a+1) ;
        n = strlen(a+1) ;
        memset(dp,0,sizeof(dp)) ;
        memset(sum,0,sizeof(sum)) ;
        for( i = n ; i >= 1 ;i--)
            for( j = i ; j <= n ;j++)
        {
            if(i==j)
            {
                dp[i][j]=1;
                sum[i][j]=1;
                continue;
            }
            dp[i][j] = dp[i][j-1] ;
            if(a[i]==a[j])
            {
                dp[i][j]++;
               // for(int k = i+1 ; k < j ;k++)
                   // dp[i][j] += dp[k][j-1];
                   dp[i][j] += sum[i+1][j-1] ;
            }
            dp[i][j] %= mod;
            sum[i][j] = (sum[i+1][j]+dp[i][j])%mod;
        }
        int ans=0;
        for(i = 1 ; i <= n ;i++)
            ans = (ans+dp[i][n])%mod ;
        printf("Case %d: %d\n",++case1,ans);
    }
    return 0;
}

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