hdu4632Palindrome subsequence (求回文数，区间DP)

4
a
aaaaa
goodafternooneveryone
welcometoooxxourproblems

Case 1: 1
Case 2: 31
Case 3: 421
Case 4: 960
#include<stdio.h>
#include<string.h>
int t,dp[1005][1005],c=0;
int main()
{
    char str[1005];
    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 r=1;r<len;r++)
        for(int i=0;i<len-r;i++)
        {
            int j=i+r;
            dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007;
            if(str[i]==str[j])
            dp[i][j]=(dp[i][j]+dp[i+1][j-1]+1)%10007;
        }
        printf("Case %d: %d\n",++c,dp[0][len-1]);
    }
}