HDU 4632 Palindrome subsequence(区间DP)

Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others)
Total Submission(s): 2830 Accepted Submission(s): 1156

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

//
//  main.cpp
//  区间dp 1001
//
//  Created by 陈永康 on 16/2/28.
//  Copyright © 2016年 陈永康. All rights reserved.
//
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;
char a[1005];
int dp[1005][1005];
int main()
{
    int t;
    scanf("%d",&t);
    int cas=0;
    while(t--)
    {
        scanf("%s",a+1);
        int len=strlen(a+1);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=len;i++)
            dp[i][i]=1;
        for(int l=1;l<len;l++)
        {
            for(int i=1;i+l<=len;i++)
            {
                int j=i+l;
//                if(a[i]==a[j])
//                {
//                    if(i==j-1)
//                        dp[i][j]+=3;
//                    else
//                        dp[i][j]=dp[i+1][j-1]*2+3;
//
//                }
//                else
//                {
//                    if(i==j-1)
//                        dp[i][j]+=2;
//                    else
//                        dp[i][j]=dp[i+1][j-1]+2;
//                }
//                for(int k=i+1;k<j;k++)
//                {
//                    if(a[i]==a[k])
//                    {
//                        if(i==k-1)
//                            dp[i][j]++;
//                        else
//                            dp[i][j]+=(dp[i+1][k-1])+1;
//                    }
//                    if(a[k]==a[j])
//                    {
//                        if(k==j-1)
//                            dp[i][j]++;
//                        else
//                            dp[i][j]+=(dp[k+1][j-1])+1;
//                    }
//                }
                dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007;
                if(a[i]==a[j])
                    dp[i][j]=(dp[i][j]+dp[i+1][j-1]+1+10007)%10007;
                //dp[i][j]%=10007;
            }
        }
        printf("Case %d: %dn",++cas,dp[1][len]);
    }
    return 0;
}