HDU 4632 Palindrome subsequence

Palindrome subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65535 K (Java/Others)

Total Submission(s): 613    Accepted Submission(s): 232

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

 

 给一个字符串，挑出里面的回文子串（子串中的子符在原串中可以了连续。

DP问题，可以推出如下状态转移方程：

　　若s[i]==s[j]，dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+dp[i+1][j-1]+1

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

　　反之若s[i]!=s[j]，dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1];

　　初始条件为dp[i][i]=1;

　　　　　　　s[i]==s[i+1]时，dp[i][i+1]=3;

　　　　　　　s[i]!=s[i+1]时，dp[i][i+1]=2;

多校第四场的第一题，比赛时不是我写的，据他们说写成递归的DP会超时……题出的没人性……

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int len;
int dp[1050][1050];
char s[2000];


int main()
{
    int t;

    scanf("%d",&t);
    getchar();

    for(int k=1;k<=t;k++)
    {
        gets(s);

        len=strlen(s);

        for(int i=0;i<len;i++)
            dp[i][i]=1;
        for(int i=0;i<len-1;i++)
            if(s[i]==s[i+1])
                dp[i][i+1]=3;
            else
                dp[i][i+1]=2;

        for(int i=0;i<len;i++)
            for(int j=i-2;j>=0;j--)
                if(s[i]==s[j])
                    dp[j][i]=(dp[j+1][i]+dp[j][i-1]+10008)%10007;
                else
                    dp[j][i]=(dp[j+1][i]+dp[j][i-1]-dp[j+1][i-1]+10007)%10007;

        printf("Case %d: %d
",k,dp[0][len-1]);
    }

    return 0;
}