Hdu4632 Palindrome subsequence 2017-01-16 11:14 51人阅读 评论(0) 收藏

Palindrome subsequence

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

 

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题目的意思是统计出一个字符串里有多少个不同的回文串的子串，不同指的是位置不同内容可以一样

区间dp，dp[i][j]表示区间[i,j]有多少个不同的回文子串。

推出状态转移方程为：

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

特别的如果s[i]==s[j]那么dp[i][j]+=dp[i+1][j-1]+1;

答案要mod10007 要注意的是 由于有减 mod是要加10007修正

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <string>
using namespace std;

int dp[1005][1005];
char s[1005];
int main()
{
int T;
while(~scanf("%d",&T))
{
    int q=1;
    while(T--)
    {
        scanf("%s",s);
        int n=strlen(s);
        memset(dp,0,sizeof dp);
        for(int j=0;j<n;j++)
        {
            for(int i=j;i>=0;i--)
            {
                dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007;
                if(s[i]==s[j])
                    dp[i][j]+=dp[i+1][j-1]+1;
                    dp[i][j]%=10007;
            }
        }
        printf("Case %d: %d\n",q++,dp[0][n-1]);
    }
}
    return 0;
}