hdu 4632 Palindrome subsequence （区间DP+容斥）

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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 = < Sx1, Sx2, …, Sxk> and Y = < Sy1, Sy2, …, Syk> , if there exist an integer i (1<=i<=k) such that xi != yi, the subsequence X and Y should be consider different even if Sxi = Syi. 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

题意：找到一个字符串里又多少个回文串子序列；
思路：区间DP，dp[i][j]表示i到j的回文串数，依次从小到大的区间开始，我们需要判断每个区间两个边缘字符是否相等+容斥原理便可以推出结果；
转移状态
**if(c[i]==c[j])
dp[i][j]=(dp[i+1][j]+dp[i][j-1]+1)%10007;
else
dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1])%10007;**
代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[1005][1005];int main(){    int T;    scanf("%d",&T);    char c[1005];    for(int t=1;t<=T;t++)    {        scanf("%s",c);        memset(dp,0,sizeof(dp));        int n=strlen(c);        for(int i=0;i<n;i++)            dp[i][i]=1;        for(int k=1;k<n;k++)            for(int i=0,j=i+k;j<n;i++,j++)            {                if(c[i]==c[j])                    dp[i][j]=(dp[i+1][j]+dp[i][j-1]+1+10007)%10007;                else                    dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007;            }        printf("Case %d: %d\n",t,dp[0][n-1]);    }}

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