HDU 4632 Palindrome subsequence（區間dp）

Palindrome subsequence

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

Total Submission(s): 2595    Accepted Submission(s): 1039

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  

/*
題意:問一個字元串的會問序列有多少個
    dp[i][j]=(dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1]);
    if(c[i]==c[j]) 那麼加上中間的dp[i+1][j-1],因為可以和i,j形成新的
    還要加上  1  （i 和  j 形成字元串)


*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define mod 10007
#define INF 0x3f3f3f3f
#define N 1005

int dp[N][N];
int len;
char c[N];

int main()
{
   int i,j,t,ca=0;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%s",c);
       len=strlen(c);
       memset(dp,0,sizeof(dp));
       for(i=0;i<len;i++)
          dp[i][i]=1;

       for(i=len-1;i>=0;i--)
         for(j=i+1;j<len;j++)
         {
            dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+mod)%mod;
            if(c[i]==c[j])
                dp[i][j]=(dp[i][j]+dp[i+1][j-1]+1+mod)%mod;
         }
         printf("Case %d: %d\n",++ca,(dp[0][len-1]+mod)%mod);
   }
   return 0;
}