HDU:4632-Palindrome subsequence

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=4632

Palindrome subsequence

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

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

---

1.这个区间dp状态转移方程推了好久，刚开始的时候只想到由dp[i][j-1]得到，但是实际上是可以有dp[i][j-1]和dp[i+1][j]一同得到，想到了之后又忘记减去dp[i+1][j-1],痛苦啊。
2.其实这个状态转移方程又两部分构成，
- dp[i][j] = dp[i][j-1]+dp[i+1][j] - dp[i+1][j-1];(要减去重合的部分，多加了一次）
- 如果新的字母形成了回文串，dp[i][j] += dp[i+1][j-1] + 1;

---

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1010;
const int MOD = 10007;
char s[maxn];
int dp[maxn][maxn];

void get_ans()
{
    memset(dp,0,sizeof(dp));
    int len = strlen(s);
    for(int i=0;i<len;i++)
        dp[i][i] = 1;
    for(int i=0;i<len;i++)
    {
        for(int j=0,k=i;k<len;k++,j++)
        {
            dp[j][k] = (dp[j][k-1] + dp[j+1][k] - dp[j+1][k-1] + MOD) % MOD;
            if(s[j] == s[k])
                dp[j][k] += dp[j+1][k-1] + 1;
            dp[j][k] %= MOD;
        }
    }
    printf("%d
",dp[0][len-1]);
}

int main()
{
    int t;
    cin>>t;
    int cas = 0;
    while(t--)
    {
        scanf("%s",s);
        printf("Case %d: ",++cas);
        get_ans();
    }
    return 0;
}