hdu4632Palindrome subsequence (求迴文數，區間DP)

標籤：dp

Problem DescriptionIn 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 sequenceis a subsequence of . (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 =and Y = , 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[sub]xi[/sub] = S[sub]yi[/sub]. Also two subsequences with different length should be considered different.
InputThe 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.
OutputFor 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

4aaaaaagoodafternooneveryonewelcometoooxxourproblems

Sample Output

Case 1: 1Case 2: 31Case 3: 421Case 4: 960
#include<stdio.h>#include<string.h>int t,dp[1005][1005],c=0;int main(){    char str[1005];    scanf("%d",&t);    while(t--)    {        scanf("%s",str);        int len=strlen(str);        memset(dp,0,sizeof(dp));        for(int i=0;i<len;i++)        dp[i][i]=1;        for(int r=1;r<len;r++)        for(int i=0;i<len-r;i++)        {            int j=i+r;            dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+10007)%10007;            if(str[i]==str[j])            dp[i][j]=(dp[i][j]+dp[i+1][j-1]+1)%10007;        }        printf("Case %d: %d\n",++c,dp[0][len-1]);    }}

hdu4632Palindrome subsequence (求迴文數，區間DP)