HDOJ 4632

 思路:设dp[i][j]表示区间[i,j]的回文串的个数,那么有dp[i][j] = dp[j+1][i] + dp[j][i-1] - dp[j+1][i-1],如果str[i] == str[j],那么dp[i][j]还要加上dp[j+1][i-1] + 1;


#include<iostream>
#include<cstdio>
#include<cstring>
#define MAX 1010
#define MOD 10007
using namespace std;
int dp[MAX][MAX];
char str[MAX];
int main(){
    int T, cnt = 0;
    //freopen("in.c", "r", stdin);
    scanf("%d", &T);
    while(T--){
        memset(str, 0, sizeof(str));
        memset(dp, 0, sizeof(dp));
        scanf("%s", str);
        int len = strlen(str);
        for(int i = 0; i< len;i ++) dp[i][i] = 1;
        for(int i = 0;i < len;i ++){
            for(int j = i-1;j >= 0;j --){
                dp[j][i] = (dp[j][i-1] + dp[j+1][i] - dp[j+1][i-1] + MOD)%MOD;
                if(str[i] == str[j]) dp[j][i] = (dp[j][i] + dp[j+1][i-1] + 1 + MOD)%MOD;
            }
        }
        printf("Case %d: %d\n", ++cnt, dp[0][len-1]);
    }
    return 0;
}


posted on 2014-04-30 18:38  wangzhili  阅读(117)  评论(0编辑  收藏  举报