hdu 4632 Palindrome subsequence（dp）

题目连接：hdu 4632 Palindrome subsequence

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1005;const int mod = 10007;char str[maxn];int N, dp[maxn][maxn];int solve () {    N = strlen(str + 1);    for (int i = 1; i <= N; i++) dp[i][i] = 1;    for (int i = 1; i < N; i++)        if (str[i] == str[i+1]) dp[i][i+1] = 3;        else dp[i][i+1] = 2;    for (int i = N; i; i--) {        for (int j = i + 2; j <= N; j++) {            dp[i][j] = ((dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]) % mod + mod) % mod;            if (str[i] == str[j])                dp[i][j] = (dp[i][j] + dp[i+1][j-1] + 1) % mod;        }    }    return dp[1][N];}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        scanf("%s", str + 1);        printf("Case %d: %d\n", kcas, solve());    }    return 0;}