uva 12034 - Race(dp计数)

题目链接：uva 12034 - Race

题目大意：有n匹马比赛，问说有多少种排名情况，可以并列。

解题思路：dp[i][j]表示i匹马，最后一名为j的情况，转移方程dp[i][j]=(dp[i−1][j]+dp[i−1][j−1])∗j

#include <cstdio>
#include <cstring>

typedef long long ll;
const int N = 1005;
const ll MOD = 10056;
ll dp[N][N], f[N];

void init () {
    memset(f, 0, sizeof(f));
    memset(dp, 0, sizeof(dp));
    dp[1][1] = 1;

    for (ll i = 1; i <= 1000; i++) {

        for (ll j = 1; j <= i; j++) {
            dp[i+1][j] = (dp[i+1][j] + dp[i][j] * j) % MOD;
            dp[i+1][j+1] = (dp[i+1][j+1] + dp[i][j] * (j + 1)) % MOD;
            f[i] = (f[i] + dp[i][j]) % MOD;
        }
    }
}

int main () {
    init();
    int cas, n;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        scanf("%d", &n);
        printf("Case %d: %lld\n", i, f[n]);
    }
    return 0;
}