hdu 4652 Dice（dp）

题目链接：hdu 4652 Dice

代码

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const int maxn = 1e6 + 5;

/* * k = 0, 设dp[i]表示连续出现i的一样的次数期望 * 设xi = dp[i] - dp[i-1] * xi = 1 + (m-1)/m * (dp[i-1]-1+xi) * xi = m + (m-1) * (dp[i-1]-1) * dp[i] = dp[i-1] + m + (m-1) * (dp[i-1] - 1) * dp[i] = dp[i-1] * m + 1; * dp[i] + 1/(m-1) = m * (dp[i-1] + 1/(m-1)) * * 令a[i] = dp[i] + 1/(m-1) * 则有a[i] = m * a[i-1] * * * k = 1, 设dp[i]表示连续出现i个两两不相同的次数期望 * 设xi = dp[i] - dp[i-1], s[i] = dp[1] + ... + dp[i] * xi = 1 + (i-1)/m * xi + ((i-1)*dp[i-1] - s[i-1])/m * xi = (m + (i-1)*dp[i-1] - s[i-1]) / (m-i+1) * dp[i] = dp[i-1] + xi * dp[i] = (m * dp[i-1] + m - s[i-1]) / (m-i+1) */

int main () {
    int cas;
    while (scanf("%d", &cas) == 1) {
        while (cas--) {
            int k, m, n;
            double ans = 0;
            scanf("%d%d%d", &k, &m, &n);
            if (k) {
                double sum = 0;
                for (int i = 1; i <= n; i++) {
                    ans = (m * ans + m - sum) / (m - i + 1);
                    sum += ans;
                }
            } else {
                if (m == 1) ans = n;
                else ans = (pow(m, n) - 1) / (m-1);
            }
            printf("%.8lf\n", ans);
        }
    }
    return 0;
}