C++11(clang++ 3.9) 解法, 执行用时: 12ms, 内存消耗: 1708K, 提交时间: 2020-10-14 22:06:33

#include<cstdio>
using namespace std;
double f[55][2505], g[55][2505];
int main(){
	int a, b, c, d, i, j, k;double ans, s;
	scanf("%d%d%d%d", &a, &b, &c, &d);
	for(i = f[0][0] = 1;i <= a;i++)
		for(j = i;j <= b * i;j++)
			for(k = 1;k <= b && k <= j;k++)
				f[i][j] += f[i - 1][j - k];
	for(i = 1;i <= a * b;i++)f[a][i] /= b;
	for(i = g[0][0] = 1;i <= c;i++)
		for(j = i;j <= d * i;j++)
			for(k = 1;k <= d && k <= j;k++)
				g[i][j] += g[i - 1][j - k];
	for(i = 1;i <= c * d;i++)g[c][i] /= d;
	for(i = 1, s = 0;i <= a * b;i++){
		g[c][i] += g[c][i - 1];
		s += g[c][i - 1] * f[a][i];
	}for(i = 2, ans = 0;i <= a * b;i++)ans += i * f[a][i] * g[c][i - 1] / s;
	printf("%lf", s < 1e-6 ? -1 : ans);
	return 0;
}