orangebird
2017-06-08 22:28:43
设n=sigma(ai)
对于每一位i,从i开始的7位中,7个魔法都不相同的概率为7! * a1/n * a2/(n-1) * a3/(n-2) * a4/(n-3) *a5/(n-4) * a6/(n-5) * a7/(n-6)
由于一共有n-6个这样的位置i,所以最后一项除以n-6可以消去
答案就是7! * a1/n * a2/(n-1) * a3/(n-2) * a4/(n-3) *a5/(n-4) * a6/(n-5) * a7
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
double a1,a2,a3,a4,a5,a6,a7,n;
int main()
{
cin>>a1>>a2>>a3>>a4>>a5>>a6>>a7;
n=a1+a2+a3+a4+a5+a6+a7;
printf("%.3lf",5040*a1/n*a2/(n-1)*a3/(n-2)*a4/(n-3)*a5/(n-4)*a6/(n-5)*a7);
}