题意:一个筛子有m个面,然后扔n次,求最大值的期望;

思路:最大值为1 有1种,2有2n-1种,  3有3n -2n 种   所以为m的时有mn -(m-1)n 种,所以分别求每一种的概率,然后乘以这个值求和就可以。

 #include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std; int n,m; int main()
{
cin>>m>>n;
double x=pow(1.0/m,n);
double ans=x;
for(int i=; i<=m; i++)
{
double xx=pow(i*1.0/m,n);
ans+=(xx-x)*i;
x=xx;
}
printf("%.12lf\n",ans);
return ;
}

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