离散概率专题

A - Little Pony and Expected Maximum

$ ans = sum_i^niP_ ext{最大值为k}$

$ = sum_i^n i((i)^k-(i-1)k) $

#include<bits/stdc++.h>
using namespace std;

double qpow(double x,int base){
	long double xt = x;
	long double ret = 1;
	while (base > 0){
		if (base & 1){
			ret *= xt;
		}
		xt *= xt;
		base /= 2;
	}
	return ret;
}
int main(){
	int n, k;
	cin >> n >> k;
	double ans = 0;
	for (int i=1;i<=n;i++){
		ans += (qpow((double)(i-1) / n, k) - qpow((double)i/n, k)) * i;
	}
	printf("%.8f",-ans);
}