CF453A Little Pony and Expected Maximum

(large{题目链接})
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(Large extbf{Solution: } large{设取n次的得到的最大值为M，那么方案数为M^{n}-(M-1)^{n}。\其中，M^n表示取值在left[ 1.M ight]的方案数，（M-1)^{n}表示取值在left[ 1,M-1 ight]的方案数。\答案即为：sum limits^{m}_{i=1}icdot dfrac {left( i^{n}-(i-1)^{n} ight) }{m^{n}}=sum limits ^{m}_{i=1}left( dfrac {i}{m} ight) ^{n}-left( dfrac {i-1}{m} ight) ^{n} imes i。})
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(Large extbf{Summary: } large{在实数域算幂直接用pow库函数即可，愚蠢的我一度想打快速幂。})
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(Large extbf{Code: })

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

int main() {
	int n, m;
	cin >> m >> n;
	if (m == 1) return cout << "1.0", 0;
	double ans = 0;
	for (int i = 1; i <= m; ++i) ans += 1.0 * i * (pow(1.0 * i / m, n) - pow(1.0 * (i - 1) / m, n));
	printf("%.12lf
", ans);
	return 0;
}