Codeforces Round #259 (Div. 1) A. Little Pony and Expected Maximum 组合数学

A. Little Pony and Expected Maximum

solution

$$res=\sum _{i=1}^m\frac{i^n-(i-1{)}^n}{m^n}=\sum _{i=1}^m(\frac{i}{m}{)}^n-(\frac{i-1}{m}{)}^n$$

code

/*Siberian Squirrel*/
/*Cute KiloFish*/
#include<bits/stdc++.h>

#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ACM_LOCAL

using namespace std;
typedef long long ll;

const double PI = acos(-1);
const double eps = 1e-4;
const int INF = 0x3f3f3f3f;
/*const int MOD = 998244353, r = 119, k = 23, g = 3;
const int MOD = 1004535809, r = 479, k = 21, g = 3;*/
const int MOD = 1e9 + 7;
const int M = 1e7 + 10;
const int N = 1e6 + 10;
const int UP = 50;
//inline int rnd(){static int seed=2333;return seed=(((seed*666666ll+20050818)%998244353)^1000000007)%1004535809;}

int n, m;

double quick_pow(double ans, int p, double res = 1) {
    
    
    for(; p; p >>= 1, ans = ans * ans)
        if(p & 1) res = res * ans;
    return res;
}

void solve(double res = 0) {
    
    
    for(int i = 1; i <= m; ++ i) {
    
    
        res += i * (quick_pow(1.0 * i / m, n) - quick_pow(1.0 * (i - 1) / m, n));
    }
    cout << fixed << setprecision(10) << res << endl;
}

int main() {
    
    
    IO;
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    int o = 1, cases = 0;
//    cin >> o;
    while(o --) {
    
    
        cin >> m >> n;
        solve();
    }
    return 0;
}