Codeforces Round #259 (Div. 2) C. Little Pony and Expected Maximum

C. Little Pony and Expected Maximum

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.

The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the m-th
face contains mdots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability

.
Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice n times.

Input

A single line contains two integers m and n (1 ≤ m, n ≤ 105).

Output

Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10  - 4.

Sample test(s)

input

6 1

output

3.500000000000

input

6 3

output

4.958333333333

input

2 2

output

1.750000000000

Note

Consider the third test example. If you've made two tosses:

You can get 1 in the first toss, and 2 in the second. Maximum equals to 2.

You can get 1 in the first toss, and 1 in the second. Maximum equals to 1.

You can get 2 in the first toss, and 1 in the second. Maximum equals to 2.

You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.

The probability of each outcome is 0.25, that is expectation equals to:



You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
题意：掷m面的骰子（1~m），每面出现概率为1/m，掷n次，问掷出最大值的期望。

思路：概率题。。按照期望的定义算就行了。其中用到了二分快速幂和等比数列的前n项和。

假设掷的n次中最大为i，分为n种情况，分别是在第1~n次出现i。然后对可能的i值（1~m）求和。公式为：

(i-1)^j*i^(n-j)/m^n
这样算会超时，先写了个二分快速幂，还是超时。仔细观察公式，是一个等比数列，于是运用等比数列前n项和公式。

#include <iostream>  
#include <stdio.h>  
#include <cmath>  
#include <algorithm>  
#include <iomanip>  
#include <cstdlib>  
#include <string>  
#include <memory.h>  
#include <vector>  
#include <queue>  
#include <stack>  
#include <map>
#include <set>
#define ll long long
#define INF 1000000

using namespace std;

inline double apowb(double a,int b){
	double ans=1;
	while(b){
		if(b&1){
			ans*=a;
		}
		a*=a;
		b>>=1;
	}
	return ans;
}

int main(){
	int m,n;
	while(cin>>m>>n){
		double ans=0;
		for(int i=1;i<=m;i++){
			
			double a1=apowb((double)i/m,n);
			double q=(double)(i-1)/i;
			double sn=a1*(1-apowb(q,n))/(1-q);
			ans+=sn;
		}
		printf("%lf\n",ans);
	}
	return 0;
}