CF 453A(Little Pony and Expected Maximum-若干次投骰,最大那次期望-推公式)

CF 453A(Little Pony and Expected Maximum-若干次掷骰,最大那次期望-推公式)

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 CF 453A(Little Pony and Expected Maximum-若干次投骰,最大那次期望-推公式) . 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:

CF 453A(Little Pony and Expected Maximum-若干次投骰,最大那次期望-推公式)

You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value

设骰子n面，掷m次

易得最大次为k的概率：k^m/n^m-(k-1)^m/n^m=(k/n)^m-最大值为k-1的概率

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1e5+10)
#define eps (1e-4)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,m;
double t;
double powe(double a,int b)
{
	if (b==1) return a;
	if (b==0) return 1;
	double t=powe(a,b/2);
	t=t*t;
	if (b&1) t=t*a;
	return t;
}
int main()
{
//	freopen("Expected Maximum.in","r",stdin);
//	freopen(".out","w",stdout);
	scanf("%d%d",&n,&m);
	t=1;For(i,m) t/=n;// cout<<t<<endl;
	double ans=0,psum=0;
	For(k,n)
	{
		double t=(double)k/(double)n;
		t=powe(t,m);
		ans+=(t-psum)*k;
		psum=t;
	}
	cout<<ans<<endl;
	
	return 0;
}

CF 453A(Little Pony and Expected Maximum-若干次投骰,最大那次期望-推公式)

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