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

A. Little Pony and Expected Maximum

time limit per test

memory limit per test

input

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.

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

Input

m and n (1 ≤ m, n ≤ 105).

Output

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:

1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2.
2. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1.
3. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2.
4. 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:

​​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;
}