Codeforces 453A Little Pony and Expected Maximum

A. 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 m dots. 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:

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:

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

先附上原题链接http://codeforces.com/problemset/problem/453/A 
题目是想求出长度为n的序列中，最大值的期望。考虑排列的总数为m^n,含i的且最大值不超过i的排列个数为i^n,即每一位都有i种可能性。需要注意的是，这些排列中会有最大值不是i的排列，即不含i且最大值小于i的排列，换句话来说，就是含i-1且最大值不超过i-1的排列。这样，最大值是i的排列种数就可以求出了。=>   i^n-(i-1)^n我们推出的公式是Sum=(m*(m^n-(m-1)^n)+.....+1.)/m^n 但是m、n的取值上限是100000，显然计算100000^100000是不可能的。我们需要简化公式。
观察到分子分母都是X^n,并且都含m，可以上下同除以m^n.展开化简得到m-(1-1/m)^n-(1-2/m)^n-....(1-(m-1)/m)^n带入计算即可。

#include<stdio.h>double quickpow(double a,int b)//快速幂{   double r=1;   while(b!=0)   {       if(b&1)          r*=a;        b=b>>1;        a=a*a;   }   return r;}int main(){   int n,m,i;   double sum,k;   scanf("%d%d",&m,&n);   sum=m;   for(i=1;i<m;i++)   {     k=quickpow((m-i)*1.0/m,n);     sum-=k;   }   printf("%.12lf\n",sum);}