神奇stl- - Little Pony and Expected Maximum

F - Little Pony and Expected Maximum
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
453A

Description

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 Input

Input

6 1

Output

3.500000000000

Input

6 3

Output

4.958333333333

Input

2 2

Output

1.750000000000

Hint

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

一开始以为用公式 ans = m - ((1/m)^n + (2/m)^n + (3/m)^n + ... + ((m-1)/m)^n)

会超时，因为以为pow时间复杂度为 O（n）；

提交：



然后百度：

所以这个题就是这么做；

附上代码：

#include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;

int main()
{
double n,m;
while(~scanf("%lf%lf",&m,&n)){
double s = 0;
if(n==1){
printf("%.5f\n",(1+m)*m/2/m);
}
else{

int len = m;

for(int i = 1;i<len;i++){
s+=pow(i/m,n);
}
printf("%.5f\n",m-s);
}
}
}


至于为什么这么做，下面是推的过程：
例如 6 3
1 - 1*1*1

2 - 2*2*2 - 1*1*1 -a

3 - 3*3*3 - 2*2*2 -b

4 - 4*4*4 - 3*3*3 -c

5 - 5*5*5 - 4*4*4 -d

6 - 6*6*6 - 5*5*5 -e

所以 ans = 6 * e  + 5*d + 4 * c + 3*b + 2*a +1 = 6*6*6*6 - (1*1*1 + 2*2*2 -+ 3*3*3 + 4*4*4 + 5*5*5) = m - (...)(同上)--;

但是n=1的时候需要注意，因为这时候是不需要减得，所以加起来除以n就行。