Code forces 453A Little Pony and Expected Maximum(概率期望)

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 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:

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

 

题意：筛子，如果有M个面的话一个面1点，一个面2点，一个面3点。。。。直到一个面M点，现在知道了M和要扔N次筛子。问能够扔到的最大点数的期望。

思路：按照第二组样例，M = 6， N = 3

那么如果扔到的最大点数是是1， 那么只有一种情况 即 1 1 1

如果扔到的最大点数是2 那么可能是 1 1 2， 1 2 1，.......每个位置都只能有1和2两种可能（不然最大点数就不是2了），但是要减去所有都是1的情况，所以最大点数是2的情况为2^n-1^n。

因为总的种数是m^n次。所以概率就是(2^n-1^n)/m^n,那么递推一下就可以了。

 

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
using namespace std;
int m, n;
#define maxn 10010
double ans, now, pre; 
int main(){
    while(~scanf("%d%d", &m, &n)){
        ans = 0;  pre = 0;
        for(int i = 1; i <= m; i++){
            now = pow((double(i)/double(m)), double(n));
            ans += (now - pre)*i;
            pre = now;
            
        }
        cout<<fixed<<setprecision(12)<<ans<<endl;
    }
    
    
    
    return 0;
}