Codeforces #259 (Div. 2) C. Little Pony and Expected Maximum

C. 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

一道比较坑的题,我也推公示了,因为细节没处理好推错了

而且当时老想着化简公式去了...

这道题的做法如下:

考虑有n个骰子, 每个m面

则当前这次投掷后最大值为m对应有m^n-(m-1)^n中情况----去除最大值不为m的情况

那这样的话公式就很容易了

代码如下:

#include <cmath>#include <cstdio>#include <iostream>#include <algorithm>#define MAXN 100010#define LL long longusing namespace std;int main(void) {    int m, n;    scanf("%d%d", &m, &n);    double ans = 0.0;    double sum = 0.0;    for(int i=1; i<=m; ++i) {        sum = pow((i-1.0)/m, n*1.0);        ans += (pow(i*1.0/m, n*1.0)-sum)*i;    }    cout.setf(ios::fixed);    cout.precision(12);    cout << ans << endl;    return 0;}