codeforces Round #259(div2) C解題報告

標籤：codeforces   acm   math   

C. Little Pony and Expected Maximumtime limit per test 1 secondmemory limit per test 256 megabytesinput standard inputoutput 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

題目大意：

一個m個面的骰子，拋擲n次，求這n次裡最大值的期望是多少。

解法：

數學題，有m個面的骰子，拋n次，那麼總共的情況就有m^n。我們從m=1開始推起。

m = 1， 只有一種情況，

m = 2，新增了2^n-1^n種情況，這些新增的情況裡面，最大值均是 2，

m = 3，新增了3^n-2^n種情況，這些新增的情況裡面，最大值均是 3，

我們就可以推出數學期望公式：  ans = 1/(m^n) * [1 + (2^n-1^n)*2 + (3^n-2^n)*3 .... + (m^n - (m-1)^n)*m]

但m^n太大，我們得改變一下式子， ans = (1/m)^n - (0/m)^n + [(2/m)^n - (1/m)^n] * 2 ...... + [(m/m)^n - ((m-1)/m)^n] * m。

代碼：

#include <cstdio>#include <cmath>using namespace std;double n, m, ans;int main() {scanf("%lf%lf", &m, &n);ans = pow(1.0/m, n);for (int i = 2; i <= m; i++)ans += (pow(i/m, n) - pow((i-1)/m, n)) * i;printf("%lf", ans);}

codeforces Round #259(div2) C解題報告