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  • 嘴巴题9 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.

    Examples

    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:

    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

    题目大意

    给你一个(m)个面的骰子,第(i)个面上的数为(i),投(n)次,问这(n)次中最大值的期望。

    题解

    考虑枚举最大值(i),直接算不太好算,考虑容斥。
    最大值为(i)的方案 = 所有数小于等于(i)的方案 - 不包含(i)的方案,即为所有数小于等于(i)且包含(i)的方案,即

    [Ans_i = i^n - (i-1)^n ]

    总方案数除以(m^n)即可
    由于太大可能会溢出,要边计算边除,即

    [frac{Ans_i}{m^n} = frac{i^n - (i-1)^n}{m^n} = frac{i}{m}^n - frac{i-1}{m}^n ]

    答案即为$$sum_{i=1}^{m} Ans_i$$

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  • 原文地址:https://www.cnblogs.com/huibixiaoxing/p/8609499.html
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