• 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$$sum_{i=1}^{m} Ans_i$$

  • 相关阅读:
    【JZOJ 4274】【NOIP2015模拟10.28B组】终章-剑之魂
    【JZOJ 4281】【NOIP2015模拟10.29A组】三色树
    【Luogu P2824】[HEOI2016/TJOI2016]排序
    【Luogu P5490】【模板】扫描线
    【Luogu P2502】[HAOI2006]旅行
    【Luogu P1629】 邮递员送信
    【Luogu P4047】[JSOI2010]部落划分
    【Luogu P4071】[SDOI2016]排列计数
    【Luogu P2508】 [HAOI2008]圆上的整点
    【Luogu P1102】A-B 数对
  • 原文地址:https://www.cnblogs.com/huibixiaoxing/p/8609499.html
Copyright © 2020-2023  润新知