CF618G.Combining Slimes

普及/提高-

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题目描述

Your friend recently gave you some slimes for your birthday. You have a very large amount of slimes with value 11 and 22 , and you decide to invent a game using these slimes.

You initialize a row with nn empty spaces. You also choose a number pp to be used in the game. Then, you will perform the following steps while the last space is empty.

  1. With probability , you will choose a slime with value 11 , and with probability , you will choose a slime with value 22 . You place the chosen slime on the last space of the board.
  2. You will push the slime to the left as far as possible. If it encounters another slime, and they have the same value vv , you will merge the slimes together to create a single slime with value v+1v+1 . This continues on until the slime reaches the end of the board, or encounters a slime with a different value than itself.

You have played the game a few times, but have gotten bored of it. You are now wondering, what is the expected sum of all values of the slimes on the board after you finish the game.

输入格式

The first line of the input will contain two integers n,pn,p ( 1<=n<=10^{9},1<=p<10^{9} ).

输出格式

Print the expected sum of all slimes on the board after the game finishes. Your answer will be considered correct if its absolute or relative error does not exceed 10410^{-4} .

Namely, let's assume that your answer is aa and the answer of the jury is bb . The checker program will consider your answer correct, if .

输入输出样例

  • 输入#1

    2 500000000
    

    输出#1

    3.562500000000000
    
  • 输入#2

    10 1
    

    输出#2

    64.999983360007620
    
  • 输入#3

    100 123456789
    

    输出#3

    269.825611298854770
    

说明/提示

In the first sample, we have a board with two squares, and there is a 0.50.5 probability of a 1 appearing and a 0.50.5 probability of a 2 appearing.

Our final board states can be 1 2 with probability 0.250.25 , 2 1 with probability 0.3750.375 , 3 2 with probability 0.18750.1875 , 3 1 with probability 0.18750.1875 . The expected value is thus (1+2)0.25+(2+1)0.375+(3+2)0.1875+(3+1)0.1875=3.5625(1+2)·0.25+(2+1)·0.375+(3+2)·0.1875+(3+1)·0.1875=3.5625 .

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