CF1543C.Need for Pink Slips

普及/提高-

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题目描述

After defeating a Blacklist Rival, you get a chance to draw 11 reward slip out of xx hidden valid slips. Initially, x=3x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are cc , mm , and pp , respectively. There is also a volatility factor vv . You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the xx valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was aa . Then,

  • If the item was a Pink Slip, the quest is over, and you will not play any more races.
  • Otherwise,
    1. If ava\leq v , the probability of the item drawn becomes 00 and the item is no longer a valid item for all the further draws, reducing xx by 11 . Moreover, the reduced probability aa is distributed equally among the other remaining valid items.
    2. If a>va > v , the probability of the item drawn reduces by vv and the reduced probability is distributed equally among the other valid items.

For example,

  • If (c,m,p)=(0.2,0.1,0.7)(c,m,p)=(0.2,0.1,0.7) and v=0.1v=0.1 , after drawing Cash, the new probabilities will be (0.1,0.15,0.75)(0.1,0.15,0.75) .
  • If (c,m,p)=(0.1,0.2,0.7)(c,m,p)=(0.1,0.2,0.7) and v=0.2v=0.2 , after drawing Cash, the new probabilities will be (Invalid,0.25,0.75)(Invalid,0.25,0.75) .
  • If (c,m,p)=(0.2,Invalid,0.8)(c,m,p)=(0.2,Invalid,0.8) and v=0.1v=0.1 , after drawing Cash, the new probabilities will be (0.1,Invalid,0.9)(0.1,Invalid,0.9) .
  • If (c,m,p)=(0.1,Invalid,0.9)(c,m,p)=(0.1,Invalid,0.9) and v=0.2v=0.2 , after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0)(Invalid,Invalid,1.0) .

You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.

输入格式

The first line of input contains a single integer tt ( 1t101\leq t\leq 10 ) — the number of test cases.

The first and the only line of each test case contains four real numbers cc , mm , pp and vv ( 0<c,m,p<10 < c,m,p < 1 , c+m+p=1c+m+p=1 , 0.1v0.90.1\leq v\leq 0.9 ).

Additionally, it is guaranteed that each of cc , mm , pp and vv have at most 44 decimal places.

输出格式

For each test case, output a single line containing a single real number — the expected number of races that you must play in order to draw a Pink Slip.

Your answer is considered correct if its absolute or relative error does not exceed 10610^{-6} .

Formally, let your answer be aa , and the jury's answer be bb . Your answer is accepted if and only if abmax(1,b)106\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6} .

输入输出样例

  • 输入#1

    4
    0.2 0.2 0.6 0.2
    0.4 0.2 0.4 0.8
    0.4998 0.4998 0.0004 0.1666
    0.3125 0.6561 0.0314 0.2048

    输出#1

    1.532000000000
    1.860000000000
    5.005050776521
    4.260163673896

说明/提示

For the first test case, the possible drawing sequences are:

  • P with a probability of 0.60.6 ;
  • CP with a probability of 0.20.7=0.140.2\cdot 0.7 = 0.14 ;
  • CMP with a probability of 0.20.30.9=0.0540.2\cdot 0.3\cdot 0.9 = 0.054 ;
  • CMMP with a probability of 0.20.30.11=0.0060.2\cdot 0.3\cdot 0.1\cdot 1 = 0.006 ;
  • MP with a probability of 0.20.7=0.140.2\cdot 0.7 = 0.14 ;
  • MCP with a probability of 0.20.30.9=0.0540.2\cdot 0.3\cdot 0.9 = 0.054 ;
  • MCCP with a probability of 0.20.30.11=0.0060.2\cdot 0.3\cdot 0.1\cdot 1 = 0.006 .

So, the expected number of races is equal to 10.6+20.14+30.054+40.006+20.14+30.054+40.006=1.5321\cdot 0.6 + 2\cdot 0.14 + 3\cdot 0.054 + 4\cdot 0.006 + 2\cdot 0.14 + 3\cdot 0.054 + 4\cdot 0.006 = 1.532 .For the second test case, the possible drawing sequences are:

  • P with a probability of 0.40.4 ;
  • CP with a probability of 0.40.6=0.240.4\cdot 0.6 = 0.24 ;
  • CMP with a probability of 0.40.41=0.160.4\cdot 0.4\cdot 1 = 0.16 ;
  • MP with a probability of 0.20.5=0.10.2\cdot 0.5 = 0.1 ;
  • MCP with a probability of 0.20.51=0.10.2\cdot 0.5\cdot 1 = 0.1 .

So, the expected number of races is equal to 10.4+20.24+30.16+20.1+30.1=1.861\cdot 0.4 + 2\cdot 0.24 + 3\cdot 0.16 + 2\cdot 0.1 + 3\cdot 0.1 = 1.86 .

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