# #P1543C. Need for Pink Slips

ID: 7141 Type: RemoteJudge 1000ms 256MiB 尝试: 1 已通过: 1 难度: 10 上传者: 标签>bitmasksbrute forcedfs and similardpimplementationmathprobabilities

# Need for Pink Slips

## Description

After defeating a Blacklist Rival, you get a chance to draw $1$ reward slip out of $x$ hidden valid slips. Initially, $x=3$ 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 $c$, $m$, and $p$, respectively. There is also a volatility factor $v$. 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 $x$ valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was $a$. Then,

• If the item was a Pink Slip, the quest is over, and you will not play any more races.
• Otherwise,
1. If $a\leq v$, the probability of the item drawn becomes $0$ and the item is no longer a valid item for all the further draws, reducing $x$ by $1$. Moreover, the reduced probability $a$ is distributed equally among the other remaining valid items.
2. If $a > v$, the probability of the item drawn reduces by $v$ and the reduced probability is distributed equally among the other valid items.

For example,

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

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.

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

The first and the only line of each test case contains four real numbers $c$, $m$, $p$ and $v$ ($0 < c,m,p < 1$, $c+m+p=1$, $0.1\leq v\leq 0.9$).

Additionally, it is guaranteed that each of $c$, $m$, $p$ and $v$ have at most $4$ decimal places.

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 $10^{-6}$.

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

## Input

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

The first and the only line of each test case contains four real numbers $c$, $m$, $p$ and $v$ ($0 < c,m,p < 1$, $c+m+p=1$, $0.1\leq v\leq 0.9$).

Additionally, it is guaranteed that each of $c$, $m$, $p$ and $v$ have at most $4$ decimal places.

## Output

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 $10^{-6}$.

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

## Samples

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

1.532000000000
1.860000000000
5.005050776521
4.260163673896


## Note

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

• P with a probability of $0.6$;
• CP with a probability of $0.2\cdot 0.7 = 0.14$;
• CMP with a probability of $0.2\cdot 0.3\cdot 0.9 = 0.054$;
• CMMP with a probability of $0.2\cdot 0.3\cdot 0.1\cdot 1 = 0.006$;
• MP with a probability of $0.2\cdot 0.7 = 0.14$;
• MCP with a probability of $0.2\cdot 0.3\cdot 0.9 = 0.054$;
• MCCP with a probability of $0.2\cdot 0.3\cdot 0.1\cdot 1 = 0.006$.
So, the expected number of races is equal to $1\cdot 0.6 + 2\cdot 0.14 + 3\cdot 0.054 + 4\cdot 0.006 + 2\cdot 0.14 + 3\cdot 0.054 + 4\cdot 0.006 = 1.532$.

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

• P with a probability of $0.4$;
• CP with a probability of $0.4\cdot 0.6 = 0.24$;
• CMP with a probability of $0.4\cdot 0.4\cdot 1 = 0.16$;
• MP with a probability of $0.2\cdot 0.5 = 0.1$;
• MCP with a probability of $0.2\cdot 0.5\cdot 1 = 0.1$.

So, the expected number of races is equal to $1\cdot 0.4 + 2\cdot 0.24 + 3\cdot 0.16 + 2\cdot 0.1 + 3\cdot 0.1 = 1.86$.