CF1543C.Need for Pink Slips

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 $a \gt v$, 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.

击败一名黑名单对手后，你将获得一次从 $x$ 张隐藏的有效奖券中抽取 $1$ 张的机会。初始时，$x=3$，这三张隐藏的有效奖券分别为：现金券（Cash Slip）、扣押解除标记（Impound Strike Release Marker）以及对手车辆的粉红券（Pink Slip of Rival's Car）。初始时，随机抽取这三张奖券的概率分别为 $c$、$m$ 和 $p$。此外还存在一个波动因子 $v$。你可以无限次地参加对手竞速赛，前提是尚未抽到粉红券。假设你每场比赛均获胜，并因此获得一次抽取奖券的机会。每次抽取时，你将从当前 $x$ 张有效奖券中按其各自对应概率抽取一张。假设你抽中了某张奖券，且抽中前其概率为 $a$，则：

• 若该奖券为粉红券，则任务结束，你将不再进行任何后续比赛；
• 否则，
  1. 若 $a \leq v$，则该奖券的概率变为 $0$，且此后所有抽取中该奖券均不再视为有效奖券，从而使 $x$ 减少 $1$；同时，原概率 $a$ 将被平均分配给其余尚存的有效奖券；
  2. 若 $a > v$，则该奖券的概率减少 $v$，而减少的这部分概率将被平均分配给其余所有当前有效的奖券。

例如：

• 若 $(c,m,p)=(0.2,0.1,0.7)$ 且 $v=0.1$，在抽中现金券后，新的概率为 $(0.1,0.15,0.75)$；
• 若 $(c,m,p)=(0.1,0.2,0.7)$ 且 $v=0.2$，在抽中现金券后，新的概率为 $(\text{Invalid},0.25,0.75)$；
• 若 $(c,m,p)=(0.2,\text{Invalid},0.8)$ 且 $v=0.1$，在抽中现金券后，新的概率为 $(0.1,\text{Invalid},0.9)$；
• 若 $(c,m,p)=(0.1,\text{Invalid},0.9)$ 且 $v=0.2$，在抽中现金券后，新的概率为 $(\text{Invalid},\text{Invalid},1.0)$。

你需要获取对手的车辆，因此你需要计算：为抽中一张粉红券所需进行的比赛场数的期望值。

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 $v$ ($0 \lt c,m,p \lt 1$, $c+m+p=1$, $0.1\leq v\leq 0.9$).

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

输入的第一行包含一个整数 $t$（$1\leq t\leq 10$）—— 表示测试用例的数量。

每个测试用例仅有一行，包含四个实数 $c$、$m$、$p$ 和 $v$（$0 \lt c,m,p \lt 1$，$c+m+p=1$，$0.1\leq v\leq 0.9$）。

此外，保证 $c$、$m$、$p$ 和 $v$ 均最多有 $4$ 位小数。

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 $\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$.

对于每个测试用例，输出一行，包含一个实数——即为了抽到粉红色收据（Pink Slip）所需进行的比赛场数的期望值。

若你的答案的绝对误差或相对误差不超过 $10^{-6}$，则该答案视为正确。

形式化地，设你的答案为 $a$，评测组的答案为 $b$。当且仅当 $\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$ 时，你的答案被接受。

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 $1\cdot 0.4 + 2\cdot 0.24 + 3\cdot 0.16 + 2\cdot 0.1 + 3\cdot 0.1 = 1.86$.

对于第一个测试用例，可能的抽取序列如下：

• P，概率为 $0.6$；
• CP，概率为 $0.2\cdot 0.7 = 0.14$；
• CMP，概率为 $0.2\cdot 0.3\cdot 0.9 = 0.054$；
• CMMP，概率为 $0.2\cdot 0.3\cdot 0.1\cdot 1 = 0.006$；
• MP，概率为 $0.2\cdot 0.7 = 0.14$；
• MCP，概率为 $0.2\cdot 0.3\cdot 0.9 = 0.054$；
• MCCP，概率为 $0.2\cdot 0.3\cdot 0.1\cdot 1 = 0.006$。

因此，比赛场数的期望值等于 $1\cdot 0.6 + 2\cdot 0.14 + 3\cdot 0.054 + 4\cdot 0.006 + 2\cdot 0.14 + 3\cdot 0.054 + 4\cdot 0.006 = 1.532$。

对于第二个测试用例，可能的抽取序列如下：

• P，概率为 $0.4$；
• CP，概率为 $0.4\cdot 0.6 = 0.24$；
• CMP，概率为 $0.4\cdot 0.4\cdot 1 = 0.16$；
• MP，概率为 $0.2\cdot 0.5 = 0.1$；
• MCP，概率为 $0.2\cdot 0.5\cdot 1 = 0.1$。

因此，比赛场数的期望值等于 $1\cdot 0.4 + 2\cdot 0.24 + 3\cdot 0.16 + 2\cdot 0.1 + 3\cdot 0.1 = 1.86$。