# HDU 5984.Pocky

## 题目

### 题面

Let’s talking about something of eating a pocky. Here is a Decorer Pocky, with colorful decorative stripes in the coating, of length L.
While the length of remaining pocky is longer than d, we perform the following procedure.
We break the pocky at any point on it in an equal possibility and this will divide the remaining pocky into two parts.
Take the left part and eat it. When it is not longer than d, we do not repeat this procedure.
Now we want to know the expected number of times we should repeat the procedure above.
Round it to 6 decimal places behind the decimal point.

### 输入解释

The first line of input contains an integer N which is the number of test cases. Each of the N lines contains two float-numbers L and d respectively with at most 5 decimal places behind the decimal point where 1 ≤ d, L ≤ 150.

### 输出解释

For each test case, output the expected number of times rounded to 6 decimal places behind the decimal point in a line.

### 样例输入

6
1.0 1.0
2.0 1.0
4.0 1.0
8.0 1.0
16.0 1.0
7.00 3.00


### 样例输出

0.000000
1.693147
2.386294
3.079442
3.772589
1.847298


## 题解

### 证明

$\varphi$ 表示从长度为 x 的线段上取到一个点的概率,有 $\int_{0}^{x} \varphi dt = 1$

\begin{aligned} f(x) &= \begin{cases} 0 && ,x\leq d\\ 1+ \int_{0}^{d} \varphi f(t)dt + \int_{d}^{x}\varphi f(t)dt && ,x > d \end{cases} \\ &= \begin{cases} 0 && ,x\leq d\\ 1+ \frac{\int_{0}^{d} f(t)dt}{x} + \frac{\int_{d}^{x} f(t)dt}{x} && ,x > d \end{cases} \\ &= \begin{cases} 0 && ,x\leq d\\ 1+ \frac{\int_{d}^{x} f(t)dt}{x} && ,x > d \end{cases} \end{aligned}

$\lim_{x->d_-} f(x) = 0$

$\lim_{x->d_+} f(x) = 1$

\begin{aligned} \because f(x) &= 1 + \frac{\int_{d}^{x}f(t)dt}{x} \\ &= 1 + \frac{F(x) - F(d)}{x} \end{aligned}

\begin{aligned} \therefore f{}'(x) &= \frac{xF{}'(x) - F(x) + F(d)}{x^2} \\ &= \frac{xf(x) - \int_{d}^{x}f(t)dt}{x^2} \\ &= \frac{xf(x) - x(f(x)-1)}{x^2} \\ &= \frac{1}{x} \end{aligned}

\begin{aligned} \therefore & f(x)=ln(x)+C \\ \because & \lim_{x->d_+} f(x) = 1 \\ \therefore & f(d) = ln(d) + C = 1 \\ \therefore & C = 1 - ln(d) \\ \therefore & f(x) = ln(x) + 1 - ln(d) \\ \therefore & f(x) = ln(\frac{x}{d}) + 1 \end{aligned}

\begin{aligned} f(x) = \begin{cases} 0 && ,x\leq d\\ ln(\frac{x}{d})+1 && ,x > d \end{cases} \end{aligned}

## 代码

/*//
#define debug
#include <ctime>
//*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <iomanip>
using namespace std;

int main() {
#ifdef debug
freopen("in.txt", "r", stdin);
int START = clock();
#endif
cin.tie(0);
cin.sync_with_stdio(false);

int T;
cin >> T;
while (T--) {
double a, b;
cin >> a >> b;
if (a <= b)
cout << "0.000000" << endl;
else
cout << fixed << setprecision(6) << 1.0 + log(a / b) << endl;
}

#ifdef debug
printf("Time:%.3fs.\n", double(clock() - START) / CLOCKS_PER_SEC);
#endif
return 0;
}