第一天的集训主要学习一些基本的知识，涉及贪心、二分、和一些简单的思想。
对于单调的连续数组，求某一元素出现的次数：
在这之前，先介绍两个函数lower_bound（地址，地址，元素）upper_bound（），头文件是<algorithm> （之前一直知道却很少用，这次一提，后面要记牢）
lower_bound（）查找大于等于X的第一个位置（是一个地址，可以通过减去首地址得到下标）
upper_bound（）查找大于X的第一个位置
那么元素个数为 upper_bound() - lower_bound() 。
得到1~N内由0和1构成的数：
采用二进制的思想，十进制0对应二进制0，1对应1，2对应10，3对应11，如此只用枚举十进制整数，变成二进制即可。
再写一道题结束这次尬写。
D - Pocky
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.
Input
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.
Output
For each test case, output the expected number of times rounded to 6 decimal places behind the decimal point in a line.
Sample Input
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
Sample Output
0.000000 1.693147 2.386294 3.079442 3.772589 1.847298
分析：
ln2=0.693147 可以推断结果与对数有关，根据样例可以猜出结果ln(L/d)+1.
具体的证明过程参考传送门 仰望高端玩家。