HDU 5984.Pocky(2016 CCPC 青岛 C)

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

设f(x)为长度为x的木棒的期望。
(1)当x<=d时，已经不用分割了，故期望f(x)=0;
(2)当x>d时，f(x)=1+f(0~d)+f(d~x);
①1:表示在长度为x的木棒上分割一次
②f(0~d):分割点在0~d上的期望，由(1)f(0~d)=0
③f(d~x):分割点在d~x上的期望,关键就是求f(d~x)

f(d~x)的求解:在某个点上分割的概率是1/x,则在t处分割的期望:(1/x)*f(t),那么在d~x上将该式积分起来就是f(d~x); 

大佬博客：https://www.oyohyee.com/post/HDU/5984/
代码：

#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        double a, b;
        scanf("%lf %lf", &a, &b);
        if(a<=b) printf("0.000000\n");
        else
        {
            double ans = log(a) - log(b) + 1.0;
            printf("%.6f\n", ans);
        }
    }
    return 0;
}