【HDU 5984 Pocky】+ 数学

Pocky
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 269 Accepted Submission(s): 132

Problem Description
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

Source
2016ACM/ICPC亚洲区青岛站-重现赛（感谢中国石油大学）

把样例输出减一后～会发现是0.693147是ln2～～2.079442是0.693147的3倍～而8.0 / 1.0 是 2.0/1.0的 3次方～这样容易联系到ln～观察样例～=.=

AC代码：

#include<cstdio>#include<cmath>using namespace std;int main(){    int T;    double a,b,ans;    scanf("%d",&T);    while(T--){        scanf("%lf %lf",&a,&b);        if(a <= b) printf("0.000000\n");        else printf("%.6lf\n",log(a / b) + 1.0);    }    return 0;}