【HDU 5984】Pocky 数学期望

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                                             Pocky

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 247    Accepted Submission(s): 119

 

Problem Description

Let’s talkingabout something of eating a pocky. Here is a Decorer Pocky, with colorfuldecorative stripes in the coating, of length L.
While the length of remaining pocky is longer than d, we perform the followingprocedure. We break the pocky at any point on it in an equal possibility andthis will divide the remaining pocky into two parts. Take the left part and eatit. 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 procedureabove. Round it to 6 decimal places behind the decimal point.

Input

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

Output

For each testcase, output the expected number of times rounded to 6 decimal places behindthe decimal point in a line.

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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

题意：将一段长度为n的绳子多次切割，直到绳子长度不大于m，求切割次数的数学期望。

很尴尬的一道题，核心代码只有两行，但是不懂概率论无从下手。数学真的是软肋。

另外通过找规律可以发现：

(1.693147-1)*2=2.386294-1

(2.386294-1)*2=3.772589-1

可以退出跟2的对数相关，再加上一些细节，得到公式log(n/m)+1;

搬运大神的解析： 原址

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define closeio std::ios::sync_with_stdio(false)
 
int main()
{
	int t;
	double n,m;
	cin>>t;
	while(t--)
	{
		cin>>n>>m;
		if(n<=m)
			printf("0.000000\n");
		else
			printf("%.6lf\n",log(n/m)+1);
	}
	return 0;
}