Pocky HDU - 5984（有毒题）

最新推荐文章于 2019-10-11 20:35:40 发布

Guuuuuu老师儿

最新推荐文章于 2019-10-11 20:35:40 发布

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分类专栏： # 概率期望

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/81567568

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本文探讨了PockyHDU-5984问题，这是一个涉及概率论与数学期望的趣味算法题目。主要内容是如何通过数学方法求解在特定条件下重复吃掉并折断Pocky棒的平均次数。

Pocky HDU - 5984

 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

这道题一直推不出来，结果结束后搜题解都是这个样的

第二个样例的结果减去1是ln2=0.693147，同理第三个样例减去1是ln4，所以猜测公式是1+ln（L/d）
？？？？
有毒 FUCK

网上找到一个推导过程，结果看不懂了。。。。高数啊～～～～

code:

#include <bits/stdc++.h>
using namespace std;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        double L,d;
        scanf("%lf%lf",&L,&d);
        if(L <= d){
            printf("0.000000\n");
            continue;
        }
        printf("%.6f\n",1+log(L/d));
    }
    return 0;
}