HDU 5984 Pocky (数学)

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本文探讨了一种关于Pocky棒的有趣数学问题：在特定条件下，不断将Pocky棒折断并吃掉较短部分的过程中，求出期望的操作次数。通过数学公式和编程实现，给出了不同长度下Pocky棒的期望操作次数。

Pocky

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 556 Accepted Submission(s): 302

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

Source

2016ACM/ICPC亚洲区青岛站-重现赛（感谢中国石油大学）

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jiangzijing2015

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>

using namespace std;
const int INF=1e9+7;
long long int fact[101];

void init(){
    fact[0]=1;
    for(int i=1;i<=20;i++){
        fact[i]=fact[i-1]*i;
    }
}

int main(){
    double x,y;
    int cases;
    init();
    scanf("%d",&cases);
    while(cases--){
        double res=0;
        scanf("%lf%lf",&x,&y);
        if(x<=y){
            printf("%.6lf\n",0);
            continue;
        }
        double tmp1=log(x)-log(y);
        double tmp2=y*1.0/x;
        for(int i=1;i<=20;i++){
            double tmp=1;
            tmp*=tmp2;
            tmp*=pow(tmp1,i-1);
            tmp*=i;
            tmp/=fact[i-1];
            res+=tmp;
        }
        printf("%.6lf\n",res);
    }
    return 0;
}
/*
1
1 4 1 4
2 1 2 1
3 2 3 2
4 3 4 3
5 5 5 5
6 6 6 6

*/