HDU 5984 Pocky (数学)

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

61.0 1.02.0 1.04.0 1.08.0 1.016.0 1.07.00 3.00

 

Sample Output

0.0000001.6931472.3862943.0794423.7725891.847298

 

Source

2016ACM/ICPC亚洲区青岛站-重现赛（感谢中国石油大学）

 

Recommend

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;}/*11 4 1 42 1 2 13 2 3 24 3 4 35 5 5 56 6 6 6*/