C - Pocky

​​传送门​​​ 题意：一根木棒，不断折断，等概率从中选取一点。然后再选择去吃，直到吃的部份小于等于d就停止，问吃的次数的数学期望。
思路：猜测题意，一看结果是小数位，又是公式推导，必然是lnx方面的，盲猜ln(l/d)+1; ln2=0.693147
ln4=2*ln2=1.386294.

/**
* From:
* Qingdao Agricultural University
* Created by XiangwangAcmer
* Date : 2019-09-28-16.13.17
* Talk is cheap.Show me your code.
*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<cmath>
#include<cctype>
#include<stack>
#include<map>
#include<string>
#include<cstdlib>
#define ll long long
using namespace std;
const ll maxn = 1e6 + 5;
const ll minn = 1e9 + 5;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LIMIT = 4294967295LL;
vector<int>v[maxn];
bool row[maxn], col[maxn];
bool flag = 0;
queue<int>q;
int main() {
ios::sync_with_stdio(false);
int n;
cin>>n;
double l,d;
while(n--){
cin>>l>>d;
if(l<=d)
cout<<0.000000<<endl;
else
printf("%.6f\n",log(l/d)+1);
}
return 0;
}