CF 312B(Archer-等比数列极限求和)

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本文介绍了一个射手对决中获胜概率的计算方法。通过给定两个射手每次射击命中目标的概率，利用等比数列无限求和公式计算出了先手射手赢得比赛的概率。

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SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is $\text{[math]}$ for SmallR while $\text{[math]}$ for Zanoes. The one who shoots in the target first should be the winner.

Output the probability that SmallR will win the match.

Input

A single line contains four integers $\text{[math]}$ .

Output

Print a single real number, the probability that SmallR will win the match.

The answer will be considered correct if the absolute or relative error doesn't exceed 10^- 6.

Sample test(s)

input
1 2 1 2

output
0.666666666667

等比数列无限求和公式 a+ap+ap^2+... 0<p<1 --> S=a/(1-p)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	double a,b,c,d;
	cin>>a>>b>>c>>d;
	printf("%.12lf\n",a/b/(1-(1-a/b)*(1-c/d)));
	
	return 0;
}