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 for SmallR while 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 .
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.
Examples
Input
1 2 1 2
Output
0.666666666667
据说这是一道概率水题,但是是我的盲点,体现的思想可能是事件发生次数越多就越精确(maybe0.0)。确定精度是让前后两个答案相减即可。
#include <bits/stdc++.h>
using namespace std;
int main()
{
double a,b,c,d;
cin>>a>>b>>c>>d;
double p=a/b,q=c/d;
double ans=p,pre=0,temp=1;
while(abs(ans-pre)>1e-9)
{
pre=ans;
temp*=(1-p)*(1-q);
ans+=temp*p;
}
cout<<fixed<<setprecision(10)<<ans<<endl;
return 0;
}
本文介绍了一个简单的概率射击游戏问题,两名射手轮流射击目标,先射中者获胜。文章提供了计算射手SmallR胜率的C++代码,并解释了如何通过迭代方法求解此问题,确保计算精度。
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