Changchun University fourteenth Programming Contest (to reproduce the race) B

B Bowling Game

Topic links: https://ac.nowcoder.com/acm/contest/912/B

topic

CUST players after the game finished province, with a small r we go bowling.

Bowling is a very high degree of difficulty of the game, but this is simply difficult to live school team members, they all are very powerful ( fried and ) out of 10 bottles are inverted. Especially small r, eyes closed once every time to throw down 10 bottles. Among them there is not so much a sewer players, every time the ball thrown into the sewers, causing a ball not hit the bottle are in.

 

 

Rounds down, we find back the ball less and less, and finally left a few numbers 9 ball. They do not like to lose 9-ball, because too light.

After you work for little sister, that: let's Songjiang bowling club technology is not so advanced, so the background is manually operated to retrieve the ball, the ball did not come back now, cause fewer ball is a ball stuck, dropped into sewer it could lead to this situation now.

School team members know are a few, they are at least per person fried and once, only one sewer players. . .

We learned that the background is a square box, about this time will be stuck Bowling, shown in blue area ${\mathrm{S\; 1}}^{}$

 

    Entry

Input common line by s1, s2 two positive integers, ${\mathrm{S\; .\; 1}}^{}$ and ${\mathrm{S\; 2}}^{}$ is a drawing area to ensure $\mathrm{S\; .\; 1\; ,\; S\; 2\; \le \; 10\; .\; 9}$ and graphical method) 

Output

输出一行，即保龄球的直径 D。

你的答案与标准答案误差在±0.001范围以内都算正确。

样例
input 
6 25
693 2853
output
2
21.586519

思路

将s1分为三个小三角形，两个直角边的边长之和就是sqrt(4*s1+s2),斜边边长就是sqrt(s2),列一元二次方程即可

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int s1,s2;
    while(cin>>s1>>s2)
    {
        double len=sqrt(s2+4*s1);
       
        double _1=(len+sqrt(len*len-8*s1))/2;
       
        double _2=(len-sqrt(len*len-8*s1))/2;
       
        double xie=sqrt(s2);
        
        double sum=xie+_1+_2;
      
        cout<<fixed<<setprecision(6)<<(4*s1)/sum<<endl;
 
    }
    return 0;
}