#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
int a,b,c1,c2;
double ans,x,y;
scanf("%d %d", &a, &b);
if(a<b){
puts("-1");
}
else if(a==b){
printf("%d
", a);
}else{
double x = a - b, y = a + b;
int c1 = x/b/2, c2 = y/b/2;
ans = min(x / c1 / 2, y / c2 / 2);
printf("%.10lf
", ans);
}
return 0;
}
Description
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Sample Input
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Hint
You can see following graphs for sample 1 and sample 3.
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
double x;
double a,b;
while(cin>>a>>b)
{
if(b>a)
{
x=-1;
}
else if(b==a)
{
x=a;
}
else
{
double r=(a+b)/2;
double d=(a-b)/2;
//cout<<r<<" "<<d<<endl;
int rr=r/b,dd=d/b;
while(r/rr<b)
{
rr--;
}
while(d/dd<b)
{
dd--;
}
//cout<<rr<<" "<<dd<<endl;
if((double)r/rr>=(double)d/dd)
{
x=(double)d/dd;
}
else
{
x=(double)r/rr;
}
}
printf("%.12f
",x);
}
return 0;
}