[Codeforces Round #320 (Div. 2) C. A Problem about Polyline （数学）

[Codeforces Round #320 (Div. 2) C. A Problem about Polyline （数学）

C. A Problem about Polyline

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

Copy

3 1

output

Copy

1.000000000000

input

Copy

1 3

output

Copy

-1

input

Copy

4 1

output

Copy

1.250000000000

Note

You can see following graphs for sample 1 and sample 3.

思路：

当(a<b)时，不存在答案，输出“-1”，下面考虑有答案的情况。

如果点((a,b))在折线的上坡上，那么折线一定经过点((a-b,0))，如果在折线的下坡上，那么折线一定经过点((a+b,0))。设$mathit c=(a+b)or(a-b) $

我们要求解的答案(mathit x)，一定满足这两个条件：

1️⃣、(x/(2*x))为正数，因为(2*x)是折线的最小周期，而((c,0))又是一个周期的结尾点。

2️⃣、(b leq x) ，这也是显然的。

设(y=c/(2*x))代表到点$(c,0) (时经过了多少个最小周期，显然当)mathit y$ 最大时，(mathit x)最小。

则(x=c/(2*y) ,bleq x) ，则答案为：

代码：

int a, b;
int main()
{
#if DEBUG_Switch
    freopen("C:\code\input.txt", "r", stdin);
#endif
    //freopen("C:\code\output.txt","r",stdin);
    cin >> a >> b;
    if (a < b)
    {
        cout << -1 << endl;
    } else
    {
        double ans = a + b;
        ans = ans / (2.0 * (int)(ans / (2.0 * b)));
        cout << fixed << setprecision(9) << ans << endl;
    }
    return 0;
}