Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] C. A Problem about Polyline 精度控制

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.

Sample test(s)

input

3 1

output

1.000000000000

input

1 3

output

-1

input

4 1

output

1.250000000000

Note

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

由于是斜率为1，所以，a - b a + b都在x轴上，只要满足不等式x 为要求的结果

x * k = (a - b) / 2; x * k = (a+b)/2; x >=b;即可。

先求出k即为(a - b) /2 / b ;(a + b) /2 / b ;x也就求出来了。

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000007
int a,b;
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
     while(scanf("%d%d",&a,&b)!=EOF)
    {
        if(a < b){
            printf("-1\n");
            continue;
        }
        if(a == b){
            printf("%d\n",b);
            continue;
        }
        double x1 = (double)(a - b) / 2.0 /((a - b) / 2 / b);
        double x2 = (double)(a + b) / 2.0 /((a + b) / 2 / b);
        //cout<<min(x1,x2)<<endl;
        printf("%.12f\n",min(x1,x2));
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}