Codeforces 579 C. A Problem about Polyline(Codeforces Round #320 (Div. 2) )

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.

题目大意：就是给你一个坐标（a，b），看是否在一个锯齿形的折线上，如果有输出最小的x，否则输出-1；解题思路：就是画一下图，然后根据点在折线上，（斜率只有1和-1），设比较的斜率为a/b，然后当k是奇数的时候要加1；然后ans =  (a+b)/k;上代码：

/**2015 - 09 - 22 晚上Author: ITAKMotto:今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。**/#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;typedef long long LL;typedef unsigned long long ULL;const int maxn = 1000;const int mod = 1e9+7;const double eps = 1e-7;struct node{    int x, y, node;} arr[maxn*maxn];bool cmp(node a, node b){    return a.node > b.node;}int main(){    int a, b;    cin>>a>>b;    int k = a/b;    if(a < b)        puts("-1");    else    {        if(k%2 == 1)            k++;        double ans = 1.0*(a+b)/(double)k;        printf("%.10lf\n",ans);    }    return 0;}