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

解决复杂几何问题：寻找特定多边形通过点的最小x值

最新推荐文章于 2020-04-03 11:37:48 发布

原创最新推荐文章于 2020-04-03 11:37:48 发布 · 432 阅读

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本文探讨了一个涉及计算特定多边形通过指定点的最小x值的问题，该多边形由一系列点定义。通过提供输入（点的坐标a和b），本文提供了解决方案并解释了如何使用数学公式来找到答案。包括实例输入输出以展示解题过程。

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.

$\text{[math]}$ $\text{[math]}$

刚开始一看题目，感觉很难，其实仔细想想并不是很难
根据点在直线上，可以得到x=(a+b)/2*k
又因为x<=b,
所以解得2*k<=(a+b)/b;
当22*k为奇数是减一，为偶数时不变

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include<queue>
#include<cmath>
using namespace std;
int main()
{
    int a,b;
    while(scanf("%d%d",&a,&b)==2)
    {
        int t=a/b+1;
        if(t<=1) printf("-1\n");
        else
        {
            if(t%2==1) t--;
            double ans=(double)(a+b)/(double)t;
            printf("%.12lf\n",ans);
        }
    }
    return 0;
}