Mysterious Light

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探讨在一个由三面等边三角形镜子构成的空间内，一束从特定位置发出的神秘光线如何反射并最终返回起点的问题。通过数学和编程方法，计算光线总行程长度。

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问题 B:

时间限制: 2 Sec 内存限制: 256 MB
提交: 209 解决: 55
[ 提交][ 状态][ 讨论版][命题人: admin]

题目描述

Snuke is conducting an optical experiment using mirrors and his new invention, the rifle of Mysterious Light.

Three mirrors of length N are set so that they form an equilateral triangle. Let the vertices of the triangle be a,b and c.

Inside the triangle, the rifle is placed at the point p on segment ab such that ap=X. (The size of the rifle is negligible.) Now, the rifle is about to fire a ray of Mysterious Light in the direction of bc.

The ray of Mysterious Light will travel in a straight line, and will be reflected by mirrors, in the same ways as "ordinary" light. There is one major difference, though: it will be also reflected by its own trajectory as if it is a mirror! When the ray comes back to the rifle, the ray will be absorbed.

The following image shows the ray's trajectory where N=5 and X=2.

It can be shown that the ray eventually comes back to the rifle and is absorbed, regardless of the values of N and X. Find the total length of the ray's trajectory.

Constraints
2≦N≦1012
1≦X≦N−1
N and X are integers.
Partial Points
300 points will be awarded for passing the test set satisfying N≦1000.
Another 200 points will be awarded for passing the test set without additional constraints.

输入

The input is given from Standard Input in the following format:N X

输出

Print the total length of the ray's trajectory.

样例输入

5 2

样例输出

提示

Refer to the image in the Problem Statement section. The total length of the trajectory is 2+3+2+2+1+1+1=12.

#include<iostream>
#define LL long long
using namespace std;
int main()
{
    LL n,x;
    while(cin>>n>>x)
    {
        LL t=x;
        LL sum=n;
        x=min(t,n-t);//分析可知x于n-x答案相同（相当于旋转了）
        LL y=max(t,n-t);//这里只计算x小的那种情况
        while(x>=1)
        {
            if(y%x==0)//如果y能整除x刚好能射出
            {
                sum+=(y/x)*2*x;
                sum-=x;
                break;
            }
            else{              //否则的话更新反射后的x，y的值。
               sum+=(y/x)*2*x;
               LL te=x;
               x=y%x;
               y=te;
            }
        }
        cout<<sum<<endl;
    }
}