Mysterious Light——AT

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

12

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 <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
typedef long long ll;
#define read read()
///const ll inf = 1e15;
///const int maxn = 2e5 + 7;
const ll mod=1e9+7;
const ll inf=0x3f3f3f3f;
const int maxn=1e6+9;
char ss[maxn];
ll n,x;
int main(){
    n=read,x=read;
    ll ans=n;
    ll a=max(x,n-x),b=min(x,n-x);
    while(1){
        ll temp=a/b;
        ll temp1=a%b;
        ans+=2*b*temp;
        if(temp1==0)
            ans-=b;
        a=b,b=temp1;
        if(b==0)
            break;
    }
    printf("%lld\n",ans);
    return 0;
}