time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vasya is studying number theory. He has denoted a function f(a, b) such that:

f(a, 0) = 0;
f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b.
Vasya has two numbers x and y, and he wants to calculate f(x, y). He tried to do it by himself, but found out that calculating this function the way he wants to do that might take very long time. So he decided to ask you to implement a program that will calculate this function swiftly.

Input
The first line contains two integer numbers x and y (1 ≤ x, y ≤ 1012).

Output
Print f(x, y).

Examples
input
3 5
output
3
input
6 3
output
1

題目大意：已知f(a,0)=0，f(a,b)=1+f(a,b−gcd(a,b))
給出x,y，求f(x,y)
解題思路：設x=A1gcdold,y=B1gcdold，若A1,B1互質，則繼續減gcdold，直至y=(B1−m)gcdold，此時A1和B1−m有公因數k（m=B1%

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
 

using namespace std;
typedef long long LL;
const LL INF=1e18;
set<LL> s;

LL gcd(LL a,LL b) {return b==0?a:gcd(b,a%b);}

void getfactors(LL x)
{
    s.clear();
    LL rx=sqrt(x);
    for(int i=2;i<=rx;i++)
    {
        if(x%i==0)
        {
            x/=i;
            s.insert(i);
            while(x%i==0) x/=i;
        }
    }
    if(x>1) s.insert(x);
}

int main()
{
    LL x,y;
    while(scanf("%lld%lld",&x,&y)!=EOF)
    {
        getfactors(x);
        LL ans=0,m;
        while(y)
        {
            LL g=gcd(x,y);
            x/=g;y/=g;
            m=y;
            for(auto p:s)
                if(x%p==0)  m=min(m,y%p);
            ans+=m;
            y-=m;
        }
        printf("%lld\n",ans);
    }
    return 0;
}
//2000000018 2000000017
//1 1000000000000
//3 135415909531