寻找a和b的互质公因子：
a和b的公因子小于等于gcd（a，b）
然后互质等价于筛gcd内的质数（公因子是gcd的因数）
复杂度O（sqrt（n））
#include<bits/stdc++.h>
#define int long long
#define pb push_back
#define fi first
#define se second
#define bg begin()
#define ed end()
#define rbg rbegin()
#define all(x) x.bg,x.ed
using namespace std;
const long long inf=2e18;
typedef long long ll;
typedef pair<ll,ll>pii;
typedef vector<ll>vi;
const double eps=1e-8;
const long double pi=acos(-1.0);
const int mod=998244353;
const int N=2e5+5;
void solve(){
int a,b;
cin>>a>>b;
int tmp=__gcd(a,b);
int ans=0;
for(int i=2;i<=tmp/i;i++){
if(tmp%i==0){
ans++;
while(tmp%i==0)tmp/=i;
}
}
if(tmp>1)ans++;
cout<<ans+1<<endl;
}
signed main(){
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
//int t;
//cin>>t;
//while(t--){
solve();
//}
return 0;
}
标签：abc142,const,gcd,int,个数,long,互质,define From： https://www.cnblogs.com/Candyk8d9/p/16795455.html