题解

直接用__gcd

#include<bits/stdc++.h>
using namespace std;
long long a,b,lcm;
signed main(){
	cin>>a>>b;
    lcm=a*b/__gcd(a,b);
    cout<<lcm;
	return 0;
}

用辗转相除

#include<bits/stdc++.h>
using namespace std;
long long a,b,lcm;
long long gcd(long long a,long long b){
	if(a%b==0) return b;
	return gcd(b,a%b);
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>a>>b;
    lcm=a*b/gcd(a,b);
    cout<<lcm;
	return 0;
}

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