【模板】最小公倍数LCM

#include <cstdio>using namespace std;#define ll long longll a,b;ll gcd(ll a,ll b){    ll t;    while (b!=0)    {        t=b;        b=a % b;        a=t;    }    return a;}ll lcm(ll a,ll b){    return a*b / gcd(a,b);}int main(){    scanf("%lld%lld",&a,&b);    printf("%lld",lcm(a,b));    return 0;}

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