gcd lcm模板

#include<bits/stdc++.h>
int gcd(int a,int b) {   //最大公约数
    return !b?a:gcd(b,a%b);     
}
int lcm(int a,int b){      //最小公倍数
    return a/gcd(a,b)*b;    //由gcd求出
}
int main() {
    int a,b;
    cin >> a >> b;
    cout << gcd(a,b) << endl;
    cout << lcm(a,b) << endl;
    return 0;
}

猜你喜欢

转载自blog.csdn.net/Sclong0218/article/details/83068945