从基础不好的角度来进行gcd和lcm函数应用

解题思路:

注意事项:

参考代码:

#include <iostream>

#include <iomanip>

using namespace std;

int gcd(int a, int b) {

//求最大公约数 

int gcdn;

if (b == 0) {   

//如果 3 0 就会直接让a做做大公约数 

return a;

} else {

gcdn = gcd(b, a % b); 

}

return gcdn;

}

int lcm(int a, int b, int gcdn) {

//求最小公倍数 

int lcmn = a * b / gcdn; 

return lcmn;

}

int main() {

int a, b;

cin >> a >> b;

int gcdn = gcd(a, b);

int lcmn = lcm(a, b, gcdn);

cout << gcdn << " " << lcmn;

return 0;

}