One line of code to solve the greatest common divisor and least common multiple

int m, n; //Two integers
int mmd; //Greatest common divisor
int mmu; //Least common multiple
while((n%=m)&&(m%=n)); //One line of code refers to this The most important line in a row
mmu=n+m;//The least common multiple, after the euclidean division, one of m and n must be 0, and the one that is not zero is the least common multiple
mmd=(n*m)/mmu;//Maximum The common divisor is equal to the lowest common multiple of two numbers multiplied together.

Here is the code that works:

#include <stdio.h>
int main() { 
    int m,n;
    int m1,n1,rem;
    int mmd,mmu;
    scanf ("%d%d", &m, &n); 
    m1=m;
    n1=n;
    if(n1)
        while((m1 %= n1) && (n1 %= m1));
    mmu = m1+n1;
    mmd = m*n/mmu;
    printf ("最大公约数：%d\n", mmu);
    printf ("最小公倍数：%d", mmd);
    return 0;
}