运用函数Gcd来求数m和n的最小公倍数

程序如下：

#include<stdio.h>

int main()
{
int m, n, lcm;
printf(“input m n”);
scanf_s(“%d%d”, &m, &n);
lcm = m * n / Gcd(m, n);
printf(“common multiple is %d\n”, lcm);
return 0;
}
int Gcd(int x,int y)
{
int r;
r = x % y;
while (r != 0)
{
x = y;
y = r;
r = x % y;
}

return y;

}

最大公约数：

程序如下：

#include<stdio.h>

int main()
{
int m, n, lcm;
printf(“input m n”);
scanf_s(“%d%d”, &m, &n);
lcm = Gcd(m, n);
printf(“common multiple is %d\n”, lcm);
return 0;
}
int Gcd(int x,int y)
{
int r;
r = x % y;
while (r != 0)
{
x = y;
y = r;
r = x % y;
}

return y;

}