数如n, m,输出最小公倍数
辗转相除法
#include <bits/stdc++.h>
using namespace std;
long long lcm(long long a, long long b) {
long long x, y, temp;
if (a > b) {
x = b, y = a;
}
else {
x = a, y = b;
}
temp = y % x;
while (temp > 0) {
y = x;
x = temp;
temp = y % x;
}
return a * b / x;
}
int main() {
long long m, n;
while (cin >> m >> n) {
long long ans = lcm(m, n);
cout << ans << endl ;
}
return 0;
}