传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2005

令F(i)表示i | gcd(x, y)的对数，f(i)表示gcd(x, y) = i的对数，那么显然

f(i) = F(i) - f(2 * i) - f(3 * i) - ...

F(i)很好求，显然，F(i) = (n / i) * (m / i)，除法向下取整。

之后就简单了~

#include 
     
      #include 
      
       const int maxn = 100005;int n, m, lmt;long long f[maxn], ans;int main(void) {	scanf("%d%d", &n, &m);	lmt = std::min(n, m);	for (int i = lmt; i; --i) {		f[i] = (long long)(n / i) * (long long)(m / i);		for (int j = i << 1; j <= lmt; j += i) {			f[i] -= f[j];		}		ans += f[i] * ((i << 1) - 1);	}	printf("%lld\n", ans);	return 0;}