我推了个式子

#include <cstdio>
using namespace std;
long long n, m, k, ans1, ans2;
int main() {
	scanf("%lld %lld", &n, &m); k = n < m? n : m;
	ans1 = k * n * m - k * (k - 1) * (n + m) / 2 + k * (k - 1) * (2 * k - 1) / 6;
	ans2 = n * (n + 1) * m * (m + 1) / 4 - ans1;
	printf("%lld %lld", ans1, ans2);
	return 0;
}