【CF 235E】Number Challenge

求    1 ≤ a, b, c ≤ 2000

#include 
#include 
#include 
#include 
#include 
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)
using namespace std;
typedef long long LL;
const int N = 2005, M = N * N, mod = 1073741824;
int A, B, C, mu[M], pri[M], pz, num[M], ab;
LL f[M], ans; bool c[M];
void Pre() {
	mu[1] = 1;
	Rep(i, 2, C) {
		if (!c[i]) pri[++ pz] = i, mu[i] = -1;
		Rep(j, 1, pz) {
			int k = pri[j] * i;
			if (k > ab) break ;
			c[k] = 1, mu[k] = -mu[i];
			if (i % pri[j] == 0) { mu[k] = 0; break ; }
		}
	}
	Rep(e, 1, C) {
		LL k = 0;
		Rep(d, 1, C) k += C / (e * d);
		for (int g = e; g <= ab; g += e) f[g] += mu[e] * k;
	}
	Rep(g, 1, ab) f[g] %= mod;
}
int main()
{
	scanf ("%d%d%d", &A, &B, &C);
	Rep(i, 1, A) Rep(j, 1, B) num[i * j] ++;
	ab = A * B;
	Pre(); // puts("fin");
	Rep(g, 1, ab) if (num[g]) {
		LL k1 = 0;
		Rep(j, 1, ab / g) (k1 += num[j * g]) %= mod;
		(ans += f[g] * k1) %= mod;
	}
	cout << ((ans + mod) % mod) << endl;

	return 0;
}