【CF 235E】Number Challenge

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求 $\text{[math]}$ 1 ≤ a, b, c ≤ 2000

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#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;}

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