思路

话不多说，直接推公式吧
$$\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} [gcd(i, j) == d]$$
$$\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} [gcd(i, j) == 1]$$
$$n = \frac{n}{d}, m = \frac{m}{d}$$
$$\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \sum_{k \mid gcd(i, j)} \mu(k)$$
$$\sum_{k = 1} ^{n} \mu(k) \sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{m}{k}}$$
到这里就结束了，只要通过$T \sqrt n$的复杂度即可得到答案

代码

/*
  Author : lifehappy
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 5e4 + 10;

int prime[N], mu[N], cnt;

bool st[N];

void init() {
    mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime[++cnt] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) mu[i] += mu[i - 1];
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T;
    init();
    scanf("%d", &T);
    while(T--) {
        int n, m, d;
        scanf("%d %d %d", &n, &m, &d);
        n /= d, m /= d;
        if(n > m) swap(n, m);
        ll ans = 0;
        for(int l = 1, r; l <= n; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ans += 1ll * (mu[r] - mu[l - 1]) * (n / l) * (m / l);
        }
        printf("%lld\n", ans);
    }
    return 0;
}