這個題的題意與 【洛谷 P2257 YY的GCD】挺相似的，只不過現在要求去gcd$gcd$為固定值的答案的數量。

由於很像，公式還是很好推的：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
#include<queue>
 

#include<vector>
using namespace std;
#define ll long long int
#define INF 0x3f3f3f3f
const int maxn = 5e4 + 10;
int a, b, d;
bool vis[maxn];
ll sum[maxn];
int prim[maxn];
int mu[maxn], g[maxn];
int cnt;
void get_mu(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!vis[i]) { mu[i] = -1
 
; prim[++cnt] = i; }
        for (int j = 1; j <= cnt && prim[j] * i <= n; j++) {
            vis[i*prim[j]] = 1;
            if (i%prim[j] == 0) break;
            else mu[prim[j] * i] = -mu[i];
        }
    }
    /*for (int j = 1; j <= cnt; j++)
        for (int i = 1; i*prim[j] <= n; i++)
            g[i*prim[j]] += mu[i];*/
 

    for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];

}
int main()
{
    int T;
    scanf("%d", &T);
    get_mu(50000);
    while (T--) {
        scanf("%d %d %d", &a, &b, &d);
        ll res, ans;
        res = min(a, b); ans = 0;
        for (ll l = 1, r; l <= res; l = r + 1) {
            r = min(a / (a / l), b / (b / l));
            ans += (a / (l*d))*(b / (l* d))*(sum[r] - sum[l - 1]);
        }
        printf("%lld\n", ans);
    }
    return 0;
}
//_CRT_SECURE_NO_WARNINGS