询问拆成四个，就像矩阵数点一样。
每一个询问的形式为 $\sum_{i=1}^n\sum_{j=1}^m[(i,j)==k]$。
$$\sum_{i=1}^n\sum_{j=1}^m[(i,j)==k]=\sum_{i=1}^{\lfloor \frac{n}{k} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{k} \rfloor}[(i,j)==1]$$
把 $\lfloor \dfrac{n}{k} \rfloor$ 换成 $n$，把 $\lfloor \dfrac{m}{k} \rfloor$ 换成 $m$
$$\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|(i,j)}\mu(d)=\sum_d \mu(d)\sum_{i=1}^{n}[d|i]\sum_{j=1}^{m}[d|j]$$
$$=\sum_d\mu(d)\lfloor \dfrac{n}{d}\rfloor\lfloor \dfrac{m}{d}\rfloor$$
求个 $\mu$ 的前缀和加数论分块。

#include <bits/stdc++.h>

const int N = 5e4 + 7;
int prime[N], prin, mu[N];
bool vis[N];

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

int solve(int n, int m) {
    int res = 0;
    for (int i = 1, j; i <= std::min(n, m); i = j + 1) {
        j = std::min(n / (n / i), m / (m / i));
        res += (mu[j] - mu[i - 1]) * (n / i) * (m / i);
    }
    return res;
}

int main() {
    init();
    int n;
    scanf("%d", &n);
    while (n--) {
        int a, b, c, d, k;
        scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
        printf("%d\n", solve(b / k, d / k) - solve(b / k, (c - 1) / k) - solve((a - 1) / k, d / k) + solve((a - 1) / k, (c - 1) / k));
    }
    return 0;
}

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