思路：

莫比乌斯反演+整除分块

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
    
     using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb emplace_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
     
      #define pli pair
      
       #define pii pair
       
        #define piii pair
        
         #define pdd pair
         
          #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int N = 5e4 + 10;int sum[N], mu[N], prime[N], cnt;bool not_p[N];void seive() { mu[1] = 1; for (int i = 2; i < N; ++i) { if(!not_p[i]) prime[++cnt] = i, mu[i] = -1; for (int j = 1; j <= cnt && i*prime[j] < N; ++j) { not_p[i*prime[j]] = true; if(i%prime[j] == 0) { mu[i*prime[j]] = 0; break; } mu[i*prime[j]] = -mu[i]; } } for (int i = 1; i < N; ++i) sum[i] = sum[i-1] + mu[i];}inline LL solve(int n, int m) { if(n <= 0 || m <= 0) return 0; LL ans = 0; int up = min(n, m); for (int l = 1, r; l <= up; l = r+1) { r = min(n/(n/l), m/(m/l)); ans += (n/l)*1LL*(m/l)*(sum[r]-sum[l-1]); } return ans;}int t, a, b, c, d, k;int main() { seive(); scanf("%d", &t); while(t--) { scanf("%d %d %d %d %d", &a, &b, &c, &d, &k); printf("%lld\n", solve(b/k, d/k)-solve((a-1)/k, d/k)-solve(b/k, (c-1)/k)+solve((a-1)/k, (c-1)/k)); } return 0;}