1 /*** 2 对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数 3 **/ 4 #include <iostream> 5 #include <cstdio> 6 #include <algorithm> 7 8 using namespace std; 9 const int maxn = 50010; 10 int isprime[maxn],prime[maxn],mu[maxn],sum[maxn]; 11 12 int t,a,b,c,d,k; 13 int cnt; 14 void mobius(int n){ 15 int i,j; 16 cnt =0; 17 mu[1] =1; 18 for(i=2;i<=n;i++){ 19 if(!isprime[i]){ 20 prime[cnt++] = i; 21 mu[i] = -1; 22 } 23 for(j=0;j<cnt&&i*prime[j]<=n;j++){ 24 isprime[i*prime[j]] = 1; 25 if(i%prime[j]) 26 mu[i*prime[j] ] = -mu[i]; 27 else{ 28 mu[i*prime[j]] = 0; 29 break; 30 } 31 } 32 } 33 } 34 35 long long solve(int n,int m){ 36 int i,la; 37 long long ret =0; 38 if(n>m) 39 swap(n,m); 40 for(i=1,la=0;i<=n;i=la+1){ 41 la = min(n/(n/i),m/(m/i)); 42 ret += (long long )(sum[la]-sum[i-1])*(n/i)*(m/i); 43 } 44 return ret; 45 } 46 47 int main(){ 48 int i,j; 49 mobius(50000); 50 for(i=1;i<50000;i++) 51 sum[i] = sum[i-1]+mu[i]; 52 scanf("%d",&t); 53 while(t--){ 54 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); 55 long long ans; 56 ans = solve(b/k,d/k)-solve((a-1)/k,d/k) 57 -solve((c-1)/k,b/k)+solve((a-1)/k,(c-1)/k); 58 printf("%lld ",ans); 59 } 60 return 0; 61 }