HYSBZ 2301

 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\n",ans);

59     }

60     return 0;

61 }

HYSBZ 2301