题解:https://blog.csdn.net/lixuepeng_001/article/details/50577932
题意:给定范围1-b和1-d求(i,j)=k的数对的数量
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std; const int MAXN = ;
bool check[MAXN+];
long long mu[MAXN+];
long long a,b,c,d,k;
const int N = 1e6 + ;
long long mob[N], vis[N], prime[N];
int tot;//用来记录prime的个数 void Mobius(int n){ //求得莫比乌斯函数值
memset(prime,,sizeof(prime));
memset(mob,,sizeof(mob));
memset(vis,,sizeof(vis));
tot = , mob[] = ;
for(int i = ; i <=n; i ++){
if(!vis[i]){
prime[tot++] = i;
mob[i] = -;
}
for(int j = ; j < tot && i * prime[j] <=n ; j ++){
vis[i * prime[j]] = ;
if(i % prime[j]) mob[i * prime[j]] = -mob[i];
else{
mob[i * prime[j]] = ;
break;
}
}
}
} int main()
{
int T;
Mobius(N);
cin>>T;
int Case=;
while(T--)
{
cin>>a>>b>>c>>d>>k;
cout<<"Case "<<++Case<<": ";
if(k==){
cout<<""<<endl;
continue;
}
b=b/k;
d=d/k;
long long ans1=;
long long ans=;
for(long long i=;i<=min(b,d);i++)
{
ans+=mob[i]*(b/i)*(d/i);
ans1+=mob[i]*((min(b,d)/i)*(min(b,d)/i));
}
printf("%lld\n",ans-(ans1/));
}
}