GCD2

传送门：

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
using namespace std;
#define ll long long
#define re register
const int N=5e4+10;
const int M=5e4;
inline void read(int &a)
{
    a=0;
    int d=1;
    char ch;
    while(ch=getchar(),ch>'9'||ch<'0')
        if(ch=='-')
            d=-1;
    a=ch^48;
    while(ch=getchar(),ch>='0'&&ch<='9')
        a=(a<<3)+(a<<1)+(ch^48);
    a*=d;
}
int pri[N],cnt;
ll mu[N];
bool vis[N];
inline void init()
{
    mu[1]=1;
    for(re int i=2;i<=M;i++)
    {
        if(!vis[i])
            pri[++cnt]=i,mu[i]=-1;
        for(re int j=1;j<=cnt&&i*pri[j]<=M;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j]==0)
                break;
            mu[i*pri[j]]=-mu[i];
        }
    }
    for(re int i=1;i<=M;i++)
        mu[i]+=mu[i-1];
}
ll solve(int n,int m,int k)
{
    ll ans=0;
    if(n>m)
        swap(n,m);
    n/=k;
    m/=k;
    for(int l=1,r;l<=n;l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        ans+=(mu[r]-mu[l-1])*(n/l)*(m/l);
    }
    return ans;
}
int main()
{
    init();
    int T;
    read(T);
    while(T--)
    {
        int a,b,c,d,k;
        read(a);
        read(b);
        read(c);
        read(d);
        read(k);
        if(k==0)
        {
            printf("0
");
            continue;
        }
        printf("%lld
",solve(b,d,k)-solve(b,c-1,k)-solve(a-1,d,k)+solve(a-1,c-1,k));
    }
    return 0;
}