BZOJ2301：莫比乌斯反演+二维容斥解决GCD范围计数

这个题是刚才刷的第一道反演题的拓展版，加上一个容斥就可以了

#include<cstdio>
#include<algorithm>
using std::min;
const int maxn=60005;
int cnt,a,b,c,d,k;
long long ans;
bool vis[maxn];
int mu[maxn],sum[maxn];
long long prim[maxn];
inline long long read()
{
    long long x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void get_mu(long long x)
{
    mu[1]=1;
    for(long long i=2;i<=x;i++)
    {
        if(!vis[i]){mu[i]=-1;prim[++cnt]=i;}
        for(long long j=1;j<=cnt&&i*prim[j]<=x;j++)
        {
            vis[i*prim[j]]=1;
            if(i%prim[j]==0) break;
            else mu[i*prim[j]]=-mu[i];
        }
    }
    for(long long i=1;i<=x;i++) sum[i]=sum[i-1]+mu[i];
}
long long calc(int a,int b)
{
    int max_rep=min(a,b);
    long long ret=0;
    for(int l=1,r;l<=max_rep;l=r+1)
    {
        r=min(a/(a/l),b/(b/l));
        ret+=(sum[r]-sum[l-1])*(1ll*a/(1ll*l*k))*(1ll*b/(1ll*l*k));
    }
    return ret;
}
int main()
{
    int T;
    T=read();
    get_mu(50000);
    while(T--)
    {
        a=read();b=read();c=read();d=read(),k=read();
        ans=calc(b,d)-calc(b,c-1)-calc(a-1,d)+calc(a-1,c-1);
        printf("%lld\n",ans);
    }
    return 0;
}