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=;
 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=,f=;char ch=getchar();
     while(ch<''||ch>'') {if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 void get_mu(long long x)
 {
     mu[]=;
     for(long long i=;i<=x;i++)
     {
         if(!vis[i]){mu[i]=-;prim[++cnt]=i;}
         for(long long j=;j<=cnt&&i*prim[j]<=x;j++)
         {
             vis[i*prim[j]]=;
             if(i%prim[j]==) break;
             else mu[i*prim[j]]=-mu[i];
         }
     }
     for(long long i=;i<=x;i++) sum[i]=sum[i-]+mu[i];
 }
 long long calc(int a,int b)
 {
     int max_rep=min(a,b);
     long long ret=;
     for(int l=,r;l<=max_rep;l=r+)
     {
         r=min(a/(a/l),b/(b/l));
         ret+=(sum[r]-sum[l-])*(1ll*a/(1ll*l*k))*(1ll*b/(1ll*l*k));
     }
     return ret;
 }
 int main()
 {
     int T;
     T=read();
     get_mu();
     while(T--)
     {
         a=read();b=read();c=read();d=read(),k=read();
         ans=calc(b,d)-calc(b,c-)-calc(a-,d)+calc(a-,c-);
         printf("%lld\n",ans);
     }
     return ;
 }