洛谷 P3455 [POI2007]ZAP-Queries || 洛谷P2522,bzoj2301

https://www.luogu.org/problemnew/show/P3455

就是https://www.cnblogs.com/hehe54321/p/9315244.html里面的方法2了，升级版的整除分块，可以两个变量一起搞

预处理莫比乌斯函数的前缀和之后就可以每次$O(\sqrt{n}+\sqrt{m})$回答

那篇题解里面用了一个技巧：${\lfloor}\frac{{\lfloor}\frac{a}{b}{\rfloor}}{c}{\rfloor}={\lfloor}\frac{a}{bc}{\rfloor}$

（当然a,b,c都为正整数）

证了好久。。。

这么证：

设${\lfloor}\frac{a}{bc}{\rfloor}=p$，则p为整数，且$p<=\frac{a}{bc}<p+1$

则$pc<=\frac{a}{b}<pc+c$

而$pc$与$pc+c$都为整数

因此$pc<={\lfloor}\frac{a}{b}{\rfloor}<pc+c$

所以$p<=\frac{{\lfloor}\frac{a}{b}{\rfloor}}{c}<p+1$

所以${\lfloor}\frac{{\lfloor}\frac{a}{b}{\rfloor}}{c}{\rfloor}=p={\lfloor}\frac{a}{bc}{\rfloor}$

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pii;
 #define N 50100
 ll prime[N+],len,mu[N+],dd[N+];
 bool nprime[N+];
 ll a,c,n,m,k,ans,ed;
 int main()
 {
     ll i,j,T,TT;
     mu[]=;
     for(i=;i<=N;i++)
     {
         if(!nprime[i])    prime[++len]=i,mu[i]=-;
         for(j=;j<=len&&i*prime[j]<=N;j++)
         {
             nprime[i*prime[j]]=;
             if(i%prime[j]==)    {mu[i*prime[j]]=;break;}
             else    mu[i*prime[j]]=-mu[i];
         }
     }
     for(i=;i<=N;i++)    dd[i]=dd[i-]+mu[i];
     scanf("%lld",&T);
     for(TT=;TT<=T;TT++)
     {
         scanf("%lld%lld%lld",&n,&m,&k);n/=k;m/=k;
         ans=;
         if(n>m)    swap(n,m);
         for(i=;i<=n;i=j+)
         {
             j=min(n,min(n/(n/i),m/(m/i)));
             ans+=(dd[j]-dd[i-])*(n/i)*(m/i);
         }
         printf("%lld\n",ans);
     }
     return ;
 }

---

https://www.luogu.org/problemnew/show/P2522

https://www.lydsy.com/JudgeOnline/problem.php?id=2301

这题基本一样的，就是加个容斥。。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pii;
 #define N 50100
 ll prime[N+],len,mu[N+],dd[N+];
 bool nprime[N+];
 ll a,c,n,m,k;
 ll calc(ll n,ll m)
 {
     if(n==||m==)    return ;
     ll ans=;
     if(n>m)    swap(n,m);
     n/=k;m/=k;
     for(ll i=,j;i<=n;i=j+)
     {
         j=min(n,min(n/(n/i),m/(m/i)));
         ans+=(dd[j]-dd[i-])*(n/i)*(m/i);
     }
     return ans;
 }
 int main()
 {
     ll i,j,T,TT;
     mu[]=;
     for(i=;i<=N;i++)
     {
         if(!nprime[i])    prime[++len]=i,mu[i]=-;
         for(j=;j<=len&&i*prime[j]<=N;j++)
         {
             nprime[i*prime[j]]=;
             if(i%prime[j]==)    {mu[i*prime[j]]=;break;}
             else    mu[i*prime[j]]=-mu[i];
         }
     }
     for(i=;i<=N;i++)    dd[i]=dd[i-]+mu[i];
     scanf("%lld",&T);
     for(TT=;TT<=T;TT++)
     {
         scanf("%lld%lld%lld%lld%lld",&a,&n,&c,&m,&k);
         printf("%lld\n",calc(n,m)-calc(a-,m)-calc(n,c-)+calc(a-,c-));
     }
     return ;
 }