SPOJDIVCNT2: Counting Divisors(莫比乌斯反演)

http://acm.tzc.edu.cn/acmhome/vProblemList.do?method=problemdetail&oj=SPOJ&pid=DIVCNT2

给出n求

其中是除数函数,0代表0次方.

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define ll long long
 ll n,sig[],N,a[];
 int p[];
 bool mark[];
 char mul[];
 int smu[],L;
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 ll Read(){
     ll t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void Init(int n){
     sig[]=;smu[]=;mul[]=;
     for (int i=;i<=n;i++){
         if (!mark[i]){
             p[++p[]]=i;
             mul[i]=-;
             smu[i]=;
             sig[i]=;
         }
         for (int j=;j<=p[]&&p[j]*i<=n;j++){
             mark[p[j]*i]=;
             if (i%p[j]==){
                 sig[p[j]*i]=sig[i]/(smu[i]+)*(smu[i]+);
                 smu[p[j]*i]=smu[i]+;
                 mul[p[j]*i]=;
                 break;
             }
             sig[p[j]*i]=sig[i]*;
             smu[p[j]*i]=;
             mul[p[j]*i]=-mul[i];
         }
     }
     for (int i=;i<=n;i++) sig[i]+=sig[i-],smu[i]=smu[i-]+std::abs((int)mul[i]);
 }

 ll R(ll x){
     if (x<=L) return sig[x];
     ll res=;
     for (register ll i=,j;i<=x;i=j+){
         j=(x/(x/i));
         res+=(j-i+)*(x/i);
     }
     return res;
 }
 ll Smu(ll x){
     if (x<=L) return smu[x];
     ll res=;
     for (register ll i=;i*i<=x;i++)
      if (mul[i]) res+=mul[i]*(x/(i*i));
     return res;
 }
 void solve(ll n){
     int m=sqrt(n);
     ll ans=;
     ll pre=smu[m],tt;
     for (int i=;i<=m;i++) if (mul[i]) ans+=R(n/i);
     for (ll i=m+,j;i<=n;i=j+){
         j=(n/(n/i));
         tt=Smu(j);
         ans+=(tt-pre)*(R(n/i));
         pre=tt;
     }
     printf("%lld\n",ans);
 }
 int main(){
     int T=read();
     ll mx=;
     N=;
     for (int i=;i<=T;i++)
      a[i]=Read(),mx=std::max(mx,a[i]);
     if (mx<=) L=;else L=pow(N,2.0/3.0);
     Init(L);
     for (int i=;i<=T;i++){
         solve(a[i]);
     }
 }