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解題思路

　　機房裏的神仙們一萬年前就會的東西拿出來學一學。杜教篩可以在\(O(n^{3/4})\)的時間內積性函數前綴和，做法如下。
　　首先設要求的是\(\sum\limits_{i=1}^n f(i)\)。設\(h=f*g\)，\(S(x)=\sum\limits_{i=1}^nf(i)\)，那麽可以得出\[\sum\limits_{i=1}^nh(i)=\sum\limits_{i=1}^n\sum\limits_{d|n}g(d)f(\frac{n}{d})\]
　　換一下枚舉順序：
　　\[\sum\limits_{i=1}^nh(i)=\sum\limits_{d=1}^ng(d)\sum\limits_{i=1}^{n/d}f(i)\]

代碼

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
 
using namespace std;
const int N=3000005;
typedef long long LL;

template<class T> void rd(T &x){
    x=0; char ch=getchar();
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+ch-‘0‘,ch=getchar(); 
}
template<class T> void out(T x){
    if(!x) return; out(x/10); putchar(‘0‘+x%10);
}
 
int T,prime[N],miu[N],lim=3000000,cnt,sum2[N];
LL sum1[N],n,phi[N],ans1,ans2;
bool vis[N];
map<LL,LL> mp1;
map<int,int> mp2;
 
struct Data{
    LL tmp1,tmp2;   
    Data(LL A=0,LL B=0){
        tmp1=A; tmp2=B; 
    }
};
 
void prework(){
    miu[1]=1; phi[1]=1;
    for(register int i=2;i<=lim;++i){
        if(!vis[i]) prime[++cnt]=i,miu[i]=-1,phi[i]=i-1;
        for(register int j=1;j<=cnt && prime[j]*i<=lim;++j){
            vis[prime[j]*i]=1;
            if(!(i%prime[j])) {
                vis[i*prime[j]]=1;
                phi[i*prime[j]]=phi[i]*prime[j];    
                break;
            }
            miu[i*prime[j]]=-miu[i];
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
    register int i=1;
    for(;i<=lim;i+=16){
        sum1[i]=sum1[i-1]+phi[i];
        sum1[i+1]=sum1[i]+phi[i+1];
        sum1[i+2]=sum1[i+1]+phi[i+2];
        sum1[i+3]=sum1[i+2]+phi[i+3];
        sum1[i+4]=sum1[i+3]+phi[i+4];
        sum1[i+5]=sum1[i+4]+phi[i+5];
        sum1[i+6]=sum1[i+5]+phi[i+6];
        sum1[i+7]=sum1[i+6]+phi[i+7];
        sum1[i+8]=sum1[i+7]+phi[i+8];
        sum1[i+9]=sum1[i+8]+phi[i+9];
        sum1[i+10]=sum1[i+9]+phi[i+10];
        sum1[i+11]=sum1[i+10]+phi[i+11];
        sum1[i+12]=sum1[i+11]+phi[i+12];
        sum1[i+13]=sum1[i+12]+phi[i+13];
        sum1[i+14]=sum1[i+13]+phi[i+14];
        sum1[i+15]=sum1[i+14]+phi[i+15];
        sum2[i]=sum2[i-1]+miu[i];
        sum2[i+1]=sum2[i]+miu[i+1];
        sum2[i+2]=sum2[i+1]+miu[i+2];
        sum2[i+3]=sum2[i+2]+miu[i+3];
        sum2[i+4]=sum2[i+3]+miu[i+4];
        sum2[i+5]=sum2[i+4]+miu[i+5];
        sum2[i+6]=sum2[i+5]+miu[i+6];
        sum2[i+7]=sum2[i+6]+miu[i+7];
        sum2[i+8]=sum2[i+7]+miu[i+8];
        sum2[i+9]=sum2[i+8]+miu[i+9];
        sum2[i+10]=sum2[i+9]+miu[i+10];
        sum2[i+11]=sum2[i+10]+miu[i+11];
        sum2[i+12]=sum2[i+11]+miu[i+12];
        sum2[i+13]=sum2[i+12]+miu[i+13];
        sum2[i+14]=sum2[i+13]+miu[i+14];
        sum2[i+15]=sum2[i+14]+miu[i+15];
    }
    for(;i<=lim;++i) 
        sum1[i]=sum1[i-1]+phi[i],
        sum2[i]=sum2[i-1]+miu[i];
}
 
Data solve(LL x){
    if(x<=lim) {return Data(sum1[x],sum2[x]);}
    if(mp1[x]) {return Data(mp1[x],mp2[x]);}
    Data tmp,nxt; tmp.tmp1=(LL)x*(x+1)/2;
    tmp.tmp2=1;
    for(register LL l=2,r;l<=x;l=r+1){
        r=x/(x/l); nxt=solve(x/l); 
        tmp.tmp1-=(r-l+1)*nxt.tmp1;
        tmp.tmp2-=(r-l+1)*nxt.tmp2;
    }
    mp1[x]=tmp.tmp1; mp2[x]=tmp.tmp2;
    return tmp;
}
 
signed main(){
    rd(T); prework(); Data zz;
    while(T--){
        rd(n); zz=solve(n);
        ans1=zz.tmp1; ans2=zz.tmp2;
        if(!ans1) putchar(‘0‘);
        else out(ans1); putchar(‘ ‘);
        if(!ans2) putchar(‘0‘);
        else {
            if(ans2<0) putchar(‘-‘),ans2=-ans2;
            out(ans2);  
        } putchar(‘\n‘);
    }
    return 0;   
}
 

BZOJ 3944: Sum(杜教篩)