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3944: Sum[杜教筛]

3944: Sum

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 3471  Solved: 946
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Description

 

Input

一共T+1行
第1行为数据组数T(T<=10)
第2~T+1行每行一个非负整数N,代表一组询问
 

 

 

Output

一共T行,每行两个用空格分隔的数ans1,ans2
 

 

Sample Input

6
1
2
8
13
30
2333

Sample Output

1 1
2 0
22 -2
58 -3
278 -3
1655470 2

HINT

 

Source

 

 
#include<cstdio>#include<cstring>typedef long long ll;typedef unsigned int uint;using namespace std;const int N=5.4e6+5;const int M=1e5+10;ll mu[N],phi[N];ll ans_mu[M],ans_phi[M];bool vis_mu[M],vis_phi[M];int n,m,T;ll ans1,ans2;int tot,prime[N/3];bool check[N];inline void sieve(){    phi[1]=mu[1]=1;    m=N-5;//    m=pow(n,2.0/3.0);    for(int i=2;i<=m;i++){        if(!check[i]) prime[++tot]=i,mu[i]=-1,phi[i]=i-1;        for(int j=1;j<=tot&&i*prime[j]<=m;j++){            check[i*prime[j]]=1;            if(!(i%prime[j])){mu[i*prime[j]]=0;phi[i*prime[j]]=phi[i]*prime[j];break;}            else mu[i*prime[j]]=-mu[i],phi[i*prime[j]]=phi[i]*(prime[j]-1);        }    }    for(int i=1;i<=m;i++) mu[i]+=mu[i-1],phi[i]+=phi[i-1];}inline ll get_phi(uint now){    if(now<=m) return phi[now];    int nn=n/now;    if(vis_phi[nn]) return ans_phi[nn];    vis_phi[nn]=1;    ll sav=(ll)now*(now+1)>>1;    for(uint i=2,pos;i<=now;i=pos+1){        pos=now/(now/i);        sav-=get_phi(now/i)*(pos-i+1);    }    return ans_phi[nn]=sav;}inline ll get_mu(uint now){    if(now<=m) return mu[now];    int nn=n/now;    if(vis_mu[nn]) return ans_mu[nn];    vis_mu[nn]=1;    ll sav=1;    for(uint i=2,pos;i<=now;i=pos+1){        pos=now/(now/i);        sav-=get_mu(now/i)*(pos-i+1);    }    return ans_mu[nn]=sav;}inline void clr(){    memset(vis_mu,0,sizeof vis_mu);    memset(vis_phi,0,sizeof vis_phi);}int main(){    sieve();    for(scanf("%d",&T);T--;clr()){        scanf("%d",&n);        ans1=get_phi(n);ans2=get_mu(n);        printf("%lld %lld\n",ans1,ans2);    }    return 0;}

 

3944: Sum[杜教筛]