[BZOJ3944]SUM 杜教筛

题意

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求 ∑ni=1μ(i) 和 ∑ni=1ϕ(i)

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1.
∑ni=1∑d|iμ(d)=1=∑ni=1∑⌊ni⌋j=1μ(j)

1=∑ni=1Sμ(⌊ni⌋)

Sμ(n)=1−∑ni=2Sμ(⌊ni⌋)

2.
∑ni=1∑d|iϕ(d)=∑ni=1i=∑ni=1∑⌊ni⌋j=1ϕ(j)

n∗(n+1)2=∑ni=1Sϕ(⌊ni⌋)

Sϕ(n)=n∗(n+1)2−∑ni=2Sϕ(⌊ni⌋)

套杜教筛。
题目卡常数有点坑，发现unsigned int要比int快很多。。

#include 
#include 
#include 
#include 
#include 
#include 
#define N 6500010
#define M 100011

using namespace std;

typedef long long ll;
typedef unsigned int uint;

int t,n,p[N],vp[M],vm[M];
ll phi[N],mu[N],rp[M],rm[M];

inline char C(){
    static char buf[100000],*p1=buf,*p2=buf;
    if(p1==p2){
        p2=(p1=buf)+fread(buf,1,100000,stdin);
        if(p1==p2)return EOF;
    }
    return *p1++;
}

inline void reaD(int &x){
    char Ch=C();x=0;
    for(;Ch>'9'||Ch<'0';Ch=C());
    for(;Ch>='0'&&Ch<='9';x=x*10+Ch-'0',Ch=C());
}

inline void first(const int x){
    mu[1]=phi[1]=1;
    ll i,j;
    for(i=2;i<=x;i++){
        if(!p[i]) p[++p[0]]=i,mu[i]=-1,phi[i]=i-1;
        for(j=1;i*p[j]<=x;j++)
            if(p[p[j]*i]=1,i%p[j]) mu[i*p[j]]=-mu[i],phi[i*p[j]]=phi[i]*(p[j]-1);
            else {mu[i*p[j]]=0,phi[i*p[j]]=phi[i]*p[j];break;}
    }
    for(i=1;i<=x;i++)phi[i]+=phi[i-1],mu[i]+=mu[i-1];
}

ll calcphi(uint x){
    if(x<=N-10) return phi[x];
    int nw=n/x;
    if(vp[nw]) return rp[nw];
    ll ret=1ll*x*(x+1)>>1;register uint i,j;
    for(i=2,j;i<=x;i=j+1){
        j=x/(x/i);
        ret-=(j-i+1)*calcphi(x/i);
    }
    vp[nw]=1;rp[nw]=ret;
    return ret;
}

ll calcmu(uint x){
    if(x<=N-10) return mu[x];
    ll ret=1;register uint i,j;
    int nw=n/x;
    if(vm[nw]) return rm[nw];
    for(i=2,j;i<=x;i=j+1){
        j=x/(x/i);
        ret-=(j-i+1)*calcmu(x/i);
    }
    vm[nw]=1;rm[nw]=ret;
    return ret;
}

int main(){
    #ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    #endif
    first(N-10);
    for(reaD(t);t;t--){
        memset(vm,0,sizeof(vm));memset(vp,0,sizeof(vp));
        reaD(n);
        ll Ans1=calcphi(n),Ans2=calcmu(n);;
        printf("%lld %lld\n",Ans1,Ans2);
    }return 0;
}