bzoj3944 Sum（杜教筛）

杜教筛裸题。。。
mu函数的前缀和怎么搞呢，我们有 μ∗1=ϵ
令g=1就好了。

#include 
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 2000010
#define uint unsigned int
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int n,prime[N>>3],tot=0;ll phi[N],mu[N],s1[10000],s2[10000];
bool notprime[N];
inline void init(){
    notprime[1]=1;phi[1]=1;mu[1]=1;
    for(int i=2;i<=2e6;++i){
        if(!notprime[i]) prime[++tot]=i,mu[i]=-1,phi[i]=i-1;
        for(int j=1;prime[j]*i<=2e6;++j){
            notprime[prime[j]*i]=1;
            if(i%prime[j]==0){
                mu[prime[j]*i]=0;phi[prime[j]*i]=phi[i]*prime[j];break;
            }phi[prime[j]*i]=phi[prime[j]]*phi[i];mu[prime[j]*i]=-mu[i];
        }
    }for(int i=2;i<=2e6;++i) mu[i]+=mu[i-1],phi[i]+=phi[i-1];
}
inline ll calc1(uint x){
    if(x<=2e6) return phi[x];
    if(s1[n/x]) return s1[n/x];
    ll res=(ll)x*(x+1)>>1;
    for(uint i=2,last;i<=x;i=last+1){
        last=x/(x/i);
        res-=calc1(x/i)*(last-i+1);
    }return s1[n/x]=res;
}
inline ll calc2(uint x){
    if(x<=2e6) return mu[x];
    if(s2[n/x]) return s2[n/x];
    ll res=1;
    for(uint i=2,last;i<=x;i=last+1){
        last=x/(x/i);
        res-=calc2(x/i)*(last-i+1);
    }return s2[n/x]=res;
}
int main(){
//  freopen("a.in","r",stdin);
    int tst=read();init();
    while(tst--){
        n=read();memset(s1,0,sizeof(s1));memset(s2,0,sizeof(s2));
        printf("%lld %lld\n",calc1(n),calc2(n));
    }return 0;
}