[Linear sieve of common integrable functions] "learning notes"

"Euler function"

φ (n) =n (1-1/P1) (1-1/P2) ... (1-1/PK)

Easy to find P[j]|i when Phi[i*p[j]]=phi[i]*p[j] because Phi[i] n is n*p[j]/p[j], the rest of the same

Proof: http://www.cnblogs.com/candy99/p/6200660.html

voidsieve () {phi[1]=1;  for(intI=2; i<=n;i++){        if(!Vis[i]) {p[++m]=i; Phi[i]=i-1; }         for(intj=1; j<=m&&i*p[j]<=n;j++) {vis[i*p[j]]=1; if(i%p[j]==0) {Phi[i*p[j]]=phi[i]*P[j];  Break; } phi[i*p[j]]=phi[i]* (p[j]-1); }    }     for(intI=1; i<=n;i++) s[i]=s[i-1]+phi[i];}

"About a few numbers"

According to the multiplication principle, the approximate number of n is ∏{I=1...R} (ai+1)

facnum[i*p[j]]=facnum[i]/(minfac[i]+1) * (minfac[i*p[j]]+1) When P[j]|i is available from the upper formula

"Approximate and"

N=p1^a1*p2^a2*...*pr^ar
Then its approximate and =∏{I=1...R} (Σ{J=1..AJ}PI^J)
P[j]|i, get Sumfac[i*p[j]]
Need to divide by σ (p[j]^ (0..minfac[i]))
multiplied by σ (pris[j]^ (0..minfac[k])
where minfac[k]=minfac[i]+1
This opens two auxiliary array records
T1[i]=σ (minfac[i]^ (0..a[minfac[i]))
T2[i]=mindiv[i]^a[minfac[i]]

int notp[n],p[n],mu[n],minfac[n],t1[n],t2[n],sf[n];

voidsieve () {mu[1]=1; sf[1].s=1;  for(intI=2; i<n;i++){        if(!Notp[i]) {p[++p[0]]=i,mu[i]=-1; Minfac[i]=i; Sf[i]=i+1; T1[i]=i+1; T2[i]=i; }         for(intj=1, k;j<=p[0]&& (K=i*p[j]) <n;j++) {Notp[i*p[j]]=1; MINFAC[K]=P[j]; if(i%p[j]==0) {Mu[i*p[j]]=0; T2[K]=t2[i]*P[j]; T1[K]=t1[i]+T2[k]; SF[K]=sf[i]/t1[i]*T1[k];  Break; } mu[i*p[j]]=-Mu[i]; T1[K]=1+P[j]; T2[K]=P[j]; SF[K]=sf[i]*Sf[p[j]]; }    }}

[Linear sieve of common integrable functions] "learning notes"