BZOJ 3944: Sum

Description

求\(\sum_{i=1}^n\varphi(i),\sum_{i=1}^n\mu(i),n\leqslant 2\times 10^9\)

Solution

杜教筛...

杜教筛

总之杜教筛就是通过这样一个式子来求积性函数前缀和\(S(n)\)

因为\(\sum_{i=1}^n\sum_{d\mid i}g(d)f(\frac{i}{d})=\sum_{i=1}^ng(i)S(\lfloor\frac{n}{i}\rfloor)\)

同时将等式右边\(g(1)S(n)\)提出来就可以得到\(g(1)S(n)=\sum_{i=1}^n\sum_{d\mid i}g(d)f(\frac{i}{d})-\sum_{i=2}^ng(i)S(\lfloor\frac{n}{i}\rfloor)\)

如果能够直接算出来前面的式子,后面的式子可以分块...加上记忆化...可以做到\(O(n^{\frac{2}{3}})\)的复杂度...不太会证明...

对于\(\varphi(n)\),有一个式子...\(\sum_{d\mid n}\varphi(d)=n\)...那么令\(g(n)=1\)即可...

\(\sum_{i=1}^n\sum_{d\mid i}\varphi(d)=\sum_{i=1}^n i=\frac{n(n+1)}{2}\)

对于\(\mu(n)\),有一个式子...\(\sum_{d\mid n}\mu(d)=[n=1]\)...所以依然令\(g(n)=1\)...

那么\(\sum_{i=1}^n\sum_{d\mid i}\mu(d)=1\)...

没了...

Code

/**************************************************************
    Problem: 3944
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:7680 ms
    Memory:56128 kb
****************************************************************/
 
#include <bits/stdc++.h>
using namespace std;
 
#define mpr make_pair
 
typedef long long ll;
typedef pair<ll,ll> pr;
const int N = 2000500;
 
int b[N],pm[N],cp;
int mu[N],phi[N],sm[N];
ll sp[N];
 
void pre(int n) {
    for(int i=2;i<=n;i++) {
        if(!b[i]) pm[++cp]=i,mu[i]=-1,phi[i]=i-1;
        for(int j=1;j<=cp && i*pm[j]<=n;j++) {
            b[i*pm[j]]=1;
            if(i%pm[j]) mu[i*pm[j]]=-mu[i],phi[i*pm[j]]=phi[i]*(pm[j]-1);
            else { phi[i*pm[j]]=phi[i]*pm[j];break; }
        }
    }
    mu[1]=phi[1]=1;
    for(int i=1;i<=n;i++) sm[i]=sm[i-1]+mu[i],sp[i]=sp[i-1]+phi[i];;
     
}
map<ll,pr> mp;
 
pr operator - (const pr &a,const pr &b) { return mpr(a.first-b.first,a.second-b.second); }
pr operator * (const pr &a,const ll b) { return mpr(a.first*b,a.second*b); }
pr S(ll n) {
    if(n<=2000000) return mpr(sm[n],sp[n]);
    if(mp.count(n)) return mp[n];
    ll f1=1,f2;
    if(n&1) f2=n*((n+1)>>1);else f2=(n>>1)*(n+1);
    pr fn=mpr(f1,f2);
    for(ll i=2,j;i<=n;i=j+1) {
        j=n/(n/i);
        fn=fn-S(n/i)*(j-i+1);
    }return mp[n]=fn;
}
 
int main() {
    int T;ll n;
    for(pre(2000000),scanf("%d",&T);T--;) {
        scanf("%lld",&n);
        pr ans=S(n);
        printf("%lld %lld\n",ans.second,ans.first);
    }return 0;
}