BZOJ 3944: Sum

记F[n]为f[i]的前缀和，G[n]为g[i]的前缀和。若g[i]=∑d|n f[i],有F[n]=G[n]-∑F[n/i](i=2...n)

然后存下n^(2/3)的F[i]开个map然后记忆化搜索一下就好了。

若f[i]=φ[i]，G[n]=n(n+1)/2,若f[i]=μ[i],G[n]=1

#include<cstring>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 4000001
#define mm 1000000007
using namespace std;
typedef long long ll;
ll a[maxn],phi[maxn],mo[maxn],mx;
int n,pri[maxn/5],b[maxn],tot;
map<int,ll> mp,mp2;
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)){if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)){x=x*10+ch-'0'; ch=getchar();}
    return x*f; 
}
void init() {
    int n=mx;
    clr(b,0);
    mo[1]=1; phi[1]=1;
    rep(i,2,n) {
        if (!b[i]) pri[++tot]=i,b[i]=1,phi[i]=i-1,mo[i]=-1;
        rep(j,1,tot){
            if (pri[j]*i>mx) break;
            b[i*pri[j]]=1;
            if (i%pri[j]!=0) phi[i*pri[j]]=phi[i]*(pri[j]-1),mo[i*pri[j]]=-mo[i];
            else {
                phi[i*pri[j]]=phi[i]*pri[j]; mo[i*pri[j]]=0;
                break;
            }
        }
    }
    rep(i,1,n) phi[i]+=phi[i-1];
    rep(i,1,n) mo[i]+=mo[i-1];
}
ll dfs(ll n){
    if (n<=mx) return phi[n];
    if (mp.count(n)) return mp[n];
    ll x;
    x=n*(n+1)/2;
    for (ll i=2,pos;i<=n;i=pos+1){
        pos=n/(n/i);
        x-=dfs(n/i)*(pos-i+1);
    }
    return mp[n]=x;
}
ll dfs2(ll n){
    if (n<=mx) return mo[n];
    if (mp2.count(n)) return mp2[n];
    ll x=1;
    for (ll i=2,pos;i<=n;i=pos+1){
        pos=n/(n/i);
        x-=dfs2(n/i)*(pos-i+1);
    }
    return mp2[n]=x;
}
int main(){
    n=read();
    mx=4000000;
    init(); 
    rep(i,1,n){
        ll x=read();
        printf("%lld %lld
",dfs(x),dfs2(x));
    }   
    return 0;
}