BZOJ 4805: 欧拉函数求和

Description

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

Solution

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Code

/**************************************************************
    Problem: 4805
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:1100 ms
    Memory:48172 kb
****************************************************************/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
const int N = 2000050;
const ll p = 1000000007;

ll Pow(ll a,ll b,ll r=1) { for(;b;b>>=1,a=a*a%p) if(b&1) r=r*a%p;return r; }
ll mul(ll a,ll b) { return (a*b-((ll)((ld)a/p*b+1e-3))*p+p)%p; }

int b[N],pr[N],cp;
ll phi[N],sp[N],inv2=Pow(2,p-2);

void pre(int n) {
    for(int i=2;i<=n;i++) {
        if(!b[i]) pr[++cp]=i,phi[i]=i-1;
        for(int j=1;j<=cp && i*pr[j]<=n;j++) {
            b[i*pr[j]]=1;
            if(i%pr[j]) phi[i*pr[j]]=phi[i]*(pr[j]-1);
            else { phi[i*pr[j]]=phi[i]*pr[j];break; }
        }
    }phi[1]=1;
    for(int i=1;i<=n;i++) sp[i]=(sp[i-1]+phi[i]);
}
map<ll,ll> mp;
ll S(ll n) {
    if(n<=2000000) return sp[n];
    if(mp.count(n)) return mp[n];
    ll fn;
    if(n&1) fn=n*((n+1)>>1);
    else fn=(n>>1)*(n+1);
    for(ll i=2,j;i<=n;i=j+1) {
        j=n/(n/i);
        fn=(fn-(j-i+1)*S(n/i));
    }return mp[n]=fn;
}

int main() {
    pre(2000000);
    ll n;
    scanf("%lld",&n);
    printf("%lld\n",S(n));
    return 0;
}