BZOJ 4804: 欧拉心算

Description

求\(\sum_{i=1}^n\sum_{i=1}^n\varphi(gcd(i,j)),T\leqslant 5\times 10^3,n\leqslant 10^7\)

Solution

数论分块+莫比乌斯反演.

化式子

\(\sum_{i=1}^n\sum_{i=1}^n\varphi(gcd(i,j))\)
\(=\sum_d\varphi(d)\sum_{i=1}^n\sum_{j=1}^n[(i,j)=d]\)
\(=\sum_d\varphi(d)(\sum_{i=1}^{\lfloor \frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \frac{n}{d}\rfloor}[(i,j)=1])\)
\(=\sum_d\varphi(d)(\sum_{p}\mu(p)\sum_{i=1}^{\lfloor \frac{n}{pd}\rfloor}\sum_{j=1}^{\lfloor \frac{n}{pd}\rfloor})\)
\(\text{Let T=pd}\)
\(=\sum_{T}\lfloor \frac{n}{T}\rfloor\lfloor \frac{n}{T}\rfloor\sum_{p}\mu(p)\varphi(\frac{T}{p})\)

因为积性函数的狄利克雷前缀和也是积性函数，并且因为\(\mu\)的存在这个式子还是很好筛的.

Code

/**************************************************************
    Problem: 4804
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:4272 ms
    Memory:128240 kb
****************************************************************/
 
#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
const int N = 10000050;
 
int pr[N],cp;
bool b[N];
LL f[N];
 
void pre() {
    f[1]=1;
    for(int i=2;i<N;i++) {
        if(!b[i]) pr[++cp]=i,f[i]=i-2;
        for(int j=1;j<=cp && (LL)i*pr[j]<N;j++) {
            b[i*pr[j]]=1;
            if(i%pr[j]) f[i*pr[j]]=f[i]*f[pr[j]];
            else {
                if(i/pr[j]%pr[j]) f[i*pr[j]]=f[i/pr[j]]*(pr[j]-1)*(pr[j]-1);
                else f[i*pr[j]]=f[i]*pr[j];
                break;
            }
        }
    }for(int i=2;i<N;i++) f[i]+=f[i-1];
}
int T,n;
int main() {
    pre();
    for(scanf("%d",&T);T--;) {
        scanf("%d",&n);
        LL ans=0;
        for(int i=1,j;i<=n;i=j+1) {
            j=n/(n/i);
            ans+=1LL*(n/i)*(n/i)*(f[j]-f[i-1]);
        }printf("%lld\n",ans);
    }
    return 0;
}