BZOJ 4804: 欧拉心算

Description

求(sum_{i=1}^nsum_{i=1}^nvarphi(gcd(i,j)),Tleqslant 5 imes 10^3,nleqslant 10^7)

Solution

数论分块+莫比乌斯反演.

化式子

(sum_{i=1}^nsum_{i=1}^nvarphi(gcd(i,j)))
(=sum_dvarphi(d)sum_{i=1}^nsum_{j=1}^n[(i,j)=d])
(=sum_dvarphi(d)(sum_{i=1}^{lfloor frac{n}{d} floor}sum_{j=1}^{lfloor frac{n}{d} floor}[(i,j)=1]))
(=sum_dvarphi(d)(sum_{p}mu(p)sum_{i=1}^{lfloor frac{n}{pd} floor}sum_{j=1}^{lfloor frac{n}{pd} floor}))
( ext{Let T=pd})
(=sum_{T}lfloor frac{n}{T} floorlfloor frac{n}{T} floorsum_{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
",ans);
    }
    return 0;
}