簡化題意可知，實際上題目求得是gcd(i,j)=1(i,j<=n)的數對數目。

線性篩出n大小的歐拉表，求和*2+1即可。需要特判1.

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include 
 
<set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-8
# define MOD 30031
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for
 
(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int
 
 x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
const int N=40005;
//Code begin...

int phi[N], prime[N], tot;
bool check[N];

void getEuler(int n){
    mem(check,0); phi[1]=1; tot=0;
    FOR(i,2,n) {
        if (!check[i]) prime[tot++]=i, phi[i]=i-1;
        FO(j,0,tot) {
            if (i*prime[j]>n) break;
            check[i*prime[j]]=true;
            if (i%prime[j]==0) {phi[i*prime[j]]=phi[i]*prime[j]; break;}
            else phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
}
int main ()
{
    LL ans=0;
    int n;
    scanf("%d",&n); 
    if (n==1) {puts("0"); return 0;}
    --n; getEuler(n);
    FOR(i,1,n) ans+=phi[i];
    ans=ans*2+1;
    printf("%lld\n",ans);
    return 0;
}

BZOJ 2190 儀仗隊(線性篩歐拉函數)