BZOJ 2190 仪仗队(线性筛欧拉函数)

简化题意可知，实际上题目求得是gcd(i,j)=1(i,j<=n)的数对数目。

线性筛出n大小的欧拉表，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-
# define MOD
# define INF
# 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<<,l,mid
# define rch p<<|,mid+,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=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
const int N=;
//Code begin...

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

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