BZOJ 2190 仪仗队(线性筛欧拉函数)

时间：2021-09-28 16:51:40

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 ;

}

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