两种方法求欧拉函数

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直接求法：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll Eular(ll n){
    ll res = n;
    for(int i=2;i*i<=n;i++){
        if(n%i == 0)  res = res/i*(i-1);	//先除后乘防止中间数据溢出
        while(n%i == 0)   n/=i;
    }
    if(n > 1)   res = res / n * (n-1);
    return res;
}

int main(){
    ll n;
    scanf("%lld",&n);
    printf("%lld\n",Eular(n));

    return 0;
}

线性筛法

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6+10;
int book[N],prime[N];
int phi[N];
int cnt;

void Eular(){	//线性筛欧拉函数
    cnt = 0;
    memset(book,0,sizeof(book));
    phi[1] = 1;
    for(int i=2;i<N;i++){
        if(!book[i]){
            prime[cnt++] = i;
            phi[i] = i-1;	//若a为质数,phi[a]=a-1;
        }
        for(int j=0;j<cnt;j++){
            if(i*prime[j] > N)
                break;
            book[i*prime[j]] = 1;
            if(i%prime[j] == 0)
                phi[i*prime[j]] = phi[i] * prime[j];	//若a为质数,b mod a=0,phi[a*b]=phi[b]*a
            else phi[i*prime[j]] = phi[i] * (prime[j]-1);
            //若a,b互质,phi[a*b]=phi[a]*phi[b](当a为质数时,if b mod a!=0 ,phi[a*b]=phi[a]*phi[b])
        }
    }
    for(int i=1;i<N;i++){
        printf("%d\n",phi[i]);
    }
}

int main(){
    Eular();

	return 0;
}