利用质因数分解定理求与n互质的数的个数

#include<bits/stdc++.h>
using namespace std;
long long fun(int x)
{
	long long ans=x;
	int t=sqrt(x);
	int cnt;
	for(int i=2;i<=t;i++)
	{
		cnt=0;
		while(x%i==0)
		{
			x/=i;
			cnt++;
		}
		if(cnt) ans=ans*(i-1)/i;
	}
	if(x>1) ans=ans*(x-1)/x;
	return ans;
}
int main()
{
	int n;
	cin>>n;
	int t;
	while(n--)
	{
		cin>>t;
		printf("%lld\n",fun(t));
	}
	return 0;
}

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1000010;
int primes[N], cnt;
int phi[N];
bool st[N];
void get_eulers(int n)
{
    phi[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!st[i])
        {
            primes[cnt++] = i;
            phi[i] = i - 1; 
        }
        for (int j = 0; primes[j] <= n / i; j++)
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0)
            {
                phi[primes[j] * i] = phi[i] * primes[j]; 
                break;
            }
            phi[primes[j] * i] = phi[i] * (primes[j] - 1);
        }
    }
}
int main()
{
    int n;
    cin >> n;
    get_eulers(n);
    LL res = 0;
    for (int i = 1; i <= n; i++) res += phi[i];
    printf("%lld\n", res);
    return 0;
}