#include<iostream>
#include<cmath>
using namespace std;
bool b[100001];
bool isprime(int i){
    for(int j = 3; j <= sqrt(i); j++){
        if(i % j == 0)return false;
    }
    return true;
}
int main(){
    ios::sync_with_stdio(false);
    for(int i = 1; i <= 100001; i++)b[i] = false;
    int n, k;
    cin >> n >> k;
    for(int j = 2; j <= n; j += 2)b[j] = !b[j];
    for(int i = 3; i <= k; i += 2){
        if(isprime(i)){
            for(int j = i; j <= n; j += i)b[j] = !b[j];
        }
    }
    for(int i = 1; i <= n; i++){
        if(b[i])cout << i << " ";
    }
    return 0;
}
    

可能用数论也能解决，但我数论太辣鸡了(我在讨论里发了)

#include<iostream>
using namespace std;
int is_prime(int n)
{
for (int i = 2; i < n; i++)
{
if (n%i == 0)
{
return 0;
break;
}
}
return 1;
}
int main()
{
int number_class;
int k;
int t;
int a[100000];
for (int x = 0; x < 100000; x++)
{
a[x] = 1;
}
cin >> number_class>>k;
for (int i = 2; i <= k; i++)
{
if (is_prime(i))
{
for (int j = 1; j <= number_class; j++)
{
if (0==j%i)
{
t = -a[j];
a[j] = t;
}
}
}
}
for (int l = 1; l <= number_class; l++)
{
if (a[l] == -1)
{
cout << l;
cout << " ";
}
}
return 0;
}