按题意模拟即可。

#include <bits/stdc++.h>
#define endl '\n'

using namespace std;

typedef long long ll;

ll a[3100];
ll b[3100];
int n, k;

bool check(ll l) {
    for (int i = 1; i <= k; i++)
        if (b[i] == l)
            return false;
    return true;
}

int main() {
    cin >> n;
    a[1] = 1;
    b[++k] = 1;
    
    for (int i = 2; i <= n; i++) {
        if (a[i - 1] - i > 0 && check(a[i - 1] - i)) {
            b[++k] = a[i - 1] - i;
            a[i] = a[i - 1] - i;
        }
            
        else {
            b[++k] = a[i - 1] + i;
            a[i] = a[i - 1] + i;
        } 
    }

    sort(a + 1, a + n + 1);

    for (int i = 1; i <= n; i++)
        cout << a[i] << " ";
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int n,a[3005];
bool b[13000];
int main(){
	cin>>n;
	a[1]=1;
	b[1]=1;
	for(int i=2;i<=n;i++){
		if(a[i-1]-i>0&&b[a[i-1]-i]==0){
			a[i]=a[i-1]-i;
			
		}
		else{
			a[i]=a[i-1]+i;
		}
		b[a[i]]=1;
	}
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++){
		cout<<a[i]<<" ";
	}
	return 0;
}

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

typedef long long ll ;

ll a[3100]={0,1};
int n;
int k = 1;//a数组的当前长度

int fd(ll m)
{
    for(int i = 1; i <= k;i++)
    {
        if(m == a[i])return i;
    }
    return -1;
}
int main()
{
    cin>>n;
    for(int i = 2; i <= n;i++)
    {
        //cout<<a[i]<<endl;
        ll t1 = a[i-1] - i;
        ll t2 = a[i-1] + i;
        if(t1 > 0 and fd(t1) == -1)a[i] = t1;
        else a[i] = t2;
        k++;
    }
    sort(a+1,a+k+1);
    for(int i = 1; i <= k;i++)
    {
        cout<<a[i]<<' ';
    }

    return 0;
}