1070 - Permutations
28 Jan 2021 — Tags: None
Click to show code.
using namespace std;
int main(void)
{
int n;
cin >> n;
if (n > 1 and n <= 3)
cout << "NO SOLUTION" << endl;
else
{
for (int i = 2; i <= n; i += 2)
cout << i << " ";
for (int i = 1; i <= n; i += 2)
cout << i << " ";
cout << endl;
}
return 0;
}