http://codeforces.com/problemset/problem/303/A
- when n is odd, A[i] = B[i] = i
- when n is even, there is no solution.
-
why? If
![]()
,
then ![]()
or
just ![]()
,
where S = 0 + 1 + ... + n - 1 = n(n - 1) / 2.
So, there must be ![]()
.
But when n is even, ![]()
.
/*92ms,0KB*/
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
if (n & 1)
{
for (int i = 0; i < n; i++) cout << i << " ";
cout << endl;
for (int i = 0; i < n; i++) cout << i << " ";
cout << endl;
for (int i = 0; i < n; i++) cout << (2 * i) % n << " ";
cout << endl;
}
else cout << -1 << endl;
return 0;
}
【額外思考】
另一種構造方法如下，並且還滿足了一個要求：
當n爲奇數時，若要求AB兩個排列不一樣，怎麼做？
可以把A序列設爲n-1,n-2,n-3,...,1,0（公差爲-1）
然後B序列設爲0,2,4,...,1,3,5,...（公差爲2）
這樣C序列就是n-1,0,1,2,...（公差爲1）