http://codeforces.com/problemset/problem/303/A
- when n is odd, A[i] = B[i] = i
- when n is even, there is no solution.
- f , 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;
}
【额外思考】
另一种构造方法如下，并且还满足了一个要求：
当n为奇数时，若要求AB两个排列不一样，怎么做？
可以把A序列设为n-1,n-2,n-3,...,1,0（公差为-1）
然后B序列设为0,2,4,...,1,3,5,...（公差为2）
这样C序列就是n-1,0,1,2,...（公差为1）
