zoj 2795

Ambiguous permutations Time Limit: 10 Seconds      Memory Limit: 32768 KB

Some programming contest problems are really tricky: not only do they require a different output format from what you might have expected, but also the sample output does not show the difference. For an example, let us look at permutations.
A permutation of the integers 1 to n is an ordering of these integers. So the natural way to represent a permutation is to list the integers in this order. With n = 5, a permutation might look like 2, 3, 4, 5, 1.
However, there is another possibility of representing a permutation: You create a list of numbers where the i-th number is the position of the integer i in the permutation. Let us call this second possibility an inverse permutation. The inverse permutation for the sequence above is 5, 1, 2, 3, 4.
An ambiguous permutation is a permutation which cannot be distinguished from its inverse permutation. The permutation 1, 4, 3, 2 for example is ambiguous, because its inverse permutation is the same. To get rid of such annoying sample test cases, you have to write a program which detects if a given permutation is ambiguous or not.

Input Specification

The input contains several test cases.
The first line of each test case contains an integer n (1 ≤ n ≤ 100000). Then a permutation of the integers 1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between 1 and n appears exactly once in the permutation.
The last test case is followed by a zero.

Output Specification

For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.

Sample Input

4
1 4 3 2
5
2 3 4 5 1
1
1
0

Sample Output

ambiguous
not ambiguous
ambiguous
#include <iostream>
using namespace std;

int main()
{
	int n;
	while(cin>>n&&n!=0)
	{
		int *a = new int[n+1];
		for(int i = 1 ; i <= n; i++)
			cin>>a[i];
		int t = 0;
		for(int j = 1 ; j <= n; j++)
			if(a[a[j]] == j)
				t++;
		if(t==n)
			cout<<"ambiguous"<<endl;
		else
			cout<<"not ambiguous"<<endl;
	}
	return 0;
}