## Algorithm

Problem Name: 2 AD-HOC - beecrowd | 1663

# Ambiguous Permutations

Local Contest, University of Ulm Germany

Timelimit: 5

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

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

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

 Sample Input Sample Output 4 1 4 3 2 5 2 3 4 5 1 1 1 0 ambiguous not ambiguous ambiguous

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <cstdio>
int vetor[100010];
int main() {
int n;
while (scanf("%d", &n) && n) {
for (int i = 1; i  < = n; i++) {
scanf("%d", &vetor[i]);
}
int ambiguo = 1;
for (int i = 1; i  < = n; i++) {
if (vetor[vetor[i]] != i) {
ambiguo = 0;
break;
}
}
printf("%s\n", ambiguo ? "ambiguous" : "not ambiguous");
}
return 0;
}
``````
Copy The Code &

Input

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

Output

cmd
ambiguous
not ambiguous
ambiguous