Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1663
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/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 |
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;
}
Input
1 4 3 2
5
2 3 4 5 1
1
1
0
Output
not ambiguous
ambiguous