Hello coders, today we are going to solve **Ambiguous Permutations Codechef Solution** whose Problem Code is **RECIPE**.

### Problem

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

**Ambiguous Permutations CodeChef Solution**

n = int(input()) while(n != 0): li = [int(i) for i in input().split()] li1 = list('0' * len(li)) for i in range(len(li)): li1[li[i] - 1] = i + 1 if(li == li1): print("ambiguous") else: print("not ambiguous") n = int(input())

#include<iostream> #include<stdio.h> using namespace std; int main() { int n,a[100001],i; while(1) { scanf("%d",&n); if(n==0)break; for(i=1;i<=n;i++) scanf("%d",&a[i]); for(i=1;i<=n;i++) { if(a[a[i]]!=i){printf("not ambiguous \n");break;} if(i==n)printf("ambiguous \n"); } } return 0; }

**Disclaimer: **The above Problem **(Ambiguous Permutations) **is generated by **CodeChef** but the solution is provided by **Chase2learn**.This tutorial is only for **Educational** and **Learning** purpose.