poj 2470 Ambiguous permutations

Ambiguous permutations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5192		Accepted: 3061

Description

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

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

Sample Output

ambiguous
not ambiguous
ambiguous

Hint

Huge input,scanf is recommended.

#include <istream>
using namespace std;
int a[100010];
int main()
{
int n;
while (scanf("%d",&n)!=EOF)
{
if(n==0)
break;
int i,p=0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for (i=1;i<=n;i++)
{
if (a[a[i]]==i)
{
p++;
}
else
break;
}
if (p==n)
{
printf("ambiguous\n");
}
else
printf("not ambiguous\n");
}
return 0;
}