poj 2470 Ambiguous permutations

Ambiguous permutations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6152		Accepted: 3627

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.

题意：给你一个数组，数组内存储的是位置转换的位置，举例说明数组a[i]存储，a[1]=3表明1这个数字存储在数组b[3]中，

所以a[]={2,3,4,5,1}

那么b[]={5,1,2,3,4};

如果a和b一样则是ambiguous；

#include <iostream>
#include <stdio.h>
using namespace std;
#define MAX 100005
int main(){
	int a[MAX],n,b[MAX];
	while(scanf("%d",&n)!=EOF){
		if (n==0)
			break;
		int k=1,index;
		for (int i=1;i<=n;i++){
			cin>>a[i];
			b[a[i]]=k++;
		}
		int flag=0;
		for (int i=1;i<=n;i++){
			if (a[i]!=b[i]){
				flag=1;
				break;
			}
		}
		if (flag==0)
			printf("ambiguous\n");
		else
			printf("not ambiguous\n");
	}
	return 0;
}