POJ 2470 || SDUT 2356 Ambiguous permutations(简单规律)

Ambiguous permutations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5319		Accepted: 3145

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.

Source

Ulm Local 2005

解题报告:这道题就是规律题，就是第i个数如果都满足i= a[a[i]]就是ambiguous，否则就不是ambiguous

代码如下：

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