ZOJ - 2795 Ambiguous permutations

Ambiguous permutations

Time Limit: 10000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu

Submit Status

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 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

41 4 3 252 3 4 5 1110

Sample Output

ambiguousnot ambiguousambiguous

Source

University of Ulm Local Contest 2005

 

 

 

分析：

水题。

ac代码：

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