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.

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.

For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.

41 4 3 252 3 4 5 1110

ambiguousnot ambiguousambiguous

Huge input,scanf is recommended.

对于一个有N个元素的队列，队列元素为[1,2,...,N-1,N]，进行一次队列变换，当前队列“数字i的位置”将成为变换后队列的第i个元素的值（下标从1开始）。

一开始拿到题目基本上又是什么都没有读懂的状态，就知道又是讨厌的英文题了。。。搞懂题意之后简单模拟就可以出解。

代码（1AC）：

#include 
    
     #include 
     
      #include 
      
       int arr[110000];int end[110000];int change(int arr[], int n){ //0 not ambiguous 1 ambiguous    int i, j;    int flag = 1;    for (i = 1; i <= n; i++){        end[arr[i]] = i;    }    for (i = 1; i <= n; i++){        if (end[i] != arr[i]){            flag = 0;        }    }    return flag;}int main(void){    int n, i;    while (scanf("%d", &n), n != 0){        memset(arr, 0, sizeof(arr));        memset(end, 0, sizeof(end));        for (i = 1; i <= n; i++){            scanf("%d", &arr[i]);        }        if (change(arr, n)){            printf("ambiguous\n");        }        else{            printf("not ambiguous\n");        }    }    return 0;}