POJ 2470 Ambiguous permutations(简单题 理解题意)

【题目简述】：事实上就是依据题目描写叙述：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.

【分析】：看我加红的部分。就是说假设叫做ambiguous permutation的话，就要通过这种变换方式后还是same.否则的话就是not ambiguous。

这样的方式就是例如以下：

数字的新旧位置互换,即是数组的下标是所谓旧位置。而这个数列本身的数字又代表一种新的位置。我们依照数字所显示的。将该数字所相应的下标变成序列中的值，便形成了新的序列。

如： 2    3    4   5    1      //  原序列

       [1]   [2]  [3]  [4] [5]    //  数组下标

经过变换后：

        [5]  [1]  [2]  [3]  [4]    //  此时新的序列就是  5  1   2  3   4。由于与原序列 2  3  4  5  1不同，所以就是not  ambiguous

         1    2    3    4     5

见代码：

注：c++的输入输出会超时！

// 524K  219Ms
#include<iostream>
using namespace std;
int a[100005];
int main()
{
	int i,j,n;
	while(scanf("%d",&n)&&n!=0)
	{
		int flag=1;
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(i=1;i<=n;i++)
		{
			if(a[a[i]]!=i)
			{
				flag=0;
				break;
			}
		}
		if(flag==0)
			printf("not ambiguous\n");
		else
			printf("ambiguous\n");
	}
	return 0;
}