ZOJ 2795-Ambiguous permutations

ZOJ Problem Set - 2795

Ambiguous permutations

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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. Withn = 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 thei-th number is the position of the integer i in the permutation. Let us call this second possibility aninverse 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 integers1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between1 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

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Source: University of Ulm Local Contest 2005

解题思路：

题目大意例如第一个数据：1是第一次输入，在输入的序列中1在第一个，4第二个输入，而2在输入的序列中的第4个。这样的可称为逆序。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int num[100000],pos[100000];int main(){int T;while(scanf("%d",&T)&&T){int i,j;for(int i=1;i<=T;i++){scanf("%d",&num[i]);pos[num[i]]=i;}int wc=1;int flag=1;for(int i=1;i<=T;i++){if(pos[i]!=num[i]){flag=0;break;}}if(flag==0){printf("not ambiguous\n");}else{printf("ambiguous\n");}}return 0;}