## page was renamed from 程序设计练习26——zju2481——Unique Ascending Array
= Unique Ascending Array =
http://acm.zju.edu.cn/show_problem.php?pid=2481

Time limit: 1 Seconds   

Memory limit: 32768K  

Given an array of integers A[N], you are asked to decide the shortest array of integers B[M], such that the following two conditions hold.

    * For all integers 0 <= i < N, there exists an integer 0 <= j < M, such that A[i] == B[j]
    * For all integers 0 =< i < j < M, we have B[i] < B[j]

Notice that for each array A[] a unique array B[] exists.


== Input ==

The input consists of several test cases. For each test case, an integer N (1 <= N <= 100) is given, followed by N integers A[0], A[1], ..., A[N - 1] in a line. A line containing only a zero indicates the end of input.


== Output ==

For each test case in the input, output the array B in one line. There should be exactly one space between the numbers, and there should be no initial or trailing spaces.


== Sample Input ==
{{{
8 1 2 3 4 5 6 7 8
8 8 7 6 5 4 3 2 1
8 1 3 2 3 1 2 3 1
0
}}}

== Sample Output ==
{{{
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3
}}}
-----
{{{#!cplusplus
/*written by proby*/
#include <iostream>
#include <list>
using namespace std;

void Add(list<int>& objList, int i){
	list<int>::iterator objIter;
	for(objIter=objList.begin(); ; objIter++){
		if(objIter==objList.end() || *objIter>i){
			objList.insert(objIter, i);
			return;
		}
		if(*objIter==i){
			return;
		}
	}
}

void MyPrint(list<int>& objList){
	list<int>::iterator objIter;
	list<int>::iterator nextIter;
	nextIter=objList.begin();
	nextIter++;
	for(objIter=objList.begin(); objIter!=objList.end(); objIter++){
		cout<<(*objIter);
		if( nextIter!=objList.end() ){
			cout<<" ";
		}
		nextIter++;
	}
	cout<<endl;
}

int main(){
	list<int> objList;
	list<int>::iterator listIter;
	int nLen, objInt, i;
	cin>>nLen;
	while(nLen!=0){
		for(i=0; i<nLen; i++){
			cin>>objInt;
			Add(objList, objInt);
		}
		MyPrint(objList);
		objList.clear();
		cin>>nLen;
	}

	return 1;
}
}}}
------