zju2481

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

/*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;
}