Unique Ascending Array

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

————————————————————————————————————

题目的意思是给出一个序列，输出去重后的排序好的序列

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <string>

#include <set>

#include <map>

#include <queue>

using namespace std;

#define inf 0x3f3f3f3f

int main()

{

    int n,a[100005];

    while(~scanf("%d",&n)&&n)

    {

        for(int  i=0; i<n; i++)

            scanf("%d",&a[i]);

        sort(a,a+n);

        printf("%d",a[0]);

        for(int i=1; i<n; i++)

            if(a[i]!=a[i-1])

                printf(" %d",a[i]);

        printf("\n");

    }

    return 0;

}

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