Unique Ascending Array

最新推荐文章于 2025-01-04 10:10:24 发布

HaifengQi

最新推荐文章于 2025-01-04 10:10:24 发布

阅读量215

点赞数

CC 4.0 BY-SA版权

分类专栏：简单题

本文链接：https://blog.youkuaiyun.com/PeopleOfVision/article/details/81778440

简单题专栏收录该内容

36 篇文章

订阅专栏

本文介绍了一种算法，用于解决给定一个整数数组A[N]，如何找到符合条件的最短整数序列B[M]的问题。对于数组A中的每个元素，B中存在一个相等的元素，并且B自身必须是一个递增序列。文章提供了完整的输入输出样例及C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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

#include<iostream>
#include<cstdio>
#include<set>
using namespace std;
int main()
{
	int n;
	
	while(cin>>n && n)
	{
	     set<int> s;
	     set<int>::iterator it;
	     int x;
	     int flag=1;
		for(int i=0;i<n;i++)
		{
			cin>>x;
			s.insert(x);
		}
		for ( it = s.begin(); it != s.end(); it++)    
        {
        	if(flag)
        	{
        		cout << *it;
        		flag=0;
			}
			else
            cout << " "<<*it;  
        }  
        cout<<endl;
	}
	return 0;
}