(hdu step 1.3.7)As Easy As A+B(排序)

最新推荐文章于 2020-04-02 23:50:37 发布

原创最新推荐文章于 2020-04-02 23:50:37 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

acm 同时被 2 个专栏收录

678 篇文章

订阅专栏

ACM——夺金之路

309 篇文章

订阅专栏

本文介绍了一个简单的排序问题，包括题目要求、输入输出样例及解决方案。通过C++实现的代码展示了如何读取整数并使用标准库函数进行升序排序。

题目：

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2678 Accepted Submission(s): 1280

Problem Description

These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.
Give you some integers, your task is to sort these number ascending (升序).
You should know how easy the problem is now!
Good luck!

Input

Input contains multiple test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains an integer N (1<=N<=1000 the number of integers to be sorted) and then N integers follow in the same line.
It is guarantied that all integers are in the range of 32-int.

Output

For each case, print the sorting result, and one line one case.

Sample Input

2
3 2 1 3
9 1 4 7 2 5 8 3 6 9

Sample Output

1 2 3
1 2 3 4 5 6 7 8 9

Author

lcy

题目分析：

排序。

代码如下：

/*
 * h.cpp
 *
 *  Created on: 2015年1月29日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <algorithm>


using namespace std;

const int maxn = 1005;
int a[maxn];


int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);
		int i;
		for(i = 0 ; i < n ; ++i){
			scanf("%d",&a[i]);
		}

		sort(a,a+n);

		for(i = 0 ; i < n-1 ; ++i){
			printf("%d ",a[i]);
		}


		printf("%d\n",a[i]);//最后一个数的后面没有空格
	}

	return 0;
}