HDU-1040 As Easy As A+B

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本文解析了HDU-1040AsEasyAsA+B题目，介绍了如何通过输入一系列整数并对其进行升序排序来解决这个问题。提供了使用C++实现的简洁代码示例。

HDU-1040 As Easy As A+B

Time Limit : 2000/1000ms (Java/Other)
Memory Limit : 65536/32768K (Java/Other)

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

题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1040

分析

题意：扯了一堆没用的话，就是让你排序

思路：妥妥的一道水题，直接sort排序过了。

代码

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
    int n,num[1005];
    cin>>n;
    for (int i=0;i<n;i++)
    {
        int m;
        cin>>m;
        for (int j=0;j<m;j++)
            cin>>num[j];
        sort(num,num+m);
        for (int j=0;j<m-1;j++)
            cout<<num[j]<<" ";
        cout<<num[m-1]<<endl;
    }
    return 0;
}