hdu 1040 As Easy As A+B

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46831    Accepted Submission(s): 20046

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

 

Recommend

We have carefully selected several similar problems for you:  1029 1062 1032 1391 2673 

 

简单的排序，用sort函数就好。

 

题意：输入一个数T限制输入T行，每行开头输入一个整数N，表示此行输入N个数，对这N个数进行排序。

 

附上代码：

 

#include <cstdio>
#include <iostream>
#include <algorithm>  //sort函数的头文件
using namespace std;
int main()
{
    int n, m, i, a[10000], j, t;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m);
        for(i=1; i<=m; i++)
            scanf("%d",&a[i]);
        sort(a+1,a+1+m);   //sort排序
        for(i=1; i<=m; i++)
        {
            printf("%d",a[i]);
            if(i!=m)     //注意输出格式，最后一个数不输出空格，直接换行
                printf(" ");
            else
                printf("
");
        }
    }
    return 0;


}