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;
}
(hdu step 1.3.7)As Easy As A+B(排序)
原文：http://blog.csdn.net/hjd_love_zzt/article/details/43278283