首页 > 代码库 > hdu1040 水题
hdu1040 水题
As Easy As A+B |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 2348 Accepted Submission(s): 1154 |
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 23 2 1 39 1 4 7 2 5 8 3 6 9 |
Sample Output 1 2 31 2 3 4 5 6 7 8 9 |
很简单一个排序问题
1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 5 int main() 6 { 7 int t,n,i,flag; 8 int a[1002]; 9 while(cin>>t&&t)10 {11 while(t--)12 {13 cin>>n;14 for(i=0;i<n;i++)15 cin>>a[i];16 flag=0;17 sort(a,a+n);18 for(i=0;i<n;i++)19 {20 if(flag==0)21 flag=1;22 else23 cout<<" ";24 cout<<a[i];25 }26 cout<<endl; 27 }28 }29 return 0;30 }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。