HDU1040 As Easy As A+B【排序】

编程基础 • 2024-12-30 14:06 • 阅读 12

As Easy As A+B

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 69843 Accepted Submission(s): 29716

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 
 

问题链接：HDU1040 As Easy As A+B。

问题描述：参见上文。

问题分析：这是一个排序问题，关键在于处理输入的循环控制。套路要掌握好，输入有多个case。

程序说明：（略）

参考链接：（略）

AC的C++语言程序：

/* HDU1040 As Easy As A+B */ #include <iostream> #include <algorithm> using namespace std; const int N = 1000; int a[N]; int main() { int t, n; while(cin >> t) { while(t--) { cin >> n; for(int i=0; i<n; i++) cin >> a[i]; sort(a, a + n); for(int i=0; i<n; i++) { if(i != 0) cout << " "; cout << a[i]; } cout << endl; } } return 0; }

今天的文章 HDU1040 As Easy As A+B【排序】分享到此就结束了，感谢您的阅读。

HDU1040 As Easy As A+B【排序】

As Easy As A+B

相关推荐