首页 > 代码库 > CDZSC_2015寒假新人（1）——基础 D
CDZSC_2015寒假新人（1）——基础 D
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
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int main(){#ifdef CDZSC_OFFLINE freopen("in.txt","r",stdin);#endif int t,n,a[1010],i; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=0; i<n; i++) { scanf("%d",&a[i]); } sort(a,a+n); for(i=0; i<n; i++) { if(i!=0) { printf(" "); } printf("%d",a[i]); } printf("\n"); } return 0;}
CDZSC_2015寒假新人（1）——基础 D
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。