今天去老校区找她，不想带电脑了，所以没时间A题了
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As Easy As A+B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45926 Accepted Submission(s): 19620
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!
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.
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
1 #include<math.h> 2 #include<stdio.h> 3 #include<string.h> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 #define N 111555 8 9 int a[N],n; 10 11 int main() 12 { 13 int t;cin>>t; 14 while(t--) 15 { 16 cin>>n; 17 for(int i=0;i<n;i++)cin>>a[i]; 18 sort(a,a+n); 19 int i; 20 for(i=0;i<n-1;i++) 21 printf("%d ",a[i]); 22 printf("%d ",a[i]); 23 24 } 25 26 return 0; 27 28 } 29 30 //freopen("1.txt", "r", stdin); 31 //freopen("2.txt", "w", stdout); 32 //**************************************