标签：des style class blog code java
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 |
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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! |
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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. |
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Output
For each case, print the sorting result, and one line
one case.
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Sample Input
2 3 2 1 3 9 1 4 7 2 5 8 3 6 9 |
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Sample Output
1 2 3 1 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 else 23 cout<<" "; 24 cout<<a[i]; 25 } 26 cout<<endl; 27 } 28 } 29 return 0; 30 }
标签：des style class blog code java
原文地址：http://www.cnblogs.com/mm-happy/p/3808005.html
