As Easy As A+B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60774 Accepted Submission(s):
26130
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
这个题目完全没有难度,学会了sort就可以了
看代码就ok了
#include <stdlib.h> #include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespace std; int a[10005]; int main() { int t,n; 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-1;i++) { printf("%d ",a[i]); } printf("%d ",a[n-1]); } return 0; }