HDU 1040 As Easy As A+B(排序)

As Easy As A+B

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

 

Answer

水题，从小到大排序。不舒服，水水更健康~

 

#include <cstdio>
#include <iostream>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#define PI acos(-1.0)
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
#define msd memset(dp,0,sizeof(dp))
using namespace std;
//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    //freopen("out.txt","w",stdout);
#endif // LOCAL
    ios::sync_with_stdio(false);
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        cin>>n;
        vector<int> v;
        v.clear();
        for(int i=0;i<n;i++)
        {
            int t;
            cin>>t;
            v.push_back(t);
        }
        sort(v.begin(),v.end());
        printf("%d",v[0]);
        for(vector<int>::iterator it=v.begin()+1;it!=v.end();it++)
        {
            printf(" %d",*it);
        }
        printf("\n");
    }
    return 0;
}