zoukankan      html  css  js  c++  java
  • hdu2637

                                                      shǎ崽 OrOrOrOrz

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3654 Accepted Submission(s): 1138
     
    Problem Description
    Acmer in HDU-ACM team are ambitious, especially shǎ崽, he can spend time in Internet bar doing problems overnight. So many girls want to meet and Orz him. But Orz him is not that easy.You must solve this problem first.
    The problem is :
    Give you a sequence of distinct integers, choose numbers as following : first choose the biggest, then smallest, then second biggest, second smallest etc. Until all the numbers was chosen .
    For example, give you 1 2 3 4 5, you should output 5 1 4 2 3
     
    Input
    There are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.
     
    Output
    Output a sequence of distinct integers described above.
     
    Sample Input
    5
    1 2 3 4 5
     
    Sample Output
    5 1 4 2 3
     
    #include<iostream>
    #include<algorithm>
    using namespace std;
    
    int main()
    {
        int n,i,flag,j;
        int a[10000];
        while(cin>>n&&n)
        {        
            for(i=0;i<n;i++)
                cin>>a[i];
            sort(a,a+n);//默认升序
            flag=0;
            for(i=0,j=n-1;i<=j;i++,j--)
            {
                if(flag==0)    flag=1;
                else
                    cout<<" ";
                if(i==j) 
                    cout<<a[i];
                else 
                    cout<<a[j]<<" "<<a[i];
            }
            cout<<endl;
        }    
        return 0;
    }
  • 相关阅读:
    查看JVM
    性能测试 -- 实际问题
    性能测试 -- 常用命令
    性能测试 -- 实际例子
    性能测试 -- 内核调优
    jmeter分布式环境
    Linux 安装配置Subversion edge
    Jenkins进阶系列之——01使用email-ext替换Jenkins的默认邮件通知
    Ubuntu下eclipse安装svn插件
    Jenkins入门系列之——03PDF文档下载
  • 原文地址:https://www.cnblogs.com/mm-happy/p/3807946.html
Copyright © 2011-2022 走看看