HDU 2673 (排序)

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 

InputThere are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.OutputOutput 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<stdio.h>
#include<algorithm>
using namespace std;

int main(){
    int a[10000],n,i,j;
    while(~scanf("%d",&n)){
        for(i=0;i<n;++i)
            scanf("%d",&a[i]);
        sort(a,a+n);
        i=0;//!!
        j=n-1;
        while(i<j){
            printf("%d %d",a[j],a[i]);
            if(i+1!=j)printf(" ");
            ++i;
            --j;//从两变填入
        }
        if(i==j)printf("%d",a[i]);
        printf("\n");
    }
    return 0;
}

  

转载于:https://www.cnblogs.com/Roni-i/p/7211435.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值