HDU 1.3.5 shǎ崽 OrOrOrOrz

最新推荐文章于 2020-03-10 22:28:43 发布

原创最新推荐文章于 2020-03-10 22:28:43 发布 · 329 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM HDU

ACM 专栏收录该内容

41 篇文章

订阅专栏

本文介绍了一个算法问题，即从一组唯一整数中选择元素，首先选择最大的，然后是最小的，接着是第二大的，第二小的，以此类推。通过提供输入示例和输出示例，详细解释了解决方法，并附上通过验证的代码实现。此教程适用于初学者，旨在提高他们对排序和数组操作的理解。

shǎ崽 OrOrOrOrz

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3988 Accepted Submission(s): 1236

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

菜鸟级的原创代码，已AC。若有可提高之处欢迎指导

//#define LOCAL  
#include<stdio.h>  
#include<string>  
#include<algorithm>  
using std::sort;  
bool cmp(int a, int b)  
{  
return a>b;  
}  
  
int main()  
{  
#ifdef LOCAL  
freopen("H://dataIn.txt", "r", stdin);  
freopen("H://dataOut.txt", "w", stdout);  
#endif  
int a[10000];  
int n;  
while(scanf("%d", &n)==1)  
{  
memset(a, 0, sizeof(a));  
for (int i = 0; i<n; i++)  
scanf("%d",&a[i]);  
sort(a, a + n, cmp);  
for (int i = 0; i < n / 2-1; i++)  
{  
printf("%d ", a[i]);  
printf("%d ", a[n-i-1]);  
}  
if (double(n) / 2 == double(n / 2))  
{  
printf("%d ", a[n/2-1]);  
printf("%d\n", a[n/2]);  
}  
else  
{  
printf("%d ", a[n / 2 - 1]);  
printf("%d ", a[n / 2 + 1]);  
printf("%d\n", a[n / 2]);  
}  
}  
return 0;  
  
}