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 |
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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 |
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Input
There are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.
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Output
Output a sequence of distinct integers described above.
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Sample Input
5 1 2 3 4 5 |
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Sample Output
5 1 4 2 3 |
菜鳥級的原創代碼，已AC。若有可提高之處歡迎指導
//#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;
}