A student of z-school found a kind of sorting called z-sort. The array a with n elements arez-sorted if two conditions hold:
ai ≥ ai - 1 for all even i,
ai ≤ ai - 1 for all odd i > 1.
For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array[1,2,3,4] isn’t z-sorted.
Can you make the array z-sorted?
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
If it’s possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word “Impossible”.
Examples
input
4
1 2 2 1
output
1 2 1 2
input
5
1 3 2 2 5
output
1 5 2 3 2
题意:给出一个Z-sort的定义,就是对于一个序列它满足ai ≥ ai - 1 for all even i ai ≤ ai - 1 for all odd i > 1 对于从1开始编号的序列,偶数位要大于等于相邻两个数。
题解:把这个序列先按照从小到大排序,然后将这个序列分为两半,后面的一半插入前面的 一半,这样就可以尽量保证是一大一小的,因为后面的数一定是大于或者等于前面数的。最后再按照定义判断一下,是否得到了正确的数列,没有则输出Impossible。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,a[1010],p[1010],ans[1010];
int main()
{
int i,t=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+1+n);
int st1=1,st2=(n+1)/2+1;
while(1)
{
t++;
if(t%2==1)
ans[t]=a[st1++];
else
ans[t]=a[st2++];
if(t==n) break;
}
for(i=1;i<=n;i++)
printf("%d ",ans[i]);
printf("\n");
return 0;
}