Educational Codeforces Round 10——B. z-sort

B. z-sort

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

A student of z-school found a kind of sorting called z-sort.
The array a with n elements
are z-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：给定一个数组是否能组成一个波动序列（这名称很贴切呃）。然后突然想到了双端队列，然后又想到了据说在两端操作时间很快的list类

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<list>
using namespace std;
int pos[1009];
int main(void)
{
list<int>lst;
int n,temp,i;
while (cin>>n)
{
lst.clear();
memset(pos,0,sizeof(pos));
int cnt=0;
for (i=0; i<n; i++)
{
cin>>temp;
lst.push_back(temp);
}
lst.sort();//list的sort是自带的方法.不能用普通sort
bool ok=1;
while (!lst.empty())//从一个排好序的序列里头与尾交替地取出一个数就可以组成波动序列
{
if(lst.front()<=lst.back())
{
pos[cnt++]=lst.front();
lst.pop_front();
if(lst.empty())//每次pop都特判一下
break;
pos[cnt++]=lst.back();
lst.pop_back();
if(lst.empty())//每次pop都特判一下
break;
}
else
{
ok=0;
break;
}
}
if(!ok)
cout<<"Impossible"<<endl;
else
{
for (i=0; i<cnt; i++)
{
printf("%d%s",pos[i],i==cnt-1?"\n":" ");
}
cout<<endl;
}
}
return 0;
}