Educational Codeforces Round 10 B. z-sort

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
Copy
4
1 2 2 1
Output
Copy
1 2 1 2
Input
Copy
5
1 3 2 2 5
Output
Copy
1 5 2 3 2

题解:这道看起来像个排序题,第偶数个数要>=前一个数,第奇数个数要<=前一个数,

事实上这道题看样例会发现只要把源数列排序后前面一个,结尾一个依次输出就行,比如 13 2 2 5 ,排序后为 1 2 2 3 5,

输出第一个 1 ,输出最后一个5,输出第二个 2 ,输出倒数第二个 3,依次类推,即可.

#include <bits/stdc++.h>
const int N=1e3+5;
using namespace std;
int a[N];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
	}
	sort(a+1,a+n+1);
	//for(int i=1;i<=n;i++) printf("%d\n",a[i]);
	if(n%2==0){
		for(int i=1;i<=n/2;i++){
			printf("%d %d ",a[i],a[n-i+1]);
		}
	}
	else{
		for(int i=1;i<=n/2;i++){
			printf("%d %d ",a[i],a[n-i+1]);
		}
		printf("%d",a[n/2+1]);
	}
	return 0;
	
}