Timofey and cubes
time limit per test 2 second memory limit per test 256 megabytes
Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number ai, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.

In this time, Timofey’s elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n-i+1)-th. He does this while i≤n-i+1.

After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

传送门:CF764B

Input

The first line contains single integer n (1≤n≤2·105) — the number of cubes. The second line contains n integers a1,2,..,an (-109≤ai≤109), where ai is the number written on the i-th cube after Dima has changed their order.

Output

Print n integers, separated by spaces — the numbers written on the cubes in their initial order. It can be shown that the answer is unique.

Sample Input

1
2
7
4 3 7 6 9 1 2

Sample Output

1
2 3 9 6 7 1 4

题解

题目不难主要是做的时候超时了 主要是Scanner和System.out太慢了 去学了大神的代码 从此以后只用FastScanner和PrintWriter了 也算是学会了点套路
## AC code:(不包含输入类)
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import java.io.*;  
import java.util.*;
public class Main {

public static void main(String[] args) throws IOException {
FastScanner sc=new FastScanner();
PrintWriter pw = new PrintWriter(System.out);
String temp;
while((temp=sc.br.readLine())!=null){
int n=Integer.parseInt(temp);
int[] cube=new int [n+1];
for(int i=1;i<=n;i++){
if(i%2!=0&&i<=n/2||i>n/2&&(n-i+1)%2!=0)
cube[n-i+1]=sc.nextInt();
else
cube[i]=sc.nextInt();
}
for(int i=1;i<=n;i++){
if(i!=n){
pw.print(cube[i]+" ");
}
else
pw.println(cube[n]);
}
pw.flush();
}

}

}