Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length n, a_1, ..., a_n.
Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append a_i to the end of b.

Reverse the order of the elements in b.

Find the sequence b obtained after these n operations.

Constraints

1 \leq n \leq 2\times 10^5

0 \leq a_i \leq 10^9

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print n integers in a line with spaces in between.
The i-th integer should be b_i.

Sample Input 1

4
1 2 3 4

Sample Output 1

4 2 1 3

After step 1 of the first operation, b becomes: 1.

After step 2 of the first operation, b becomes: 1.

After step 1 of the second operation, b becomes: 1, 2.

After step 2 of the second operation, b becomes: 2, 1.

After step 1 of the third operation, b becomes: 2, 1, 3.

After step 2 of the third operation, b becomes: 3, 1, 2.

After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.

After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.

Thus, the answer is 4 2 1 3.

Sample Input 2

3
1 2 3

Sample Output 2

3 1 2

As shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third operation. Thus, the answer is 3 1 2.

Sample Input 3

1
1000000000

Sample Output 3

1000000000

Sample Input 4

6
0 6 7 6 7 0

Sample Output 4

0 6 6 0 7 7

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

typedef pair<LL,LL>PII;

/*
题意：
给你n个数，求区间有多少对除数和被除数的对数

思路：
线段树 区间和/ 树状数组 前缀和
2e5的数的除数最多才81个
对于一个数的贡献，我最好是区间的贡献，也就是1-n的贡献。
*/

const int N=2e5+10;
int a[N],n;
vector<int>ans;

int main(){
    ans.clear();
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    if(n == 1)
    {
        printf("%d\n",a[1]);
        return 0;
    }
    if(n%2 == 0){
        int t = n;
        while(t>1){
            ans.push_back(a[t]);
            t -= 2;
        }
        t = 1;
        while(t<n)
        {
            ans.push_back(a[t]);
            t += 2;
        }
    }
    else if(n%2 == 1){
        int t = n;
        while(t>=1){
            ans.push_back(a[t]);
            t -= 2;
        }
        t = 2;
        while(t<n)
        {
            ans.push_back(a[t]);
            t += 2;
        }
    }
    int sz = ans.size();
    for(int i=0;i<sz;i++){
        if(i) printf(" ");
        printf("%d",ans[i]);
    }
    return 0;
}