Problem Description

You are given an integer sequence of length n, a1,…,an. Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append ai to the end of b. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.

Constraints

1≤n≤2×1050≤ai≤109n and ai are integers.

Input

Input is given from Standard Input in the following format:

n a1 a2 … an

Output

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

Example

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

题意： 给出 n 个数和一个空序列，要求对空序列执行 n 次操作，每次操作将第 i 个数放到序列末尾，并序列进行反转，问 n 次操作后获得的序列

思路：

简单的推导一下可以发现：

n 为奇数时：最终序列元素的下标为 n,n-2,n-4,...,2,1,3,...,n-5,n-3,n-1n 为偶数时：最终序列元素的下标为 n,n-2,n-4,...,2,1,...,n-5,n-3,n-1

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 200000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std; int a[N]; int main(){ int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); if(n%2){ for(int i=n;i>=1;i-=2) printf("%d ",a[i]); for(int i=2;i<=n;i+=2) printf("%d ",a[i]); printf("\n"); } else{ for(int i=n;i>=1;i-=2) printf("%d ",a[i]); for(int i=1;i<=n;i+=2) printf("%d ",a[i]); printf("\n"); } return 0; }