pushpush（AtCoder-2648）

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≤2×105
• 0≤ai≤109
• n 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-1
• n 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;
}