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题目描述

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:
1.Append ai to the end of b.
2.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.

输入

Input is given from Standard Input in the following format:
n
a1 a2 … an

输出

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

样例输入

4
1 2 3 4

样例输出

4 2 1 3

提示

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.

规律：a[n] . a[n-2] . a[n-4] . ...a[n-5] . a[n-3] . a[n-1]

代码如下：

#include 
using namespace std;
#define ll long long
#define pre(i,x,n) for(int i=x;i<=n;i++)
int n,a[200010],b[200010];
int main()
{
    int head=1,tail,temp=1;
    scanf("%d",&n);
    tail=n;
    for(int i=1;i<=n;i++)
      scanf("%d",&a[i]);
    for(int i=n;i>0;i--)
    {
        if(temp)
        {
           b[head]=a[i];
           head++;
           temp=!temp;
        }
        else
        {
           b[tail]=a[i];
           tail--;
           temp=!temp;
        }
    }
    printf("%d",b[1]);
    for(int i=2;i<=n;i++)
      printf(" %d",b[i]);
    putchar('\n');
    return 0;
}