AtCoder：pushpush（STL & 思维）

最新推荐文章于 2022-10-16 21:15:08 发布

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本文探讨了一个有趣的算法问题——逆序追加序列。通过给定整数序列，执行特定的追加和反转操作，最终得出序列结果。利用C++中的list数据结构，实现了简洁高效的解决方案。

C - pushpush

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

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.

Sample Input 1

Copy

4
1 2 3 4

Sample Output 1

Copy

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

Copy

3
1 2 3

Sample Output 2

Copy

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

Copy

1
1000000000

Sample Output 3

Copy

1000000000

Sample Input 4

Copy

6
0 6 7 6 7 0

Sample Output 4

Copy

0 6 6 0 7 7

题意：给n个数a和一个空序列b，每次执行两个操作，①将a[i]放到b的末尾，②翻转b序列，输出执行完n个数后的b序列。

思路：容易想到是将a[i]前后前后地放到b里面，那么用queue或者list都能方便实现。

# include <bits/stdc++.h>
using namespace std;
list<int>l;
int main()
{
    int n, m, t=1;
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        scanf("%d",&m);
        t = !t;
        if(t) l.push_back(m);
        else l.push_front(m);
    }
    if(!(n&1)) l.reverse();
    for(auto i=l.begin();i!=l.end();++i)
        printf("%d ",*i);
    return 0;
}