AtCoder Regular Contest 077-C

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

题目大意：放入尾部，再反转，求最终序列情况
解题思路：稍微模拟一下就可以发现规律

#include<iostream>using namespace std;const int MAXN=2e5+5;int a[MAXN];int main(){    int n;    while(cin>>n)    {        for(int i=1;i<=n;++i)           {            cin>>a[i];        }        if(n&1)        {            cout<<a[n];            for(int i=n-2;i>=1;i-=2)                cout<<" "<<a[i];            for(int i=2;i<=n-1;i+=2)                cout<<" "<<a[i];            cout<<endl;        }else        {            cout<<a[n];            for(int i=n-2;i>=2;i-=2)                cout<<" "<<a[i];            for(int i=1;i<=n-1;i+=2)                cout<<" "<<a[i];            cout<<endl;        }    }    return 0;}