【codeforces 764B】Timofey and cubes

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number ai, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.

In this time, Timofey’s elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n - i + 1)-th. He does this while i ≤ n - i + 1.

After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

Input
The first line contains single integer n (1 ≤ n ≤ 2·105) — the number of cubes.

The second line contains n integers a1, a2, …, an ( - 109 ≤ ai ≤ 109), where ai is the number written on the i-th cube after Dima has changed their order.

Output
Print n integers, separated by spaces — the numbers written on the cubes in their initial order.

It can be shown that the answer is unique.

Examples
input
7
4 3 7 6 9 1 2
output
2 3 9 6 7 1 4
input
8
6 1 4 2 5 6 9 2
output
2 1 6 2 5 4 9 6
Note
Consider the first sample.

At the begining row was [2, 3, 9, 6, 7, 1, 4].
After first operation row was [4, 1, 7, 6, 9, 3, 2].
After second operation row was [4, 3, 9, 6, 7, 1, 2].
After third operation row was [4, 3, 7, 6, 9, 1, 2].
At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].

【题目链接】:http://codeforces.com/contest/764/problem/B

【题解】

如果从初始数列到目标数列
操作依次为
[1..倒1]翻转
[2..倒2]翻转
[3..倒3]翻转
…
[i..倒i]翻转
前面的翻转区间覆盖了后面的翻转区间;
所以如果是偶数次翻转,那么翻转的区间的两端那个元素是不会改变顺序的;还是原来的位置;
而如果是奇数次翻转,两端的元素就会交换位置;
根据这个规则处理出原序列就好;
(reverse函数和sort函数都是左闭右开区间…..)

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 2e5+100;

int a[MAXN];
int n;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i]);
    int cnt = 0;
    for (int i = 1;i<=n;i++)
    {
        int j = n-i+1;
        if (i <= j)
        {
            cnt++;
            if (cnt&1)
                swap(a[i],a[j]);
        }
    }
    rep1(i,1,n)
        {
            printf("%d",a[i]);
            if (i==n)
                puts("");
            else
                putchar(' ');
        }
    return 0;
}