CF764 B. Timofey and cubes(水题)

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B. Timofey and cubes
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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.

  1. At the begining row was [2396714].
  2. After first operation row was [4176932].
  3. After second operation row was [4396712].
  4. After third operation row was [4376912].
  5. At fourth operation we reverse just middle element, so nothing has changed. The final row is [4376912]. So the answer for this case is row [2396714].



题目大意:对于长度为n的序列的位置i,若i


解题思路:看第一个样例解释就很清楚了,实际上就是从两头开始往中间两两进行变换,注意变换的奇偶次数决定这对数最终是否有变换到。


/* ***********************************************
┆  ┏┓   ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃       ┃ ┆
┆┃   ━   ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃       ┃ ┆
┆┃   ┻   ┃ ┆
┆┗━┓ 马 ┏━┛ ┆
┆  ┃ 勒 ┃  ┆      
┆  ┃ 戈 ┗━━━┓ ┆
┆  ┃ 壁     ┣┓┆
┆  ┃ 的草泥马  ┏┛┆
┆  ┗┓┓┏━┳┓┏┛ ┆
┆   ┃┫┫ ┃┫┫ ┆
┆   ┗┻┛ ┗┻┛ ┆
************************************************ */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
#define   mem(x,v)      memset(x,v,sizeof(x))
typedef pair pii;
typedef pair pll;
typedef vector vi;
typedef vector vll;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}



int a[200005];


int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	//ios::sync_with_stdio(0);
	//cin.tie(0);
	int n = read();
	rep(i,1,n) a[i] = read();
	int num = 0;
	for(int i=1,j=n;i<=j;i++,j--)
    {
        num++;
        int ta = a[i], tb = a[j];
        if(num%2==1)a[i] = tb, a[j] = ta;
    }
	for(int i=1;i<=n;i++)
    {
        printf("%d",a[i]);
        if(i!=n)printf(" ");
    }
	return 0;
}


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