A. Favorite Sequence
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp has a favorite sequence a[1…n] consisting of n integers. He wrote it out on the whiteboard as follows:
he wrote the number a1 to the left side (at the beginning of the whiteboard);
he wrote the number a2 to the right side (at the end of the whiteboard);
then as far to the left as possible (but to the right from a1), he wrote the number a3;
then as far to the right as possible (but to the left from a2), he wrote the number a4;
Polycarp continued to act as well, until he wrote out the entire sequence on the whiteboard.
The beginning of the result looks like this (of course, if n≥4).
For example, if n=7 and a=[3,1,4,1,5,9,2], then Polycarp will write a sequence on the whiteboard [3,4,5,2,9,1,1].
You saw the sequence written on the whiteboard and now you want to restore Polycarp’s favorite sequence.
Input
The first line contains a single positive integer t (1≤t≤300) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains an integer n (1≤n≤300) — the length of the sequence written on the whiteboard.
The next line contains n integers b1,b2,…,bn (1≤bi≤109) — the sequence written on the whiteboard.
Output
Output t answers to the test cases. Each answer — is a sequence a that Polycarp wrote out on the whiteboard.
Example
inputCopy
6
7
3 4 5 2 9 1 1
4
9 2 7 1
11
8 4 3 1 2 7 8 7 9 4 2
1
42
2
11 7
8
1 1 1 1 1 1 1 1
outputCopy
3 1 4 1 5 9 2
9 1 2 7
8 2 4 4 3 9 1 7 2 8 7
42
11 7
1 1 1 1 1 1 1 1
Note
In the first test case, the sequence a matches the sequence from the statement. The whiteboard states after each step look like this:
[3]⇒[3,1]⇒[3,4,1]⇒[3,4,1,1]⇒[3,4,5,1,1]⇒[3,4,5,9,1,1]⇒[3,4,5,2,9,1,1].
本题就是在进行相关数据的读取以后，依据数列的形成顺序的关系，对于格式进行相关的输出即可
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
int a[800];
int b[800];
while(t--)
{
int n;
cin>>n;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
for(int i=1;i<=n/2;i++)
{
cout<<a[i]<<' ';
cout<<a[n-i+1]<<' ';
}
if(n&1)
cout<<a[n/2+1]<<endl;
else
cout<<endl;
}
//system("pause");
}