array rotation

PROBLEM :

Given an array of N size. You have to rotate array by d elements.

Input:

The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of three lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements. The third line of each test case contains "d" .

Output:

Corresponding to each test case, in a new line, print the modified array.

Constraints:

1 = T = 100
1 = N = 200
1 = A[i] = 1000

Example:

Input

1
5
1 2 3 4 5
2

Output
3 4 5 1 2

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
int main()
 {
int t,no,a[100],i,j,k,temp  ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
   cin>>a[i] ;
 
   cin>>k ;
 
   for(i=0;i<k;i++)
   {
       for(j=0;j<no-1;j++)
       {
           temp=a[j] ;
           a[j]=a[j+1] ;
           a[j+1]=temp ;
       }
   }
 
   for(i=0;i<no;i++)
   cout<<a[i]<<" " ;
   cout<<endl ;
}
return 0;
}

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BETTER SOLUTION :
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#include<iostream>
using namespace std;
void rotate_array(int [],int ) ;
int main()
 {
int t,no,a[200],i,k ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
   cin>>a[i] ;
 
   cin>>k ;
 
   for(i=0;i<k;i++)
       rotate_array(a,no) ;
 
   for(i=0;i<no;i++)
   cout<<a[i]<<" " ;
 
   cout<<endl ;
}
return 0;
}

void rotate_array(int a[],int no)
{
    int i,ele ;
    ele=a[0] ;
   
    for(i=0;i<no-1;i++)
        a[i]=a[i+1] ;
       
    a[no-1]=ele ;
}

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BEST SOLUTION :
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#include<iostream>
using namespace std;
void reverse_array(int [],int ,int ) ;
int main()
 {
int t,no,a[200],i,k ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
       cin>>a[i] ;
     
  cin>>k ;
 
  reverse_array(a,0,k-1) ;
  reverse_array(a,k,no-1) ;
  reverse_array(a,0,no-1) ;
 
  for(i=0;i<no;i++)
  cout<<a[i]<<" " ;
 
  cout<<endl ;
}
return 0;
}

void reverse_array(int a[],int start,int end)
{
    int temp ;
    while(start<end)
    {
        temp=a[start] ;
        a[start]=a[end] ;
        a[end]=temp ;
       
        start++ ;
        end-- ;
    }
}

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