#include <bits/stdc++.h>
using namespace std;
int main()
{
//write your code here
int t;
cin>>t;
while(t–)
{
int n,k,b;
cin>>n>>k;
int a[n];
if(n>k)
{
b=k;
}
else
{
b=k%n;
}
for(int i=0;i<n;i++)
{
cin>>a[i];
}
if(n%k==0)
{
for(int i=0;i<n;i++)
{
cout<<a[i]<<" “;
}
}
else
{
while(b–)
{
int t=a[n-1];
for(int i=n-1;i>0;i–)
{
a[i]=a[i-1];
}
a[0]=t;
}
for(int i=0;i<n;i++)
{
cout<<a[i]<<” ";
}
}
cout<<endl;
}
return 0;
}
just take an integer i=0
if you have to rotate by k times
make i = n-k
because u eventually need to start printing from n-k digit
then just print n digits and use a[i%n] each time.
so in example
i=n-k = 3
a[i] = 4
a[4%5 ] = 5
a[5%5] = 1
a[6%5 ] = 2
a[7%5 ] = 3
what u got in the end = [4,5,1,2,3]
also make k=k%n before doing all that, because by rotating array of length n - n times, we get back the original array.