列表做法，时间复杂度为\(O(nk)\)

def MoveRight():
    n = int(input())
    nums = [int(it) for it in input().split()]
    k = int(input())

    for i in range(k):
        tmp = nums[-1]
        nums.pop()
        nums.insert(0, tmp)

    print(*nums)

if __name__ == '__main__':
    MoveRight()

使用双端队列，时间复杂度为\(O(k)\)

from collections import deque

def MoveRight():
    n = int(input())
    nums = deque([int(it) for it in input().split()])
    k = int(input())

    for i in range(k):
        tmp = nums[-1]
        nums.pop()
        nums.appendleft(tmp)

    print(*nums)

if __name__ == '__main__':
    MoveRight()

使用双端队列的内置函数rotate，该函数的底层实现就是第二个程序，故时间复杂度也为\(O(k)\)

from collections import deque

def MoveRight():
    n = int(input())
    nums = deque([int(it) for it in input().split()])
    k = int(input())

    nums.rotate(k)

    print(*nums)

if __name__ == '__main__':
    MoveRight()

#include <stdio.h>
void MoveRight(int a[],int n);
void Output(int a[],int n);
int main()
{
int a[10000],n,k,i;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
scanf("%d",&k);
for(i=0;i<k;i++)
MoveRight(a,n);
Output(a,n);
return 0;
}

void MoveRight(int a[],int n)
{
int i,temp;
temp=a[n-1];
for(i=n-2;i>=0;i--)
a[i+1]=a[i];
a[0]=temp;
}
void Output(int a[],int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",a[i]);
}

#include<iostream>
using namespace std;

class Array
{
private:
    int n;
    int size;
    int *a;
public:
    Array(int set_n)
    {
        n=set_n;
        size=n*2;
        a=new int[size];
    }

    void set_array()
    {
        for(int i=0;i<n;i++)
            cin>>a[i];
    }

    void output()
    {
        for(int i=0;i<n;i++)
            cout<<a[i]<<" ";
        cout<<endl;
    }

    void move_to_right(int k)
    {
        int tmp=0;
        for(int i=0;i<k;i++)
        {
            tmp=a[n-1];
            for(int i=n-1;i>=1;i--)
            {
                a[i]=a[i-1];
            }
            a[0]=tmp;
        }
    }

};
int main()
{
    int n,k;
    cin>>n;
    Array arr(n);
    arr.set_array();
    //arr.output();
    cin>>k;
    arr.move_to_right(k);
    arr.output();

    system("pause");
    return 0;
}