1008 circular right shift of array elements (20 points)

An array A There is N（>0） It's an integer , On the premise that no other array is allowed , Loop each integer to the right M（≥0） A place , the A The data in is from （A0​A1​⋯AN−1​） Transformation for （AN−M​⋯AN−1​A0​A1​⋯AN−M−1​）（ Last M The number loop moves to the front of M A place ）. If you need to consider the program to move data as little as possible , How to design a way to move ？

Input format :

Each input contains a test case , The first 1 Line input N（1≤N≤100） and M（≥0）; The first 2 Line input N It's an integer , Space between .

Output format :

In a row, the output loop shifts to the right M Sequence of integers after bits , Space between , There must be no extra spaces at the end of the sequence .

sample input :

6 2
1 2 3 4 5 6

sample output :

5 6 1 2 3 4

  Topic analysis ：

1. Use an intermediate variable to store the last item of the array

2. Let the last item of the array -1 Items move backward in turn

3. Let the first item = Intermediate variable

4. Output

The code is as follows ：

C Language ：

#include <stdio.h>

int main() {
	int a[101];
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 0; i < m; i++) {
		int t = a[n - 1];
		for (int j = n - 2; j >= 0; j--) {
			a[j + 1] = a[j];
		}
		a[0] = t;
	}
	printf("%d", a[0]);
	for (int i = 1; i < n; i++) {
		printf(" %d", a[i]);
	}
}

C++:

#include <iostream>
using namespace std;

int main() {
	int a[101];
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < m; i++) {
		int t = a[n - 1];
		for (int j = n - 2; j >= 0; j--) {
			a[j + 1] = a[j];
		}
		a[0] = t;
	}
	cout << a[0];
	for (int i = 1; i < n; i++) {
		cout << " " << a[i];
	}
}