K 次操作后打印数组
给定一个大小为 N 的数组 arr[] 和一个数字 K 。任务是在 K 操作后打印数组 arr[] ,以便在每次操作时,将数组中的每个元素 arr[i] 替换为max–arr[I],其中 max 是数组中的最大元素。 举例:
输入: arr[] = {4,8,12,16},K = 4 输出: 0 4 8 12 解释: 对于给定的数组 arr[] = {4,8,12,16 }。对于每个操作 K,数组的变化如下: { 12,8,4,0 }–K = 1 { 0,4,8,12 }–K = 2 { 12,8,4,0 }–K = 3 { 0,4,8,12 }–K = 4 输入: arr[] = {8,0,3,5},K = 3 输出: 0 每个操作 K 的数组变化如下: { 0,8,5,3 }–K = 1 { 8,0,3,5 }–K = 2 { 0,8,5,3 }–K = 3
进场:思路是每走一步后清晰观察阵面。
- 最初,因为我们从数组中减去最大元素,所以我们可以确定数组中至少有一个元素的值为零。这发生在第一步 K = 1 之后。让这一步后的阵为 A[] 带最大元素 M 。
- 在此第一步之后,阵列变得停滞,并且值交替地从其值 A[i] 变为M–A[I]。
- 例如,让我们取数组 arr[] = {4,8,12,16}。在这个数组中,最大值是 16,让我们假设 K = 4。
- 第一步,(即)对于 K = 1 ,数组简化为 {12,8,4,0} 。也就是说,每个元素 arr[i]都被替换为 16–arr[I]。设这个数组是 A[],这个数组 M 的最大元素是 12。
- 第一步后,该数组中的每个元素 A[i] 在 A[i] 和12–A[I]之间交替变化。也就是说,对于 K = 2 ,数组现在变成 {0,4,8,12} 。
- 同样,对于第三步,(即) K = 3 ,数组再次变为 {12,8,4,0} ,与 K = 1 相同。
- 而对于第四步,(即) K = 4 ,数组变为 {0,4,8,12} ,与 K = 2 等相同。
所以可以得出结论,我们只需要检查 K 是奇数还是偶数:
- 如果 K 是奇数:将每个元素arr【I】替换为arr【I】–min,其中 min 是数组中的最小元素。
- 如果 K 是偶数:将每个元素arr【I】替换为max–arr【I】,其中 max 是数组中最大的元素。
以下是上述方法的实现:
卡片打印处理机(Card Print Processor 的缩写)
// C++ program to print the array
// after K operations
#include <bits/stdc++.h>
using namespace std;
// Function to print the array
// after K operations
void printArray(int A[], int n, int K)
{
// Variables to store the minimum and
// the maximum elements of the array
int minEle = INT_MAX,
maxEle = INT_MIN;
// Loop to find the minimum and the
// maximum elements of the array
for (int i = 0; i < n; i++) {
minEle = min(minEle, A[i]);
maxEle = max(maxEle, A[i]);
}
// If K is not equal to 0
if (K != 0) {
// If K is odd
if (K % 2 == 1) {
// Replace every element with
// max - arr[i]
for (int i = 0; i < n; i++)
A[i] = maxEle - A[i];
}
// If K is even
else {
// Replace every element with
// A[i] - min
for (int i = 0; i < n; i++)
A[i] = A[i] - minEle;
}
}
// Printing the array after K operations
for (int i = 0; i < n; i++)
cout << A[i] << " ";
}
// Driver code
int main()
{
int arr[] = { 4, 8, 12, 16 };
int K = 4;
int N = sizeof(arr) / sizeof(arr[0]);
printArray(arr, N, K);
return 0;
}
Java 语言(一种计算机语言,尤用于创建网站)
// Java program to print the array
// after K operations
import java.io.*;
class GFG {
// Function to print the array
// after K operations
static void printArray(int[] A, int n, int K)
{
// Variables to store the minimum and
// the maximum elements of the array
int minEle = Integer.MAX_VALUE,
maxEle = Integer.MAX_VALUE;
// Loop to find the minimum and the
// maximum elements of the array
for (int i = 0; i < n; i++) {
minEle = Math.min(minEle, A[i]);
maxEle = Math.max(maxEle, A[i]);
}
// If K is not equal to 0
if (K != 0) {
// If K is odd
if (K % 2 == 1) {
// Replace every element with
// max - arr[i]
for (int i = 0; i < n; i++)
A[i] = maxEle - A[i];
}
// If K is even
else {
// Replace every element with
// A[i] - min
for (int i = 0; i < n; i++)
A[i] = A[i] - minEle;
}
}
// Printing the array after K operations
for (int i = 0; i < n; i++)
System.out.print(A[i] + " ");
}
// Driver Code
public static void main (String[] args)
{
int[] arr = { 4, 8, 12, 16 };
int K = 4;
int N = arr.length;
printArray(arr, N, K);
}
}
// This code is contributed by shivanisinghss2110
Python 3
# Python3 program to print the array
# after K operations
# Function to print the array
# after K operations
def printArray(A, n, K):
# Variables to store the minimum and
# the maximum elements of the array
minEle = 10**9
maxEle = -10**9
# Loop to find the minimum and the
# maximum elements of the array
for i in range(n):
minEle = min(minEle, A[i])
maxEle = max(maxEle, A[i])
# If K is not equal to 0
if (K != 0):
# If K is odd
if (K % 2 == 1):
# Replace every element with
# max - arr[i]
for i in range(n):
A[i] = maxEle - A[i]
# If K is even
else:
# Replace every element with
# A[i] - min
for i in range(n):
A[i] = A[i] - minEle
# Printing the array after K operations
for i in A:
print(i, end=" ")
# Driver code
if __name__ == '__main__':
arr=[4, 8, 12, 16]
K = 4
N = len(arr)
printArray(arr, N, K)
# This code is contributed by mohit kumar 29
C
// C# program to print the array
// after K operations
using System;
class GFG{
// Function to print the array
// after K operations
static void printArray(int[] A, int n, int K)
{
// Variables to store the minimum and
// the maximum elements of the array
int minEle = Int32.MaxValue,
maxEle = Int32.MinValue;
// Loop to find the minimum and the
// maximum elements of the array
for (int i = 0; i < n; i++) {
minEle = Math.Min(minEle, A[i]);
maxEle = Math.Max(maxEle, A[i]);
}
// If K is not equal to 0
if (K != 0) {
// If K is odd
if (K % 2 == 1) {
// Replace every element with
// max - arr[i]
for (int i = 0; i < n; i++)
A[i] = maxEle - A[i];
}
// If K is even
else {
// Replace every element with
// A[i] - min
for (int i = 0; i < n; i++)
A[i] = A[i] - minEle;
}
}
// Printing the array after K operations
for (int i = 0; i < n; i++)
Console.Write(A[i] + " ");
}
// Driver code
static public void Main ()
{
int[] arr = { 4, 8, 12, 16 };
int K = 4;
int N = arr.Length;
printArray(arr, N, K);
}
}
// This code is contributed by shubhamsingh10
java 描述语言
<script>
// Javascript program to print the array
// after K operations
// Function to print the array
// after K operations
function printArray(A, n, K)
{
// Variables to store the minimum and
// the maximum elements of the array
var minEle = 100000000, maxEle = -100000000;
// Loop to find the minimum and the
// maximum elements of the array
for (var i = 0; i < n; i++) {
minEle = Math.min(minEle, A[i]);
maxEle = Math.max(maxEle, A[i]);
}
// If K is not equal to 0
if (K != 0) {
// If K is odd
if (K % 2 == 1) {
// Replace every element with
// max - arr[i]
for (var i = 0; i < n; i++)
A[i] = maxEle - A[i];
}
// If K is even
else {
// Replace every element with
// A[i] - min
for (var i = 0; i < n; i++)
A[i] = A[i] - minEle;
}
}
// Printing the array after K operations
for (var i = 0; i < n; i++)
document.write(A[i] + " ");
}
// Driver code
arr = [ 4, 8, 12, 16 ];
var K = 4;
var N = arr.length;
printArray(arr, N, K);
</script>
Output:
0 4 8 12
时间复杂度: O(N),其中 N 是数组的大小。
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