给定的大小为N []数组ARR和整数K，一个正的，则任务是通过重复地选择一对索引(I，J)，使用K阵列整除的每个元素，并执行改编[I] = ARR [I ] + arr [j] 。
例子：
Input: arr[] = {3, 6, 3}, K = 6
Output: True
Explanation:
Update arr[0] = arr[0] + arr[0].
Update arr[2] = arr[2] + arr[2]
Input: arr[] = {6, 6, 8, 7, 3}, K = 6
Output: False
方法：请按照以下步骤解决问题：
- 遍历数组arr [] ：
- 检查arr [i]是否可被K整除：
- 初始化一个变量，例如temp，以在每次迭代后存储arr [i]的值。
- 重复将arr [i]乘以2，直到arr [i]
。 - 检查arr [i]％K！= 0 。如果发现为真，则打印False并返回。
- 检查arr [i]是否可被K整除：
- 完成上述步骤后，将True打印，因为所有元素现在都可以被K整除。
下面是上述方法的实现：
C++
// C++ Program for the above approach
#include
using namespace std;
// Utility function to check if all array
// elements can be made divisible by K or not
bool makeArrayDivisibleByKUtil(
int arr[], int N, int K)
{
// Traverse the array
for (int i = 0; i < N; i++) {
// Check whether arr[i] is
// divisible by K or not
if (arr[i] % K != 0) {
// Stores value of arr[i]
int temp = arr[i];
while (arr[i] < K * temp) {
// Multiply by 2 until it
// is less than K * temp
arr[i] = arr[i] * 2;
}
if (arr[i] % K != 0) {
return false;
}
}
}
// Return true as all array
// elements are divisible by K
return true;
}
// Function to check whether all array
// elements can be made divisible by K or not
void makeArrayDivisibleByK(
int arr[], int N, int K)
{
if (makeArrayDivisibleByKUtil(arr, N, K)) {
cout << "True" << endl;
}
else {
cout << "False" << endl;
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 3, 6, 3 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Given value of K
int K = 6;
makeArrayDivisibleByK(arr, N, K);
return 0;
} C
#include
// Utility function to check if all array
// elements can be made divisible by K or not
int makeArrayDivisibleByKUtil(
int arr[], int N, int K)
{
// Traverse the array
for (int i = 0; i < N; i++)
{
// Check whether arr[i] is
// divisible by K or not
if (arr[i] % K != 0)
{
// Stores value of arr[i]
int temp = arr[i];
while (arr[i] < K * temp)
{
// Multiply by 2 until it
// is less than K * temp
arr[i] = arr[i] * 2;
}
if (arr[i] % K != 0)
{
return 0;
}
}
}
// Return true as all array
// elements are divisible by K
return 1;
}
// Function to check whether all array
// elements can be made divisible by K or not
void makeArrayDivisibleByK(
int arr[], int N, int K)
{
if (makeArrayDivisibleByKUtil(arr, N, K))
{
printf("True\n");
}
else
{
printf("False\n");
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 3,6,3 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Given value of K
int K = 6;
makeArrayDivisibleByK(arr, N, K);
return 0;
}
// This code is contributed by santoshcharan. Java
import java.io.*;
class GFG
{
public static boolean makeArrayDivisibleBykUtil(int arr[],
int n,int k)
{
// Traverse the array
for(int i = 0; i < n; i++)
{
// Check whether arr[i] is
// divisible by k or not
if(arr[i] % k != 0)
{
// Stores value of arr[i]
int temp = arr[i];
while(arr[i] < k * arr[i])
{
// Multiply by 2 until it
// is less than k*temp
arr[i] = arr[i] * 2;
}
if(arr[i] % k != 0)
{
return false;
}
}
}
// Return true as all array
// elements are divisible by k
return true;
}
public static void makeArrayDivisibleByk(int arr[],
int n,int k)
{
if(makeArrayDivisibleBykUtil(arr, n, k))
{
System.out.println("True");
}
else
{
System.out.println("False");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 3 , 6 , 3};
// Size of the array
int n = 3;
int k = 6;
// Calling the function
makeArrayDivisibleByk(arr, n, k);
}
}
// This code is contributed by santoshcharan.Python3
# Python 3 Program for the above approach
# Utility function to check if all array
# elements can be made divisible by K or not
def makeArrayDivisibleByKUtil(arr, N, K):
# Traverse the array
for i in range(N):
# Check whether arr[i] is
# divisible by K or not
if (arr[i] % K != 0):
# Stores value of arr[i]
temp = arr[i]
while (arr[i] < K * temp):
# Multiply by 2 until it
# is less than K * temp
arr[i] = arr[i] * 2
if (arr[i] % K != 0):
return False
# Return true as all array
# elements are divisible by K
return True
# Function to check whether all array
# elements can be made divisible by K or not
def makeArrayDivisibleByK(arr, N, K):
if (makeArrayDivisibleByKUtil(arr, N, K)):
print("True")
else:
print("False")
# Driver Code
if __name__ == '__main__':
# Given array
arr = [3, 6, 3]
# Size of the array
N = len(arr)
# Given value of K
K = 6
makeArrayDivisibleByK(arr, N, K)
# This code is contributed by bgangwar59.C#
using System;
class GFG
{
public static bool makeArrayDivisibleBykUtil(int [] arr,
int n,int k)
{
// Traverse the array
for(int i = 0; i < n; i++)
{
// Check whether arr[i] is
// divisible by k or not
if(arr[i] % k != 0)
{
// Stores value of arr[i]
//int temp = arr[i];
while(arr[i] < k * arr[i])
{
// Multiply by 2 until it
// is less than k*temp
arr[i] = arr[i] * 2;
}
if(arr[i] % k != 0)
{
return false;
}
}
}
// Return true as all array
// elements are divisible by k
return true;
}
public static void makeArrayDivisibleByk(int []arr,
int n,int k)
{
if(makeArrayDivisibleBykUtil(arr, n, k))
{
Console.WriteLine("True");
}
else
{
Console.WriteLine("False");
}
}
// Driver Code
public static void Main(string[] args)
{
// Given array
int[] arr = { 3 , 6 , 3};
// Size of the array
int n = 3;
int k = 6;
// Calling the function
makeArrayDivisibleByk(arr, n, k);
}
}
// This code is contributed by chitranayal.输出：
True时间复杂度： O(N * logN)
辅助空间： O(1)