可被它们的乘积或数字总和整除的数组元素的计数

给定一个数组arr[] 。任务是计算数组中可以被数字乘积或数字总和整除的元素。

例子：

Input: arr[] = {123, 25, 36, 7}
Output: 2
Explanation: Following are the elements that follows the given conditions
Sum of digits of 123 is 6 and the product of digits is also 6. 
Since 123 is not divisible by 6, so it is not counted. 
The sum of digits of 36 is 9 and product of digits is 18. Since 36 is divisible by both, so it is considered in answer. 
Similarly, sum of digits and product of digits of 7 is 7 itself, hence it is also considered.
Hence, 2 is the final answer. 

Input: arr[] = {10, 22, 15}
Output: 2
Explanation:  The sum of digits in 10 is 1, and 10 is divisible by 1, hence we consider it. The product of digits of 15 is 5, and 15 is divisible by 5, we consider it in our output.

方法：这个问题是基于观察的。请按照以下步骤解决给定的问题。

• 声明一个变量说count = 0 ，以存储满足给定条件的元素的数量。
• 对于数组arr[]中的每个数字，求其数字之和及其数字之积。
• 检查数字是否可以被数字之和或数字的乘积整除。
• 如果是，请将其添加到您的计数变量中。
• 返回计数作为最终答案。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find the number of elements
// in arr[] that follows the given conditions
int solve(int arr[], int n)
{
 
    // To count the number of elements
    // that follow the given conditions
    int count = 0;
 
    // Iterate through the array arr[]
    for (int i = 0; i < n; i++) {
        int p = arr[i];
        int sum = 0, prod = 1;
        while (p > 0) {
            int r = p % 10;
            sum += r;
            prod *= r;
            p /= 10;
        }
 
        // Check if condition satisfies or not
        if ((sum > 0 && arr[i] % sum == 0)
            || (prod > 0 && arr[i] % prod == 0))
            count++;
    }
 
    // Return count as the final answer
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 123, 25, 36, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << solve(arr, N) << endl;
 
    return 0;
}

Java

// Java program for above approach
import java.util.*;
public class GFG {
 
  // Function to find the number of elements
  // in arr[] that follows the given conditions
  static int solve(int[] arr, int n)
  {
 
    // To count the number of elements
    // that follow the given conditions
    int count = 0;
 
    // Iterate through the array arr[]
    for (int i = 0; i < n; i++) {
      int p = arr[i];
      int sum = 0, prod = 1;
      while (p > 0) {
        int r = p % 10;
        sum += r;
        prod *= r;
        p /= 10;
      }
 
      // Check if condition satisfies or not
      if ((sum > 0 && arr[i] % sum == 0)
          || (prod > 0 && arr[i] % prod == 0))
        count++;
    }
 
    // Return count as the final answer
    return count;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int[] arr = { 123, 25, 36, 7 };
    int N = arr.length;
 
    // Function Call
    System.out.println(solve(arr, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 program for above approach
 
# Function to find the number of elements
# in arr[] that follows the given conditions
def solve(arr, n):
 
        # To count the number of elements
        # that follow the given conditions
    count = 0
 
    # Iterate through the array arr[]
    for i in range(0, n):
        p = arr[i]
        sum = 0
        prod = 1
        while (p > 0):
            r = p % 10
            sum += r
            prod *= r
            p //= 10
 
            # Check if condition satisfies or not
        if ((sum > 0 and arr[i] % sum == 0)
                or (prod > 0 and arr[i] % prod == 0)):
            count += 1
 
        # Return count as the final answer
    return count
 
# Driver Code
if __name__ == "__main__":
 
    arr = [123, 25, 36, 7]
    N = len(arr)
 
    # Function Call
    print(solve(arr, N))
 
    # This code is contributed by rakeshsahni

C#

// C# program for above approach
using System;
class GFG {
 
  // Function to find the number of elements
  // in arr[] that follows the given conditions
  static int solve(int[] arr, int n)
  {
 
    // To count the number of elements
    // that follow the given conditions
    int count = 0;
 
    // Iterate through the array arr[]
    for (int i = 0; i < n; i++) {
      int p = arr[i];
      int sum = 0, prod = 1;
      while (p > 0) {
        int r = p % 10;
        sum += r;
        prod *= r;
        p /= 10;
      }
 
      // Check if condition satisfies or not
      if ((sum > 0 && arr[i] % sum == 0)
          || (prod > 0 && arr[i] % prod == 0))
        count++;
    }
 
    // Return count as the final answer
    return count;
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = { 123, 25, 36, 7 };
    int N = arr.Length;
 
    // Function Call
    Console.Write(solve(arr, N));
  }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript program for above approach
 
// Function to find the number of elements
// in arr[] that follows the given conditions
function solve(arr, n)
{
 
    // To count the number of elements
    // that follow the given conditions
    let count = 0;
 
    // Iterate through the array arr[]
    for (let i = 0; i < n; i++) {
        let p = arr[i];
        let sum = 0, prod = 1;
        while (p > 0) {
            let r = p % 10;
            sum += r;
            prod *= r;
            p = Math.floor(p / 10);
        }
 
        // Check if condition satisfies or not
        if ((sum > 0 && arr[i] % sum == 0)
            || (prod > 0 && arr[i] % prod == 0))
            count++;
    }
 
    // Return count as the final answer
    return count;
}
 
// Driver Code
let arr = [123, 25, 36, 7 ];
let N = arr.length;
 
    // Function Call
document.write(solve(arr, N));
 
// This code is contributed by gfgking.

输出

2

时间复杂度： O(N)
辅助空间： O(1)