所有 Array 元素的数字乘积之和

给定一个数组arr ，任务是找到所有数组元素的数字乘积之和

例子：

Input: arr[]={11, 23, 41}
Output: 11
Explanation: 1*1 + 2*3 + 4*1 = 1 + 6 + 4 = 1111

Input: arr[]={46, 32, 78, 0}
Output: 86 

方法：为了解决这个问题，找到所有数字的数字的乘积，然后将它们相加。请按照以下步骤解决此问题：

1. 创建一个函数findProduct ，它将获取一个数字并找到其数字的乘积。
2. 创建一个变量sum来存储最终答案并将其初始化为 0。
3. 现在，遍历数组和每个元素：
   • 将其传递给函数findProduct并获取其数字的乘积。
   • 然后将它的 digit 乘积加到sum中。
4. 返回总和作为最终答案。

下面是上述方法的实现：

C++

// C++ code for the above approach
  
#include 
using namespace std;
  
// Function to find the product of the
// digits of a number N
int findProduct(int N)
{
    if (N == 0) {
        return 0;
    }
  
    int product = 1;
    while (N > 0) {
        product *= (N % 10);
        N /= 10;
    }
  
    return product;
}
  
// Function to find the sum of the product of
// digits of all array elements
int sumOfProduct(vector arr)
{
  
    int sum = 0;
    for (auto x : arr) {
        sum += findProduct(x);
    }
  
    return sum;
}
  
// Driver Code
int main()
{
    vector arr = { 46, 32, 78, 0 };
    cout << sumOfProduct(arr);
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {
  
// Function to find the product of the
// digits of a number N
static int findProduct(int N)
{
    if (N == 0) {
        return 0;
    }
  
    int product = 1;
    while (N > 0) {
        product *= (N % 10);
        N /= 10;
    }
  
    return product;
}
  
// Function to find the sum of the product of
// digits of all array elements
static int sumOfProduct(int []arr)
{
  
    int sum = 0;
    for (int i = 0; i < arr.length; i++) {
        sum += findProduct(arr[i]);
    }
  
    return sum;
}
  
// Driver code
public static void main(String args[])
{
    int []arr = { 46, 32, 78, 0 };
    System.out.println(sumOfProduct(arr));
  
}
}
  
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
  
# Function to find the product of the
# digits of a number N
def findProduct(N):
    if (N == 0):
        return 0
  
    product = 1
    while (N > 0):
        product *= (N % 10)
        N = N // 10
    return product
  
# Function to find the sum of the product of
# digits of all array elements
def sumOfProduct(arr):
  
    sum = 0
    for x in arr:
        sum += findProduct(x)
  
    return sum
  
# Driver Code
arr = [46, 32, 78, 0]
print(sumOfProduct(arr))
  
# This code is contributed by Saurabh Jaiswal

C#

// C# code to implement above approach
using System;
class GFG {
  
// Function to find the product of the
// digits of a number N
static int findProduct(int N)
{
    if (N == 0) {
        return 0;
    }
  
    int product = 1;
    while (N > 0) {
        product *= (N % 10);
        N /= 10;
    }
  
    return product;
}
  
// Function to find the sum of the product of
// digits of all array elements
static int sumOfProduct(int []arr)
{
  
    int sum = 0;
    for (int i = 0; i < arr.Length; i++) {
        sum += findProduct(arr[i]);
    }
  
    return sum;
}
  
// Driver code
public static void Main()
{
    int []arr = { 46, 32, 78, 0 };
    Console.Write(sumOfProduct(arr));
  
}
}
  
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach
  
    // Function to find the product of the
    // digits of a number N
    function findProduct(N) {
      if (N == 0) {
        return 0;
      }
  
      let product = 1;
      while (N > 0) {
        product *= (N % 10);
        N = Math.floor(N / 10)
      }
      return product;
    }
  
    // Function to find the sum of the product of
    // digits of all array elements
    function sumOfProduct(arr) {
  
      let sum = 0;
      for (let x of arr) {
        sum += findProduct(x);
      }
  
      return sum;
    }
  
    // Driver Code
    let arr = [46, 32, 78, 0];
    document.write(sumOfProduct(arr));
  
  // This code is contributed by Potta Lokesh
  

输出：

86

时间复杂度： O(NlogM)，其中 N 是数组的大小，M 是数组中的最大数
辅助空间： O(1)