数组元素的总和是给定数字的倍数

给定一个由正整数和整数N组成的数组arr[] ，任务是找到所有数组元素的总和，这些元素是N 的倍数

例子：

Input: arr[] = {1, 2, 3, 5, 6}, N = 3
Output: 9
Explanation: From the given array, 3 and 6 are multiples of 3. Therefore, sum = 3 + 6 = 9.

Input: arr[] = {1, 2, 3, 5, 7, 11, 13}, N = 5
Output: 5

方法：这个想法是遍历数组，对于每个数组元素，检查它是否是N的倍数并添加这些元素。请按照以下步骤解决问题：

1. 初始化一个变量，比如sum ，以存储所需的总和。
2. 遍历给定的数组并对每个数组元素执行以下操作。
3. 检查数组元素是否是N的倍数。
4. 如果元素是N的倍数，则将该元素添加到sum 。
5. 最后，打印sum的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the sum of array
// elements which are multiples of N
void mulsum(int arr[], int n, int N)
{
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++) {
 
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0) {
            sum = sum + arr[i];
        }
    }
 
    // Print total sum
    cout << sum;
}
 
// Driver Code
int main()
{
 
    // Given arr[]
    int arr[] = { 1, 2, 3, 5, 6 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int N = 3;
 
    mulsum(arr, n, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
 
// Function to find the sum of array
// elements which are multiples of N
static void mulsum(int arr[], int n, int N)
{
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++)
    {
 
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0)
        {
            sum = sum + arr[i];
        }
    }
 
    // Print total sum
    System.out.println(sum);
}
 
 
// Driver Code
public static void main(String[] args)
{
     
    // Given arr[]
    int arr[] = { 1, 2, 3, 5, 6 };
    int n = arr.length;
    int N = 3;
    mulsum(arr, n, N);
}
}
 
// This code is contributed by jana_sayantan.

Python

# Python3 program for the above approach
  
# Function to find the sum of array
# elements which are multiples of N
def mulsum(arr, n, N):
      
    # Stores the sum
    sums = 0
  
    # Traverse the array
    for i in range(0, n):
        if arr[i] % N == 0:
              sums = sums + arr[i]
  
    # Print total sum
    print(sums)
  
# Driver Code
if __name__ == "__main__":
  
    # Given arr[]
    arr = [ 1, 2, 3, 5, 6 ]
  
    n = len(arr)
     
    N = 3
  
    # Function call
    mulsum(arr, n, N)

C#

// C# program for the above approach
using System;
public class GFG
{
 
// Function to find the sum of array
// elements which are multiples of N
static void mulsum(int[] arr, int n, int N)
{
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++)
    {
 
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0)
        {
            sum = sum + arr[i];
        }
    }
 
    // Print total sum
    Console.Write(sum);
}
 
// Driver Code
static public void Main ()
{
    // Given arr[]
    int[] arr = { 1, 2, 3, 5, 6 };
    int n = arr.Length;
    int N = 3;
    mulsum(arr, n, N);
}
}
 
// This code is contributed by Dharanendra L V.

Javascript

 
// JavaScript program for the above approach
 
// Function to find the sum of array
// elements which are multiples of N
function mulsum(arr, n, N)
{
     
    // Stores the sum
    var sum = 0;
     
    // Traverse the given array
    for(var i = 0; i < n; i++)
    {
         
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0)
        {
            sum = sum + arr[i];
        }
    }
     
    // Print total sum
    document.write(sum);
}
 
// Driver Code
 
// Given arr[]
var arr = [ 1, 2, 3, 5, 6 ];
var n = arr.length;
var N = 3;
 
mulsum(arr, n, N);
 
// This code is contributed by rdtank
 

输出：

9

时间复杂度： O(N)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live