可被给定数 K 整除的数组中所有元素的总和

给定一个包含 N 个元素和一个数字 K 的数组。任务是找到所有可以被 K 整除的元素的总和。
例子：

Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
        K = 3
Output : 30
Explanation: As 15, 9, 6 are divisible by 3. So, sum of elements divisible by K = 15 + 9 + 6 = 30.

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
        K = 2
Output : 20

思路是遍历数组，一一检查元素。如果一个元素可被 K 整除，则将该元素的值与到目前为止的总和相加，并在到达数组末尾时继续此过程。
下面是上述方法的实现：

C++

// C++ program to find sum of all the elements
// in an array divisible by a given number K
 
#include 
using namespace std;
 
// Function to find sum of all the elements
// in an array divisible by a given number K
int findSum(int arr[], int n, int k)
{
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }
 
    // Return calculated sum
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    cout << findSum(arr, n, k);
 
    return 0;
}

Java

// Java program to find sum of all the elements
// in an array divisible by a given number K
 
import java.io.*;
 
class GFG {
 
// Function to find sum of all the elements
// in an array divisible by a given number K
static int findSum(int arr[], int n, int k)
{
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }
 
    // Return calculated sum
    return sum;
}
 
    // Driver code
    public static void main (String[] args) {
     
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.length;
    int k = 3;
 
    System.out.println( findSum(arr, n, k));
    }
}
 
// this code is contributed by anuj_67..

Python3

# Python3 program to find sum of
# all the elements in an array
# divisible by a given number K
 
# Function to find sum of all
# the elements in an array
# divisible by a given number K
def findSum(arr, n, k) :
 
    sum = 0
 
    # Traverse the array
    for i in range(n) :
 
        # If current element is divisible
        # by k add it to sum
        if arr[i] % k == 0 :
 
            sum += arr[i]
 
    # Return calculated sum
    return sum
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 15, 16, 10, 9, 6, 7, 17]
    n = len(arr)
    k = 3
 
    print(findSum(arr, n, k))
 
# This code is contributed by ANKITRAI1

C#

// C# program to find sum of all the elements
// in an array divisible by a given number K
 
using System;
 
public class GFG{
 
// Function to find sum of all the elements
// in an array divisible by a given number K
static int findSum(int []arr, int n, int k)
{
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }
 
    // Return calculated sum
    return sum;
}
 
    // Driver code
     
    static public void Main (){
     
    int []arr = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.Length;
    int k = 3;
 
    Console.WriteLine( findSum(arr, n, k));
    }
}
//This code is contributed by anuj_67..

PHP


Javascript

输出： 30

时间复杂度：O(N)，其中 N 是数组中元素的数量。辅助空间： O(1)