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📜  可被给定数 K 整除的数组中所有元素的乘积

📅  最后修改于: 2022-05-13 01:57:49.084000             🧑  作者: Mango

可被给定数 K 整除的数组中所有元素的乘积

给定一个包含 N 个元素和一个数字 K 的数组。任务是找到数组中所有可以被 K 整除的元素的乘积。

例子

Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
        K = 3
Output : 810

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
        K = 2
Output : 384 

思路是遍历数组,一一检查元素。如果一个元素可以被 K 整除,则将该元素的值乘以到目前为止的乘积,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 Product of all the elements
// in an array divisible by a given number K
 
#include 
using namespace std;
 
// Function to find Product of all the elements
// in an array divisible by a given number K
int findProduct(int arr[], int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    cout << findProduct(arr, n, k);
 
    return 0;
}


Java
// Java program to find Product of all the elements
// in an array divisible by a given number K
 
import java.io.*;
 
class GFG {
 
// Function to find Product of all the elements
// in an array divisible by a given number K
static int findProduct(int arr[], int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// 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(findProduct(arr, n, k));
    }
}
 
 
// This code is contributed by inder_verma..


Python3
# Python3 program to find Product of all
# the elements in an array divisible by
# a given number K
 
# Function to find Product of all the elements
# in an array divisible by a given number K
def findProduct(arr, n, k):
 
    prod = 1
 
    # Traverse the array
    for i in range(n):
 
        # If current element is divisible
        # by k, multiply with product so far
        if (arr[i] % k == 0):
            prod *= arr[i]
 
    # Return calculated product
    return prod
 
# Driver code
if __name__ == "__main__":
 
    arr= [15, 16, 10, 9, 6, 7, 17 ]
    n = len(arr)
    k = 3
 
    print (findProduct(arr, n, k))
 
# This code is contributed by ita_c


C#
// C# program to find Product of all
// the elements in an array divisible
// by a given number K
using System;
 
class GFG
{
 
// Function to find Product of all
// the elements in an array divisible
// by a given number K
static int findProduct(int []arr, int n, int k)
{
    int prod = 1;
 
    // Traverse the array
    for (int i = 0; i < n; i++)
    {
 
        // If current element is divisible
        // by k multiply with product so far
        if (arr[i] % k == 0)
        {
            prod *= arr[i];
        }
    }
 
    // Return calculated product
    return prod;
}
 
// Driver code
public static void Main()
{
    int []arr = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.Length;
    int k = 3;
     
    Console.WriteLine(findProduct(arr, n, k));
}
}
 
// This code is contributed by inder_verma


PHP


Javascript


输出:
810

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)