可被给定数字 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 k 的数组，任务是找出该数组中所有可被 k 整除的元素的乘积
示例:
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
这个想法是遍历数组并逐个检查元素。如果一个元素可以被 K 整除，那么就把这个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这个过程。 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 <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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..
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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
function findProduct(&$arr, $n, $k)
{
$prod = 1;
// Traverse the array
for ($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
$arr = array(15, 16, 10, 9, 6, 7, 17 );
$n = sizeof($arr);
$k = 3;
echo (findProduct($arr, $n, $k));
// This code is contributed
// by Shivi_Aggarwal
?>
java 描述语言
<script>
// Function to find Product of all the elements
// in an array divisible by a given number K
function findProduct( arr, n, k)
{
var prod = 1;
// Traverse the array
for (var 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;
}
var arr = [15, 16, 10, 9, 6, 7, 17 ];
document.write(findProduct(arr, 7, 3));
</script>
Output:
810
时间复杂度 : O(N)，其中 N 为数组中的元素个数。 辅助空间: O(1)
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