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
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）
java 描述语言
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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