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)