计算给定数组中存在的 K 的因子

给定一个数组arr[]和一个整数K ，任务是计算数组中存在的K因子的计数。
例子：

Input: arr[] = {1, 2, 4, 5, 6}, K = 6 
Output: 3 
Explanation: 
There are three numbers present in the array those are factors of K = 6 – {1, 2, 6}
Input: arr[] = {1, 2, 12, 24}, K = 20 
Output: 2 
Explanation: 
There are two numbers present in the array those are factors of K = 20 – {1, 2} 
 

朴素的方法：这个问题的一个简单解决方案是找到 K 的所有因子，然后对每个因子迭代数组并检查它是否存在于数组中。如果是，则将因子计数增加 1。
有效的方法：这个想法不是找到数字 K 的所有因子，而是在数组上迭代并检查每个元素是否是 K 的因子，或者不是在模运算符的帮助下。如果是，则增加 K 的因子计数。
下面是上述方法的实现：

C++

// C++ implementation to find the count
// of factors of K present in array
 
#include 
using namespace std;
 
// Function to find the count
// of factors of K present in array
int calcCount(int arr[], int n, int k)
{
    int count = 0;
 
    // Loop to consider every
    // element of array
    for (int i = 0; i < n; i++) {
        if (k % arr[i] == 0)
            count++;
    }
 
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 6;
 
    // Function Call
    cout << calcCount(arr, n, k);
 
    return 0;
}

Java

// Java implementation to find the count
// of factors of K present in array
class GFG{
 
// Function to find the count
// of factors of K present in array
static int calcCount(int arr[], int n, int k)
{
    int count = 0;
 
    // Loop to consider every
    // element of array
    for(int i = 0; i < n; i++)
    {
       if (k % arr[i] == 0)
           count++;
    }
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 4, 5, 6 };
    int n = arr.length;
    int k = 6;
 
    // Function Call
    System.out.print(calcCount(arr, n, k));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation to find the count
# of factors of K present in array
 
# Function to find the count
# of factors of K present in array
def calcCount(arr, n, k):
 
    count = 0
 
    # Loop to consider every
    # element of array
    for i in range(0, n):
        if (k % arr[i] == 0):
            count = count + 1
 
    return count
 
# Driver Code
arr = [ 1, 2, 4, 5, 6 ]
n = len(arr)
k = 6
 
# Function Call
print(calcCount(arr, n, k))
 
# This code is contributed by PratikBasu

C#

// C# implementation to find the count
// of factors of K present in array
using System;
 
class GFG{
 
// Function to find the count
// of factors of K present in array
static int calcCount(int []arr, int n, int k)
{
    int count = 0;
 
    // Loop to consider every
    // element of array
    for(int i = 0; i < n; i++)
    {
       if (k % arr[i] == 0)
           count++;
    }
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 4, 5, 6 };
    int n = arr.Length;
    int k = 6;
 
    // Function Call
    Console.Write(calcCount(arr, n, k));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript implementation to find the count
// of factors of K present in array
 
// Function to find the count
// of factors of K present in array
function calcCount(arr, n, k)
{
    var count = 0;
 
    // Loop to consider every
    // element of array
    for (var i = 0; i < n; i++) {
        if (k % arr[i] == 0)
            count++;
    }
 
    return count;
}
 
// Driver Code
var arr = [ 1, 2, 4, 5, 6 ];
var n = arr.length;
var k = 6;
 
// Function Call
document.write( calcCount(arr, n, k));
 

输出：

3

时间复杂度： O(N)

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