计算数组中能被 k 整除的元素个数

给定一个整数数组。任务是计算可被给定数 k 整除的元素数量。

例子：

Input: arr[] = { 2, 6, 7, 12, 14, 18 }, k = 3
Output: 3
Numbers which are divisible by k are { 6, 12, 18 }

Input: arr[] = { 2, 6, 7, 12, 14, 18 }, k = 2
Output: 5

方法一：开始遍历数组并检查当前元素是否可被 K 整除。如果是，则增加计数。当遍历所有元素时打印计数。

下面是上述方法的实现：

C++

// C++ program to Count the number of elements
// in an array which are divisible by k
#include 
using namespace std;
 
// Function to count the elements
int CountTheElements(int arr[], int n, int k)
{
    int counter = 0;
 
    for (int i = 0; i < n; i++) {
        if (arr[i] % k == 0)
            counter++;
    }
 
    return counter;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 6, 7, 12, 14, 18 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    cout << CountTheElements(arr, n, k);
 
    return 0;
}

Java

// Java program to Count the number of elements
// in an array which are divisible by k
import java.util.*;
 
class Geeks {
 
 
// Function to count the elements
static int CountTheElements(int arr[], int n, int k)
{
    int counter = 0;
 
    for (int i = 0; i < n; i++) {
        if (arr[i] % k == 0)
            counter++;
    }
 
    return counter;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 2, 6, 7, 12, 14, 18 };
    int n = arr.length;
    int k = 3;
 
    System.out.println(CountTheElements(arr, n, k));
}
}
 
// This code is contributed by ankita_saini

Python3

# Python 3 program to Count the
# number of elements in an array
# which are divisible by k
 
# Function to count the elements
def CountTheElements(arr, n, k):
    counter = 0
 
    for i in range(0, n, 1):
        if (arr[i] % k == 0):
            counter = counter+1
     
    return counter
 
# Driver code
if __name__ == '__main__':
    arr = [2, 6, 7, 12, 14, 18];
    n = len(arr)
    k = 3
 
    print(CountTheElements(arr, n, k))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to Count the number of elements
// in an array which are divisible by k
using System;
 
class Geeks {
 
 
// Function to count the elements
static int CountTheElements(int []arr, int n, int k)
{
    int counter = 0;
 
    for (int i = 0; i < n; i++) {
        if (arr[i] % k == 0)
            counter++;
    }
 
    return counter;
}
 
// Driver code
public static void Main()
{
    int []arr = { 2, 6, 7, 12, 14, 18 };
    int n = arr.Length;
    int k = 3;
 
    Console.WriteLine(CountTheElements(arr, n, k));
}
}
//This code is contributed by inder_verma..

Javascript

 
// Javascript program to Count the number
// of elements in an array which are
// divisible by k
 
// Function to count the elements
function CountTheElements(arr, n, k)
{
    let counter = 0;
 
    for(let i = 0; i < n; i++)
    {
        if (arr[i] % k == 0)
            counter++;
    }
 
    return counter;
}
 
// Driver code
let arr = [ 2, 6, 7, 12, 14, 18 ];
let n = arr.length;
let k = 3;
 
document.write(CountTheElements(arr, n, k));
 
// This code is contributed by subhammahato348
 

输出：

3

方法 2：另一种方法是使用内置函数– Count_if 。该函数接收指向包含元素和辅助函数的容器的开始和结束的指针。它将返回函数将返回 true 的元素的计数。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
int main()
{
    vector v{ 2, 6, 7, 12, 14, 18 };
 
    // Count the number elements which when passed
    // through the function (3rd argument) returns true
    int res = count_if(v.begin(), v.end(),
                       [](int i, int k = 3) { return i % k == 0; });
    cout << res;
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
public static void main(String[] args)
{
   int []v = { 2, 6, 7, 12, 14, 18 };
 
    // Count the number elements which when passed
    // through the function (3rd argument) returns true
    int res = 0;
    for(int i = 0; i < v.length; i++) {
        if(v[i] % 3 == 0)
            res++;
    }
     System.out.print(res);
 
}
}
 
// This code is contributed by gauravrajput1

Python3

# Java implementation of the above approach
v = [ 2, 6, 7, 12, 14, 18 ];
 
# Count the number elements which when passed
# through the function (3rd argument) returns true
res = 0
for i in range(0,len(v)):
    if(v[i] % 3 == 0):
        res+=1
     
print(res)
# This code is contributed by shivanisinghss2110

C＃

// C# implementation of the above approach
using System;
 
class GFG
{
public static void Main(String[] args)
{
   int []v = { 2, 6, 7, 12, 14, 18 };
 
    // Count the number elements which when passed
    // through the function (3rd argument) returns true
    int res = 0;
    for(int i = 0; i < v.Length; i++) {
        if(v[i] % 3 == 0)
            res++;
    }
     Console.Write(res);
 
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
// JavaScript implementation of the above approach
var v = [ 2, 6, 7, 12, 14, 18 ];
 
    // Count the number elements which when passed
    // through the function (3rd argument) returns true
    var res = 0;
    for(var i = 0; i < v.length; i++) {
        if(v[i] % 3 == 0)
            res++;
    }
     document.write(res);
 
// This code is contributed by shivanisinghss2110
 

输出：

3

时间复杂度： O(N)