检查数字偶数位的数字乘积是否可被K整除

给定数字N，任务是检查数字的偶数位的数字乘积是否可被K整除。如果可整除，则输出“是”，否则输出“否”。
例子：

Input: N = 5478, K = 5
Output: YES
Since, 5 * 7 = 35, which is divisible by 5

Input: N = 19270, K = 2
Output: NO 

方法：

1. 在从右到左的偶数位置查找数字的乘积。
2. 然后通过将其取模为“ K”来检查可除性
3. 如果取模为0，则输出YES，否则输出NO

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// below function checks whether
// product of digits at even places
// is divisible by K
bool productDivisible(int n, int k)
{
    int product = 1, position = 1;
    while (n > 0) {
 
        // if position is even
        if (position % 2 == 0)
            product *= n % 10;
        n = n / 10;
        position++;
    }
 
    if (product % k == 0)
        return true;
    return false;
}
 
// Driver code
int main()
{
    int n = 321922;
    int k = 3;
 
    if (productDivisible(n, k))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java

// JAVA implementation of the above approach
class GFG {
// below function checks whether
// product of digits at even places
// is divisible by K
 
    static boolean productDivisible(int n, int k) {
        int product = 1, position = 1;
        while (n > 0) {
 
            // if position is even
            if (position % 2 == 0) {
                product *= n % 10;
            }
            n = n / 10;
            position++;
        }
 
        if (product % k == 0) {
            return true;
        }
        return false;
    }
 
// Driver code
    public static void main(String[] args) {
        int n = 321922;
        int k = 3;
 
        if (productDivisible(n, k)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}

Python3

# Python3 implementation of the
# above approach
 
# below function checks whether
# product of digits at even places
# is divisible by K
def productDivisible(n, k):
    product = 1
    position = 1
    while n > 0:
         
        # if position is even
        if position % 2 == 0:
            product *= n % 10
        n = n / 10
        position += 1
    if product % k == 0:
        return True
    return False
 
# Driver code
n = 321922
k = 3
if productDivisible(n, k) == True:
    print("YES")
else:
    print("NO")
 
# This code is contributed
# by Shrikant13

C#

// C# implementation of the above approach
using System;
 
class GFG
{
// below function checks whether
// product of digits at even places
// is divisible by K
static bool productDivisible(int n, int k)
{
    int product = 1, position = 1;
    while (n > 0)
    {
 
        // if position is even
        if (position % 2 == 0)
            product *= n % 10;
        n = n / 10;
        position++;
    }
 
    if (product % k == 0)
        return true;
    return false;
}
 
// Driver code
public static void Main()
{
    int n = 321922;
    int k = 3;
 
    if (productDivisible(n, k))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

 0)
    {
 
        // if position is even
        if ($position % 2 == 0)
            $product *= $n % 10;
        $n = (int)($n / 10);
        $position++;
    }
 
    if ($product % $k == 0)
        return true;
    return false;
}
 
// Driver code
$n = 321922;
$k = 3;
 
if (productDivisible($n, $k))
    echo "YES";
else
    echo "NO";
 
// This code is contributed by mits
?>

输出：

YES

方法2：使用字符串()方法：

1. 将整数转换为字符串，然后遍历字符串并通过将其存储在乘积中来乘以所有偶数索引。
2. 如果乘积可被k整除，则返回True否则为False。

下面是实现：

Python3

# Python3 implementation of the
# above approach
 
# Function checks whether
# product of digits at even places
# is divisible by K
def productDivisible(n, k):
    product = 1
     
    # Converting integer to string
    num = str(n)
     
    # Traversing the string
    for i in range(len(num)):
        if(i % 2 == 0):
            product = product*int(num[i])
 
    if product % k == 0:
        return True
    return False
 
 
# Driver code
n = 321922
k = 3
if productDivisible(n, k) == True:
    print("YES")
else:
    print("NO")
 
# This code is contributed by vikkycirus

输出：

YES