检查偶数位数的乘积是否可被数字奇数位数的和除

给定数字N和N中的数字位数，任务是检查是否可以用奇数个位数的总和除掉数字偶数个位数的乘积。如果可以整除，则输出“ TRUE”，否则输出“ FALSE”。
例子：

Input: N = 2157 
Output: TRUE
Since, 1 * 7 = 7, which is divisible by 2+5=7

Input: N = 1234
Output: TRUE
Since, 2 * 4 = 8, which is divisible by 1 + 3 = 4   

方法：

1. 在从右到左的偶数位置查找数字的乘积。
2. 从右到左找到奇数位的数字总和。
3. 然后通过对乘积求模来检查乘积的可除性
4. 如果取模为0，则输出TRUE，否则输出FALSE

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 sum of digits at odd places
bool productSumDivisible(int n, int size)
{
    int sum = 0, product = 1;
    while (n > 0) {
 
        // if size is even
        if (size % 2 == 0) {
            product *= n % 10;
        }
 
        // if size is odd
        else {
            sum += n % 10;
        }
        n = n / 10;
        size--;
    }
 
    if (product % sum == 0)
        return true;
    return false;
}
 
// Driver code
int main()
{
    int n = 1234;
    int len = 4;
 
    if (productSumDivisible(n, len))
        cout << "TRUE";
    else
        cout << "FALSE";
 
    return 0;
}

Java

// JAVA implementation of the above approach
 
class GFG {
 
    // below function checks whether
    // product of digits at even places
    // is divisible by sum of digits at odd places
    static boolean productSumDivisible(int n, int size)
    {
        int sum = 0, product = 1;
        while (n > 0) {
 
            // if size is even
            if (size % 2 == 0) {
                product *= n % 10;
            }
 
            // if size is odd
            else {
                sum += n % 10;
            }
            n = n / 10;
            size--;
        }
 
        if (product % sum == 0) {
            return true;
        }
        return false;
    }
    // Driver code
 
    public static void main(String[] args)
    {
        int n = 1234;
        int len = 4;
 
        if (productSumDivisible(n, len)) {
            System.out.println("TRUE");
        }
        else {
            System.out.println("FALSE");
        }
    }
}

Python 3

# Python 3 implementation of the above approach
 
# Below function checks whether product
# of digits at even places is divisible
# by sum of digits at odd places
def productSumDivisible(n, size):
    sum = 0
    product = 1
    while (n > 0) :
 
        # if size is even
        if (size % 2 == 0) :
            product *= n % 10
 
        # if size is odd
        else :
            sum += n % 10
         
        n = n // 10
        size -= 1
 
    if (product % sum == 0):
        return True
    return False
 
# Driver code
if __name__ == "__main__":
    n = 1234
    len = 4
 
    if (productSumDivisible(n, len)):
        print("TRUE")
    else :
        print("FALSE")
 
# This code is contributed by ChitraNayal

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 productSumDivisible(int n, int size)
    {
        int sum = 0, product = 1;
        while (n > 0) {
 
            // if size is even
            if (size % 2 == 0) {
                product *= n % 10;
            }
 
            // if size is odd
            else {
                sum += n % 10;
            }
            n = n / 10;
            size--;
        }
 
        if (product % sum == 0) {
            return true;
        }
        return false;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 1234;
        int len = 4;
 
        if (productSumDivisible(n, len))
            Console.WriteLine("TRUE");
        else
            Console.WriteLine("FALSE");
    }
}

PHP

 0)
    {
 
        // if size is even
        if ($size % 2 == 0)
        {
            $product *= $n % 10;
        }
 
        // if size is odd
        else
        {
            $sum += $n % 10;
        }
        $n = $n / 10;
        $size--;
    }
 
    if ($product % $sum == 0)
        return true;
    return false;
}
 
// Driver code
$n = 1234;
$len = 4;
 
if (productSumDivisible($n, $len))
    echo "TRUE";
else
    echo "FALSE";
 
// This code is contributed by anuj_67..
?>

Javascript

 
// Javascript implementation of the above approach
 
// below function checks whether
// product of digits at even places
// is divisible by sum of digits at odd places
function productSumDivisible(n, size)
{
    var sum = 0, product = 1;
    while (n > 0) {
 
        // if size is even
        if (size % 2 == 0) {
            product *= n % 10;
        }
 
        // if size is odd
        else {
            sum += n % 10;
        }
        n = parseInt(n / 10);
        size--;
    }
 
    if (product % sum == 0)
        return true;
    return false;
}
 
// Driver code
var n = 1234;
var len = 4;
if (productSumDivisible(n, len))
    document.write("TRUE");
else
    document.write("FALSE");
 
// This code is contributed by noob2000.

输出：

TRUE

方法2：使用字符串()方法

将整数转换为字符串，然后遍历字符串并执行两个操作

• 将所有奇数索引相乘并将其存储在产品中
• 将所有偶数索引相加并将其总和存储
• 如果乘积可被总和整除，则返回True否则为False

下面是实现：

Python3

# Python 3 implementation of the above approach
 
# Below function checks whether product
# of digits at even places is divisible
# by sum of digits at odd places
def productSumDivisible(n):
    sum = 0
    product = 1
     
    # Converting integer to string
    num = str(n)
     
    # Traveersing the string
    for i in range(len(num)):
        if(i % 2 != 0):
            product = product*int(num[i])
        else:
            sum = sum+int(num[i])
 
    if (product % sum == 0):
        return True
    return False
 
 
# Driver code
if __name__ == "__main__":
    n = 1234
 
    if (productSumDivisible(n)):
        print("TRUE")
    else:
        print("FALSE")
 
# This code is contributed by vikkycirus

输出：

TRUE