检查数字在偶数和奇数位的乘积是否相等

给定整数N ，任务是检查数字的偶数和奇数位的数字乘积是否相等。如果相等，则打印“是”，否则打印“否” 。

例子：

Input: N = 2841 
Output: Yes 
Product of digits at odd places = 2 * 4 = 8 
Product of digits at even places = 8 * 1 = 8

Input: N = 4324 
Output: No 
Product of digits at odd places = 4 * 2 = 8 
Product of digits at even places = 3 * 4 = 12   

方法：

• 在偶数位置查找数字乘积并将其存储在prodEven中。
• 在奇数位找到数字乘积并将其存储在prodOdd中。
• 如果prodEven = prodOdd，则打印“是”，否则打印“否” 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if the product
// of even positioned digits is equal to
// the product of odd positioned digits in n
bool productEqual(int n)
{
 
    // If n is a single digit number
    if (n < 10)
        return false;
    int prodOdd = 1, prodEven = 1;
 
    while (n > 0) {
 
        // Take two consecutive digits
        // at a time
        // First digit
        int digit = n % 10;
        prodOdd *= digit;
        n /= 10;
 
        // If n becomes 0 then
        // there's no more digit
        if (n == 0)
            break;
 
        // Second digit
        digit = n % 10;
        prodEven *= digit;
        n /= 10;
    }
 
    // If the products are equal
    if (prodEven == prodOdd)
        return true;
 
    // If products are not equal
    return false;
}
 
// Driver code
int main()
{
    int n = 4324;
 
    if (productEqual(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
     
    // Function that returns true
    // if the product of even positioned
    // digits is equal to the product of
    // odd positioned digits in n
    static boolean productEqual(int n)
    {
     
        // If n is a single digit number
        if (n < 10)
            return false;
        int prodOdd = 1, prodEven = 1;
     
        while (n > 0)
        {
     
            // Take two consecutive digits
            // at a time
            // First digit
            int digit = n % 10;
            prodOdd *= digit;
            n /= 10;
     
            // If n becomes 0 then
            // there's no more digit
            if (n == 0)
                break;
     
            // Second digit
            digit = n % 10;
            prodEven *= digit;
            n /= 10;
        }
     
        // If the products are equal
        if (prodEven == prodOdd)
            return true;
     
        // If products are not equal
        return false;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int n = 4324;
     
        if (productEqual(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
    // This code is contributed by Ryuga
}

Python3

# Python implementation of the approach
 
# Function that returns true if the product
# of even positioned digits is equal to
# the product of odd positioned digits in n
def productEqual(n):
    if n < 10:
        return False
    prodOdd = 1; prodEven = 1
 
    # Take two consecutive digits
    # at a time
    # First digit
    while n > 0:
        digit = n % 10
        prodOdd *= digit
        n = n//10
 
        # If n becomes 0 then
        # there's no more digit
        if n == 0:
            break;
        digit = n % 10
        prodEven *= digit
        n = n//10
 
    # If the products are equal
    if prodOdd == prodEven:
        return True
 
    # If the products are not equal
    return False
 
# Driver code
n = 4324
if productEqual(n):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Shrikant13

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function that returns true
// if the product of even positioned
// digits is equal to the product of
// odd positioned digits in n
static bool productEqual(int n)
{
 
    // If n is a single digit number
    if (n < 10)
        return false;
    int prodOdd = 1, prodEven = 1;
 
    while (n > 0)
    {
 
        // Take two consecutive digits
        // at a time
        // First digit
        int digit = n % 10;
        prodOdd *= digit;
        n /= 10;
 
        // If n becomes 0 then
        // there's no more digit
        if (n == 0)
            break;
 
        // Second digit
        digit = n % 10;
        prodEven *= digit;
        n /= 10;
    }
 
    // If the products are equal
    if (prodEven == prodOdd)
        return true;
 
    // If products are not equal
    return false;
}
 
// Driver code
static void Main()
{
    int n = 4324;
 
    if (productEqual(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by mits

PHP

 0)
    {
 
        // Take two consecutive digits
        // at a time
         
        // First digit
        $digit = $n % 10;
        $prodOdd *= $digit;
        $n /= 10;
 
        // If n becomes 0 then
        // there's no more digit
        if ($n == 0)
            break;
 
        // Second digit
        $digit = $n % 10;
        $prodEven *= $digit;
        $n /= 10;
    }
 
    // If the products are equal
    if ($prodEven == $prodOdd)
        return true;
 
    // If products are not equal
    return false;
}
 
// Driver code
$n = 4324;
if (productEqual(!$n))
    echo "Yes";
else
    echo "No";
 
// This code is contributed by jit_t
?>

输出：

No

方法2：使用字符串()方法：

1. 将整数转换为字符串。遍历字符串并将所有偶数索引乘积存储在一个变量中，将所有奇数索引乘积存储在另一个变量中。
2. 如果两者相等，则打印是，否则，否

下面是实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
void getResult(int n)
{
     
    // To store the respective product
    int proOdd = 1;
    int proEven = 1;
 
    // Converting integer to string
    string num = to_string(n);
 
    // Traversing the string
    for(int i = 0; i < num.size(); i++)
        if (i % 2 == 0)
            proOdd = proOdd * (num[i] - '0');
        else
            proEven = proEven * (num[i] - '0');
 
    if (proOdd == proEven)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver code
int main()
{
    int n = 4324;
     
    getResult(n);
     
    return 0;
}
 
// This code is contributed by sudhanshugupta2019a

Python3

# Python3 implementation of the approach
 
def getResult(n):
 
    # To store the respective product
    proOdd = 1
    proEven = 1
     
    # Converting integer to string
    num = str(n)
     
    # Traversing the string
    for i in range(len(num)):
        if(i % 2 == 0):
            proOdd = proOdd*int(num[i])
        else:
            proEven = proEven*int(num[i])
 
    if(proOdd == proEven):
        print("Yes")
    else:
        print("No")
 
 
# Driver code
if __name__ == "__main__":
    n = 4324
    getResult(n)
 
# This code is contributed by vikkycirus

输出：

No