要添加的最小元素，以便可以将两个矩阵相乘

给定两个矩阵A和B，分别为pxq(或q X p)和rxs(或s X r)。任务是找到要添加到两个矩阵中的任何一个以使它们相乘的最小数量。
如果一个矩阵中的列数等于另一个矩阵中的行数，则两个矩阵是可乘的。
例子：

Input: p = 2, q = 3, r = 5, s = 6 
Output: 4 
Two columns can be added to make q = 5. 
To add two columns, the number of elements will be added 2 * no. of rows i.e. 2 * 2 = 4
Input: p = 11, q = 5, r = 10, s = 11 
Output: 0 
 

方法：

• 如果一个矩阵中的列数等于另一矩阵中的行数，则不需要添加额外的元素。
• 如果它们不相等，则必须在一个矩阵中添加一些额外的列，或者在另一个矩阵中添加一些额外的行。

因此，计算两种情况下的额外元素数并打印最少的元素。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to calculate the minimum
// number of extra elements is to add
int minExtraElements(int p, int q, int r, int s)
{
    // num1 will store minimum number of
    // extra elements required
    // to make A x B possible
 
    // num2 will store minimum number of
    // extra elements required
    // to make B x A possible
    int num1, num2;
 
    // if either A x B or B x A is possible,
    // it will return 0
    if (q == r || p == s)
        return 0;
 
    else {
        // it will calculate minimum number of
        // extra elements required
        // to make A x B possible
        if (q < r)
            num1 = (r - q) * p;
        else
            num1 = (q - r) * s;
 
        // it will calculate minimum number of
        // extra elements required
        // to make B x A possible
        if (p < s)
            num2 = (s - p) * r;
        else
            num2 = (p - s) * q;
    }
 
    // return minimum of both
    return min(num1, num2);
}
 
// Driver code
int main()
{
 
    int p = 2, q = 3, r = 5, s = 6;
    cout << minExtraElements(p, q, r, s) << endl;
 
    return 0;
}

Java

// Java implementation of the above approach
 
class GFG
{
    // Function to calculate the minimum
    // number of extra elements is to add
    static int minExtraElements(int p, int q, int r, int s)
    {
        // num1 will store minimum number of
        // extra elements required
        // to make A x B possible
     
        // num2 will store minimum number of
        // extra elements required
        // to make B x A possible
        int num1, num2;
     
        // if either A x B or B x A is possible,
        // it will return 0
        if (q == r || p == s)
            return 0;
     
        else {
            // it will calculate minimum number of
            // extra elements required
            // to make A x B possible
            if (q < r)
                num1 = (r - q) * p;
            else
                num1 = (q - r) * s;
     
            // it will calculate minimum number of
            // extra elements required
            // to make B x A possible
            if (p < s)
                num2 = (s - p) * r;
            else
                num2 = (p - s) * q;
        }
     
        // return minimum of both
        return Math.min(num1, num2);
    }
     
    // Driver code
    public static void main(String []rags)
    {
     
        int p = 2, q = 3, r = 5, s = 6;
        System.out.println(minExtraElements(p, q, r, s));
     
         
    }
}
 
// This code is contributed by ihritik

Python

# Python implementation of the above approach
 
 
# Function to calculate the minimum
# number of extra elements is to add
def minExtraElements(p,  q, r, s):
 
    # num1 will store minimum number of
    # extra elements required
    # to make A x B possible
 
    # num2 will store minimum number of
    # extra elements required
    # to make B x A possible
 
 
    # if either A x B or B x A is possible,
    # it will return 0
    if (q == r or p == s):
        return 0
 
    else :
        # it will calculate minimum number of
        # extra elements required
        # to make A x B possible
        if (q < r):
            num1 = (r - q) * p
        else:
            num1 = (q - r) * s
 
        # it will calculate minimum number of
        # extra elements required
        # to make B x A possible
        if (p < s):
            num2 = (s - p) * r
        else:
            num2 = (p - s) * q
     
 
    # return minimum of both
    return min(num1, num2)
 
 
# Driver code
 
     
p = 2
q = 3
r = 5
s = 6
print(minExtraElements(p, q, r, s))
 
# This code is contributed by ihritik

C#

// C# implementation of the above approach
 
using System;
class GFG
{
    // Function to calculate the minimum
    // number of extra elements is to add
    static int minExtraElements(int p, int q, int r, int s)
    {
        // num1 will store minimum number of
        // extra elements required
        // to make A x B possible
     
        // num2 will store minimum number of
        // extra elements required
        // to make B x A possible
        int num1, num2;
     
        // if either A x B or B x A is possible,
        // it will return 0
        if (q == r || p == s)
            return 0;
     
        else {
            // it will calculate minimum number of
            // extra elements required
            // to make A x B possible
            if (q < r)
                num1 = (r - q) * p;
            else
                num1 = (q - r) * s;
     
            // it will calculate minimum number of
            // extra elements required
            // to make B x A possible
            if (p < s)
                num2 = (s - p) * r;
            else
                num2 = (p - s) * q;
        }
     
        // return minimum of both
        return Math.Min(num1, num2);
    }
     
    // Driver code
    public static void Main()
    {
     
        int p = 2, q = 3, r = 5, s = 6;
        Console.WriteLine(minExtraElements(p, q, r, s));
     
         
    }
}
 
// This code is contributed by ihritik

Javascript

 
// Javascript implementation of the above approach
 
 
// Function to calculate the minimum
// number of extra elements is to add
function minExtraElements( p, q, r, s)
{
    // num1 will store minimum number of
    // extra elements required
    // to make A x B possible
 
    // num2 will store minimum number of
    // extra elements required
    // to make B x A possible
    let num1, num2;
 
    // if either A x B or B x A is possible,
    // it will return 0
    if (q == r || p == s)
        return 0;
 
    else {
        // it will calculate minimum number of
        // extra elements required
        // to make A x B possible
        if (q < r)
            num1 = (r - q) * p;
        else
            num1 = (q - r) * s;
 
        // it will calculate minimum number of
        // extra elements required
        // to make B x A possible
        if (p < s)
            num2 = (s - p) * r;
        else
            num2 = (p - s) * q;
    }
 
    // return minimum of both
    return Math.min(num1, num2);
}
 
 
 
    // Driver Code
     
    let p = 2, q = 3, r = 5, s = 6;
    document.write(minExtraElements(p, q, r, s) + "</br>");
     

输出：

4

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