Frobenius coin problem - GeeksforGeeks

Frobenius coin problem - GeeksforGeeks
Given two coins of denominations "X" and "Y" respectively, find the largest amount that cannot be obtained using these two coins (assuming infinite supply of coins) followed by the total number of such non obtainable amounts, if no such value exists print "NA".

Input : X=2, Y=5  
Output: Largest amount = 3
        Total count  = 2
We cannot represent 1 and 3 from infinite supply
of given two coins. The largest among these 2 is 3.
We can represent all other amounts for example 13
can be represented 2*4 + 5.

Input : X=5, Y=10
Output: NA
There are infinite number of amounts that cannot
be represented by these two coins.

One important observation is, if GCD of X and Y is not one, then all values that can be formed by given two coins are multiples of GCD. For example if X = 4 and Y = 6. Then all values are multiple of 2. So all values that are not multiple of 2, cannot be formed by X and Y. Thus there exist infinitely many values that cannot be formed by 4 and 6, and our answer becomes “NA”.

This general problem for n coins is known as classic Forbenius coin problem.

When the number of coins is two, there is 
explicit formula if GCD is not 1. The formula
is:
  Largest amount A = (X * Y) - (X + Y)
  Total amount = (X -1) * (Y - 1) /2 
 

Hence, we can now easily answer the above question by following the below steps:

1. Calculate GCD of X and Y
2. If GCD is 1 then required largest amount is (X*Y)-(X+Y) and total count is (X-1)*(Y-1)/2
3. Else print “NA”

int gcd(int a, int b)

{

    int c;

    while (a != 0)

    {

        c = a;

        a = b%a;

        b = c;

    }

    return b;

}

 

// Function to print the desired output

void forbenius(int X,int Y)

{

    // Solution doesn't exist if GCD is

    // not 1

    if (gcd(X,Y) != 1)

    {

        cout << "NA\n";

        return;

    }

 

    // Else apply the formula

    int A = (X*Y)-(X+Y);

    int N = (X-1)*(Y-1)/2;

 

    cout << "Largest Amount = " << A << endl;

    cout << "Total Count = " << N << endl;

}

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