# Frobenius coin problem

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”.

**Examples : **

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.

## We strongly recommend that you click here and practice it, before moving on to the solution.

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:

- Calculate GCD of X and Y
- If GCD is 1 then required largest amount is (X*Y)-(X+Y) and total count is (X-1)*(Y-1)/2
- Else print “NA”
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Below is a the program based on the same.

## C++

`// C++ program to find the largest number that ` `// cannot be formed from given two coins ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Utility function to find gcd ` `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; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `X = 2,Y = 5; ` ` ` `forbenius(X,Y); ` ` ` ` ` `X = 5, Y = 10; ` ` ` `cout << endl; ` ` ` `forbenius(X,Y); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find the largest ` `// number that cannot be formed ` `// from given two coins ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` `// Utility function to find gcd ` ` ` `static` `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 ` ` ` `static` `void` `forbenius(` `int` `X, ` ` ` `int` `Y) ` ` ` `{ ` ` ` `// Solution doesn't exist ` ` ` `// if GCD is not 1 ` ` ` `if` `(gcd(X, Y) != ` `1` `) ` ` ` `{ ` ` ` `System.out.println( ` `"NA"` `); ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Else apply the formula ` ` ` `int` `A = (X * Y) - (X + Y); ` ` ` `int` `N = (X - ` `1` `) * (Y - ` `1` `) / ` `2` `; ` ` ` ` ` `System.out.println(` `"Largest Amount = "` `+ A ); ` ` ` `System.out.println(` `"Total Count = "` `+ N ); ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `X = ` `2` `,Y = ` `5` `; ` ` ` `forbenius(X, Y); ` ` ` `X = ` `5` `; ` ` ` `Y = ` `10` `; ` ` ` `System.out.println(); ` ` ` `forbenius(X, Y); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

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## Python3

`# Python3 program to find the largest ` `# number that cannot be formed ` `# from given two coins ` ` ` `# Utility function to find gcd ` `def` `gcd(a, b): ` ` ` `while` `(a !` `=` `0` `): ` ` ` `c ` `=` `a; ` ` ` `a ` `=` `b ` `%` `a; ` ` ` `b ` `=` `c; ` ` ` ` ` `return` `b; ` ` ` `# Function to print the desired output ` `def` `forbenius(X, Y): ` ` ` ` ` `# Solution doesn't exist ` ` ` `# if GCD is not 1 ` ` ` `if` `(gcd(X, Y) !` `=` `1` `): ` ` ` `print` `(` `"NA"` `); ` ` ` `return` `; ` ` ` ` ` `# Else apply the formula ` ` ` `A ` `=` `(X ` `*` `Y) ` `-` `(X ` `+` `Y); ` ` ` `N ` `=` `(X ` `-` `1` `) ` `*` `(Y ` `-` `1` `) ` `/` `/` `2` `; ` ` ` ` ` `print` `(` `"Largest Amount ="` `, A); ` ` ` `print` `(` `"Total Count ="` `, N); ` ` ` `# Driver Code ` `X ` `=` `2` `; ` `Y ` `=` `5` `; ` `forbenius(X, Y); ` ` ` `X ` `=` `5` `; ` `Y ` `=` `10` `; ` `print` `(""); ` `forbenius(X, Y); ` ` ` `# This code is contributed by mits ` |

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## C#

`// C# program to find the largest ` `// number that cannot be formed ` `// from given two coins ` `using` `System; ` ` ` `class` `GFG ` `{ ` `// Utility function to find gcd ` ` ` `static` `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 ` ` ` `static` `void` `forbenius(` `int` `X, ` `int` `Y) ` ` ` `{ ` ` ` `// Solution doesn't exist ` ` ` `// if GCD is not 1 ` ` ` `if` `(gcd(X,Y) != 1) ` ` ` `{ ` ` ` `Console.WriteLine( ` `"NA"` `); ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Else apply the formula ` ` ` `int` `A = (X * Y) - (X + Y); ` ` ` `int` `N = (X - 1) * (Y - 1) / 2; ` ` ` ` ` `Console.WriteLine(` `"Largest Amount = "` `+ A ); ` ` ` `Console.WriteLine(` `"Total Count = "` `+ N ); ` ` ` `} ` ` ` ` ` ` ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `X = 2,Y = 5; ` ` ` `forbenius(X,Y); ` ` ` `X = 5; ` ` ` `Y = 10; ` ` ` `Console.WriteLine(); ` ` ` `forbenius(X,Y); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

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## PHP

`<?php ` `// php program to find the largest ` `// number that cannot be formed ` `// from given two coins ` ` ` `// Utility function to find gcd ` `function` `gcd(` `$a` `, ` `$b` `) ` `{ ` ` ` `$c` `; ` ` ` `while` `(` `$a` `!= 0) ` ` ` `{ ` ` ` `$c` `= ` `$a` `; ` ` ` `$a` `= ` `$b` `% ` `$a` `; ` ` ` `$b` `= ` `$c` `; ` ` ` `} ` ` ` ` ` `return` `$b` `; ` `} ` ` ` `// Function to print the desired output ` `function` `forbenius(` `$X` `, ` `$Y` `) ` `{ ` ` ` `// Solution doesn't exist ` ` ` `// if GCD is not 1 ` ` ` `if` `(gcd(` `$X` `, ` `$Y` `) != 1) ` ` ` `{ ` ` ` `echo` `"NA\n"` `; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Else apply the formula ` ` ` `$A` `= (` `$X` `* ` `$Y` `) - (` `$X` `+ ` `$Y` `); ` ` ` `$N` `= (` `$X` `- 1) * (` `$Y` `- 1) / 2; ` ` ` ` ` `echo` `"Largest Amount = "` `, ` `$A` `, ` `"\n"` `; ` ` ` `echo` `"Total Count = "` `, ` `$N` `, ` `"\n"` `; ` `} ` ` ` `// Driver Code ` ` ` ` ` `$X` `= 2; ` `$Y` `= 5; ` ` ` `forbenius(` `$X` `, ` `$Y` `); ` ` ` ` ` `$X` `= 5; ` `$Y` `= 10; ` ` ` `echo` `"\n"` `; ` ` ` `forbenius(` `$X` `, ` `$Y` `); ` ` ` `// This code is contributed by ajit. ` `?> ` |

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**Output :**

Largest Amount = 3 Total Count = 2 NA

**References: **

https://en.wikipedia.org/wiki/Coin_problem

This article is contributed by Ashutosh Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above