Frobenius硬币问题

分别给定两个面额为“ X”和“ Y”的硬币，找到使用这两个硬币无法获得的最大金额(假设无限量供应硬币)，然后找到无法获得的金额的总数，如果不存在这样的值，则打印出来“ NA”。

例子 ：

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.

我们强烈建议您单击此处并进行实践，然后再继续解决方案。

一个重要的观察结果是，如果X和Y的GCD不为1，则可以由给定的两个硬币形成的所有值都是GCD的倍数。例如，如果X = 4并且Y =6。那么所有值都是2的倍数。因此，不是2的倍数的所有值都不能由X和Y形成。因此，存在无限多个不能由4和D形成的值。 6，我们的答案变为“不适用”。

n个硬币的一般问题称为经典Forbenius硬币问题。

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 
 

因此，我们现在可以按照以下步骤轻松地回答上述问题：

1. 计算X和Y的GCD
2. 如果GCD为1，则所需的最大数量为(X * Y)-(X + Y)，总计数为(X-1)*(Y-1)/ 2
3. 其他打印“ NA”

   下面是基于相同的程序。

   C++

   // C++ program to find the largest number that
   // cannot be formed from given two coins
   #include 
   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;
   }

   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

   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

   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

   输出 ：

   Largest Amount = 3
   Total Count = 2
   
   NA
   

   参考：
   https://zh.wikipedia.org/wiki/硬币问题