计算所需的纸币数量

您有不限量的面值A和B美元的钞票(A不等于B) 。您想使用正好N张钞票总共支付S元。任务是找到您需要的价值A澳元的纸币数量。如果没有解决方案，则返回-1 。

例子：

Input: A = 1, B = 2, S = 7, N = 5
Output: 3
3 * A + 2 * B = S
3 * 1 + 2 * 2 = 7

Input: A = 2, B = 1, S = 7, N = 5
Output: 2

Input: A = 2, B = 1, S = 4, N = 5
Output: -1

Input: A = 2, B = 3, S = 20, N = 8
Output: 4

方法：设x为需要的价值A的票据的数目，然后其余票据，即N – x必须为价值B的票据。由于它们的总和要求为S，因此必须满足以下方程式：

S = (A * x) + (B * (N – x))
Solving the equation further,
S = (A * x) + (B * N) – (B * x)
S – (B * N) = (A – B) * x
x = (S – (B * N)) / (A – B)
After solving for x, it is the number of notes of value A required
only if the value of x is an integer else the result is not possible.

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
  
// Function to return the amount of notes
// with value A required
int bankNotes(int A, int B, int S, int N)
{
    int numerator = S - (B * N);
    int denominator = A - B;
  
    // If possible
    if (numerator % denominator == 0)
        return (numerator / denominator);
    return -1;
}
  
// Driver code
int main()
{
    int A = 1, B = 2, S = 7, N = 5;
    cout << bankNotes(A, B, S, N) << endl;
}
      
// This code is contributed by mits

Java

// Java implementation of the approach
class GFG {
  
    // Function to return the amount of notes
    // with value A required
    static int bankNotes(int A, int B, int S, int N)
    {
        int numerator = S - (B * N);
        int denominator = A - B;
  
        // If possible
        if (numerator % denominator == 0)
            return (numerator / denominator);
        return -1;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int A = 1, B = 2, S = 7, N = 5;
        System.out.print(bankNotes(A, B, S, N));
    }
}

Python3

# Python3 implementation of the approach
  
# Function to return the amount of notes
# with value A required
def bankNotes(A, B, S, N):
  
    numerator = S - (B * N)
    denominator = A - B
  
    # If possible
    if (numerator % denominator == 0):
        return (numerator // denominator)
    return -1
  
# Driver code
A, B, S, N = 1, 2, 7, 5
print(bankNotes(A, B, S, N))
  
# This code is contributed 
# by mohit kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
    // Function to return the amount of notes
    // with value A required
    static int bankNotes(int A, int B,
                         int S, int N)
    {
        int numerator = S - (B * N);
        int denominator = A - B;
  
        // If possible
        if (numerator % denominator == 0)
            return (numerator / denominator);
        return -1;
    }
  
    // Driver code
    public static void Main()
    {
        int A = 1, B = 2, S = 7, N = 5;
        Console.Write(bankNotes(A, B, S, N));
    }
}
  
// This code is contributed by Akanksha Rai

输出：

3