将两个给定数的正乘积减至最少N个减量

给出三个整数X，Y，和N，任务是找到可以通过降低或者X或Y的由1至多N次值来获得的X和Y的最小可能正产品。

例子：

Input: X = 5, Y= 6, N = 4
Output: 6
Explanation:
Decrease the value of X by 4, X = 5 – 4 = 1 and Y = 6.
Therefore, the minimized product = X * Y = 1 * 6 = 6

Input: X = 49, Y = 4256, N = 10
Output: 165984

方法：可以根据以下观察结果解决给定问题：

If X ≤ Y: Reducing X minimizes the product. 
If Y ≤ X: Reducing Y minimizes the product.

Mathematical Proof:
If (X – 2) * Y < (X – 1) * (Y – 1) 
=> X * Y – 2 * Y < X * Y – X – Y + 1 
=> – 2 × Y < -X – Y + 1 
=> Y > X – 1

请按照以下步骤解决问题：

• 如果X≤Y：请执行以下步骤：
  • 如果N 打印Y *(X – N)作为答案，因为减少X会使乘积最小。
  • 否则，将X减少到1，并将剩余的N从Y减少到最小。因此，将Y – max(1，N – X + 1))打印为所需的最小化产品。
• 否则，如果N ，则将X *(Y – N)打印为最小乘积。如果N≥Y，Y降低到1和打印最大(X – (N – Y + 1)，1)作为最小化的产物。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to minimize
// the product of two numbers
int minProd(int X, int Y,
            int N)
{
    if (X <= Y) {
 
        if (N < X)
 
            // Reducing X, N times,
            // minimizes the product
            return (X - N) * Y;
        else {
 
            // Reduce X to 1 and reduce
            // remaining N from Y
            return max(Y - (N - X + 1), 1);
        }
    }
 
    if (Y >= N)
 
        // Reducing Y, N times,
        // minimizes the product
        return (Y - N) * X;
 
    // Reduce Y to 1 and reduce
    // remaining N from X
    return max(X - (N - Y + 1), 1);
    ;
}
 
// Driver Code
int main()
{
    int X = 47, Y = 42, N = 167;
    cout << minProd(X, Y, N);
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to minimize
// the product of two numbers
static int minProd(int X, int Y,
                   int N)
{
    if (X <= Y)
    {
        if (N < X)
   
            // Reducing X, N times,
            // minimizes the product
            return (X - N) * Y;
        else
        {
             
            // Reduce X to 1 and reduce
            // remaining N from Y
            return Math.max(Y - (N - X + 1), 1);
        }
    }
   
    if (Y >= N)
   
        // Reducing Y, N times,
        // minimizes the product
        return (Y - N) * X;
   
    // Reduce Y to 1 and reduce
    // remaining N from X
    return Math.max(X - (N - Y + 1), 1);
}
   
// Driver Code
public static void main (String[] args)
{
    int X = 47, Y = 42, N = 167;
     
    System.out.println(minProd(X, Y, N));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program to implement
# the above approach
 
# Function to minimize
# the product of two numbers
def minProd(X, Y, N):
    if (X <= Y):
 
        if (N < X):
 
            # Reducing X, N times,
            # minimizes the product
            return (X - N) * Y
        else:
 
            # Reduce X to 1 and reduce
            # remaining N from Y
            return max(Y - (N - X + 1), 1)
 
    if (Y >= N):
 
        # Reducing Y, N times,
        # minimizes the product
        return (Y - N) * X
 
    # Reduce Y to 1 and reduce
    # remaining N from X
    return max(X - (N - Y + 1), 1)
 
# Driver Code
if __name__ == "__main__":
    X = 47
    Y = 42
    N = 167
    print (minProd(X, Y, N))
 
# This code is contributed by Chitranayal

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to minimize
// the product of two numbers
static int minProd(int X,
                   int Y, int N)
{
  if (X <= Y)
  {
    if (N < X)
 
      // Reducing X, N times,
      // minimizes the product
      return (X - N) * Y;
    else
    {
      // Reduce X to 1 and reduce
      // remaining N from Y
      return Math.Max(Y - (N -
                      X + 1), 1);
    }
  }
 
  if (Y >= N)
 
    // Reducing Y, N times,
    // minimizes the product
    return (Y - N) * X;
 
  // Reduce Y to 1 and reduce
  // remaining N from X
  return Math.Max(X - (N -
                  Y + 1), 1);
}
   
// Driver Code
public static void Main(String[] args)
{
  int X = 47, Y = 42, N = 167;
  Console.WriteLine(minProd(X, Y, N));
}
}
 
// This code is contributed by Rajput-Ji

输出：

1

时间复杂度： O(1)
辅助空间： O(1)