计算将C分为两部分并加到A和B以使A严格大于B的方法

给定三个整数A，B和C ，任务是计算将C分为两部分并加到A和B的方式的数量，以使A严格大于B。

例子：

Input: A = 5, B = 3, C = 4
Output: 3
The possible values of A and B after dividing C are:
A = 7, B = 5 where C is divided into 2 and 2.
A = 8, B = 4 where C is divided into 3 and 1.
A – 9, B = 3 where C is divided into 4 and 0.

Input: A = 3, B = 5, C = 5
Output: 2

方法：仔细观察后，针对该问题形成以下关系。

• 令ADDA和ADDB分别添加到A和B.
• 因此， addA + addB = C且应满足不等式A + addA> B + addB 。
• 现在，由于addB = C – addA并将其放在不等式中：

  A + addA > B + (C - addA)
  or, 2addA > C + B - A
  or, 2addA >= C + B - A + 1
  or, addA >= (C + B - A + 1) / 2
  

• 由于addA必须为非负数，因此addA = max(0，(C + B – A + 1)/ 2) 。
• 该划分应为上限划分，因此我们可以将其重写为addA = max(0，(C + B – A + 2)/ 2)。
• 令该值等于minAddA 。由于[minAddA，C]中的所有整数值addA都满足关系A + addA> B + addB ，因此所需的方式数等于max(0，C – minAddA + 1) 。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
  
#include 
using namespace std;
  
// Function to count the number of ways to divide
// C into two parts and add to A and B such
// that A is strictly greater than B
int countWays(int A, int B, int C)
{
    // Minimum value added to A to satisfy
    // the given relation
    int minAddA = max(0, (C + B - A + 2) / 2);
  
    // Number of different values of A, i.e.,
    // number of ways to divide C
    int count_ways = max(C - minAddA + 1, 0);
  
    return count_ways;
}
  
// Driver code
int main()
{
    int A = 3, B = 5, C = 5;
  
    cout << countWays(A, B, C);
  
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
  
class GFG{
  
    // Function to count the number of ways to divide
    // C into two parts and add to A and B such
    // that A is strictly greater than B
    static int countWays(int A, int B, int C)
    {
        // Minimum value added to A to satisfy
        // the given relation
        int minAddA = Math.max(0, (C + B - A + 2) / 2);
      
        // Number of different values of A, i.e.,
        // number of ways to divide C
        int count_ways = Math.max(C - minAddA + 1, 0);
      
        return count_ways;
    }
      
    // Driver code
    public static void main(String args[])
    {
        int A = 3, B = 5, C = 5;
      
        System.out.println(countWays(A, B, C));
    }
}
  
// This code is contributed by AbhiThakur

Python3

# Python3 implementation of the above approach
  
# Function to count the number of ways to divide
# C into two parts and add to A and B such
# that A is strictly greater than B
def countWays(A, B, C):
      
    # Minimum value added to A to satisfy
    # the given relation
    minAddA = max(0, (C + B - A + 2) // 2)
      
    # Number of different values of A, i.e.,
    # number of ways to divide C
    count_ways = max(C - minAddA + 1, 0)
      
    return count_ways
  
# Driver code
A = 3
B = 5
C = 5
print(countWays(A, B, C))
  
# This code is contributed by shivanisingh

C#

// C# implementation of the above approach
using System;
   
class GFG
{
  
// Function to count the number of ways to divide
// C into two parts and add to A and B such
// that A is strictly greater than B
static int countWays(int A, int B, int C)
{
    // Minimum value added to A to satisfy
    // the given relation
    int minAddA = Math.Max(0, (C + B - A + 2) / 2);
  
    // Number of different values of A, i.e.,
    // number of ways to divide C
    int count_ways = Math.Max(C - minAddA + 1, 0);
  
    return count_ways;
}
  
// Driver Code
public static void Main(String[] args)
{
    int A = 3, B = 5, C = 5;
  
    Console.Write(countWays(A, B, C));
}
  
}
  
// This code is contributed by shivanisinghss2110

输出：

2