Count ways to divide C in two parts and add to A and B to make A strictly greater than B

Given three integers A, B and C, the task is 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.

Examples:

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

Approach: On observing carefully, the following relation is formed for this problem.



  • Let addA and addB be added to A and B respectively.
  • Therefore, addA + addB = C and it should satisfy the inequality A + addA > B + addB.
  • Now, since addB = C – addA and put it in the inequality:
    A + addA > B + (C - addA)
    or, 2addA > C + B - A
    or, 2addA >= C + B - A + 1
    or, addA >= (C + B - A + 1) / 2
    
  • Since addA must be non negative, addA = max(0, (C + B – A + 1) / 2).
  • The division should be ceiling division, thus we can rewrite as addA = max(0, (C + B – A + 2) / 2).
  • Let this value be equal to minAddA. Since all integer values addA from [minAddA, C], satisfies the relation A + addA > B + addB, so the required number of ways is equal to max(0, C – minAddA + 1).

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

2

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