Contar formas de dividir C en dos partes y sumar a A y B para hacer que A sea estrictamente mayor que B

Dados tres números enteros A, B y C , la tarea es contar el número de formas de dividir C en dos partes y sumar A y B de modo que A sea estrictamente mayor que B. 
Ejemplos: 
 

Entrada: A = 5, B = 3, C = 4 
Salida: 3 
Los posibles valores de A y B después de dividir C son: 
A = 7, B = 5 donde C se divide en 2 y 2. 
A = 8, B = 4 donde C se divide en 3 y 1. 
A – 9, B = 3 donde C se divide en 4 y 0.
Entrada: A = 3, B = 5, C = 5 
Salida: 2 
 

Planteamiento: Al observar detenidamente, se forma la siguiente relación para este problema. 
 

• Sume addA y addB a A y B respectivamente .
• Por lo tanto, addA + addB = C y debería satisfacer la desigualdad A + addA > B + addB .
• Ahora, ya que addB = C – addA y ponlo en la desigualdad: 
   

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

• Dado que addA debe ser no negativo, addA = max(0, (C+ B – A + 1) / 2) .
• La división debe ser una división máxima, por lo que podemos reescribirla como addA = max(0, (C+ B – A + 2) / 2).
• Sea este valor igual a minAddA . Dado que todos los valores enteros addA de [minAddA, C] , satisface la relación A + addA > B + addB , por lo que el número requerido de formas es igual a max(0, C – minAddA + 1) .

A continuación se muestra la implementación del enfoque anterior:
 

// 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;
}

// 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 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# 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

<script>
 
// Javascript 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
function countWays(A, B, C)
{
    // Minimum value added to A to satisfy
    // the given relation
    var minAddA = Math.max(0, parseInt((C + B - A + 2) / 2));
 
    // Number of different values of A, i.e.,
    // number of ways to divide C
    var count_ways = Math.max(C - minAddA + 1, 0);
 
    return count_ways;
}
 
// Driver code
var A = 3, B = 5, C = 5;
document.write( countWays(A, B, C));
 
// This code is contributed by rutvik_56.
</script>

2

Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)