# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

• 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++

 `// 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 `

Output:

```2
```

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