# Count operations of the given type required to reduce N to 0

Given an integer **n**. The task is to count the number of operations required to reduce **n** to **0**. In every operation, **n** can be updated as **n = n – d** where **d** is the smallest prime divisor of **n**.

**Examples:**

Input:n = 5

Output:1

5 is the smallest prime divisor, thus it gets subtracted and n gets reduced to 0.

Input:n = 25

Output:11

5 is the smallest prime divisor, thus it gets subtracted and n gets reduced to 20. Then 2 is the smallest divisor and so on.

Input:n = 4

Output:2

**Approach:**

- When
**n**is**even**then the smallest prime divisor of**n**will be**2**and subtracting**2**from**n**will again give an even integer i.e. gain**2**will be chosen as the smallest prime divisor and these steps will repeat until**n**gets reduced to**0**. - When
**n**is**odd**then the smallest prime divisor of**n**will also be odd and subtracting an odd integer from another odd integer will give an even integer as the result and then the result can be found out by repeating step 1 for the current even integer. - Thus, the task is to find the smallest divisor
**d**, subtract it,**n = n – d**and print**1 + ((n – d) / 2)**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the required ` `// number of operations ` `int` `countOperations (` `int` `n) ` `{ ` ` ` `int` `i = 2; ` ` ` ` ` `// Finding the smallest divisor ` ` ` `while` `((i * i) < n && (n % i)) ` ` ` `i += 1; ` ` ` ` ` `if` `((i * i) > n) ` ` ` `i = n; ` ` ` ` ` `// Return the count of operations ` ` ` `return` `(1 + (n - i)/2); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5; ` ` ` `cout << countOperations(n); ` `} ` ` ` `//This code is contributed by Shivi_Aggarwal ` |

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## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` `// Function to return the required ` `// number of operations ` `static` `int` `countOperations (` `int` `n) ` `{ ` ` ` `int` `i = ` `2` `; ` ` ` ` ` `// Finding the smallest divisor ` ` ` `while` `((i * i) < n && (n % i) > ` `0` `) ` ` ` `i += ` `1` `; ` ` ` ` ` `if` `((i * i) > n) ` ` ` `i = n; ` ` ` ` ` `// Return the count of operations ` ` ` `return` `(` `1` `+ (n - i) / ` `2` `); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `n = ` `5` `; ` ` ` `System.out.println(countOperations(n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to return the required ` `# number of operations ` `def` `countOperations(n): ` ` ` `i ` `=` `2` ` ` ` ` `# Finding the smallest divisor ` ` ` `while` `((i ` `*` `i) < n ` `and` `(n ` `%` `i)): ` ` ` `i ` `+` `=` `1` ` ` ` ` `if` `((i ` `*` `i) > n): ` ` ` `i ` `=` `n ` ` ` ` ` `# Return the count of operations ` ` ` `return` `(` `1` `+` `(n ` `-` `i)` `/` `/` `2` `) ` ` ` `# Driver code ` `n ` `=` `5` `print` `(countOperations(n)) ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` `class` `GFG ` `{ ` ` ` `// Function to return the required ` `// number of operations ` `static` `int` `countOperations (` `int` `n) ` `{ ` ` ` `int` `i = 2; ` ` ` ` ` `// Finding the smallest divisor ` ` ` `while` `((i * i) < n && (n % i) > 0) ` ` ` `i += 1; ` ` ` ` ` `if` `((i * i) > n) ` ` ` `i = n; ` ` ` ` ` `// Return the count of operations ` ` ` `return` `(1 + (n - i) / 2); ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ` `int` `n = 5; ` ` ` `Console.WriteLine(countOperations(n)); ` `} ` `} ` ` ` `// This code is contributed by mits ` |

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## PHP

`<?php ` `// PHP implementation of the approach ` ` ` `// Function to return the required ` `// number of operations ` `function` `countOperations(` `$n` `) ` `{ ` ` ` `$i` `= 2; ` ` ` ` ` `# Finding the smallest divisor ` ` ` `while` `((` `$i` `* ` `$i` `) < ` `$n` `and` `(` `$n` `% ` `$i` `)) ` ` ` `$i` `+= 1; ` ` ` ` ` `if` `((` `$i` `* ` `$i` `) > ` `$n` `) ` ` ` `$i` `= ` `$n` `; ` ` ` ` ` `# Return the ` `count` `of operations ` ` ` `return` `1 + ` `floor` `((` `$n` `- ` `$i` `) / 2); ` `} ` ` ` `// Driver code ` `$n` `= 5 ; ` `echo` `countOperations(` `$n` `); ` ` ` `// This code is contributed by Ryuga ` `?> ` |

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**Output:**

1

**Time Complexity:**

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