Given a large integer **N**, the task is to find all the possible remainders when **N** is divided by all the positive integers from **1** to **N + 1**.

**Examples:**

Input:N = 5Output:0 1 2 5

5 % 1 = 0

5 % 2 = 1

5 % 3 = 2

5 % 4 = 1

5 % 5 = 0

5 % 6 = 5

Input:N = 11Output:0 1 2 3 5 11

**Naive approach:** Run a loop from **1** to **N + 1** and return all the unique remainders found when dividing **N** by any integer from the range. But this approach is not efficient for larger values of **N**.

**Efficient approach:** It can be observed that one part of the answer will always contain numbers between **0** to **ceil(sqrt(n))**. It can be proven by running the naive algorithm on smaller values of **N** and checking the remainders obtained or by solving the equation **ceil(N / k) = x** or **x ≤ (N / k) < x + 1** where **x** is one of the remainders for all integers **k** when **N** is divided by **k** for **k** from **1** to **N + 1**.

The solution to the above inequality is nothing but integers **k** from **(N / (x + 1), N / x]** of length **N / x – N / (x + 1) = N / (x ^{2} + x)**. Therefore, iterate from

**k = 1**to

**ceil(sqrt(N))**and store all the unique

**N % k**. What if the above

**k**is greater than

**ceil(sqrt(N))**? They will always correspond to values

**0 ≤ x < ceil(sqrt(N))**. So, again start storing remainders from

**N / (ceil(sqrt(N)) – 1**to

**0**and return the final answer with all the possible remainders.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `typedef` `long` `long` `int` `ll;` ` ` `// Function to find all the distinct` `// remainders when n is divided by` `// all the elements from` `// the range [1, n + 1]` `void` `findRemainders(ll n)` `{` ` ` ` ` `// Set will be used to store` ` ` `// the remainders in order` ` ` `// to eliminate duplicates` ` ` `set<ll> vc;` ` ` ` ` `// Find the remainders` ` ` `for` `(ll i = 1; i <= ` `ceil` `(` `sqrt` `(n)); i++)` ` ` `vc.insert(n / i);` ` ` `for` `(ll i = n / ` `ceil` `(` `sqrt` `(n)) - 1; i >= 0; i--)` ` ` `vc.insert(i);` ` ` ` ` `// Print the contents of the set` ` ` `for` `(` `auto` `it : vc)` ` ` `cout << it << ` `" "` `;` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `ll n = 5;` ` ` ` ` `findRemainders(n);` ` ` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` ` ` `class` `GFG` `{` ` ` `// Function to find all the distinct` `// remainders when n is divided by` `// all the elements from` `// the range [1, n + 1]` `static` `void` `findRemainders(` `long` `n)` `{` ` ` ` ` `// Set will be used to store` ` ` `// the remainders in order` ` ` `// to eliminate duplicates` ` ` `HashSet<Long> vc = ` `new` `HashSet<Long>();` ` ` ` ` `// Find the remainders` ` ` `for` `(` `long` `i = ` `1` `; i <= Math.ceil(Math.sqrt(n)); i++)` ` ` `vc.add(n / i);` ` ` `for` `(` `long` `i = (` `long` `) (n / Math.ceil(Math.sqrt(n)) - ` `1` `); ` ` ` `i >= ` `0` `; i--)` ` ` `vc.add(i);` ` ` ` ` `// Print the contents of the set` ` ` `for` `(` `long` `it : vc)` ` ` `System.out.print(it+ ` `" "` `);` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `long` `n = ` `5` `;` ` ` ` ` `findRemainders(n);` `}` `}` ` ` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 implementation of the approach` `from` `math ` `import` `ceil, floor, sqrt` ` ` `# Function to find all the distinct` `# remainders when n is divided by` `# all the elements from` `# the range [1, n + 1]` `def` `findRemainders(n):` ` ` ` ` `# Set will be used to store` ` ` `# the remainders in order` ` ` `# to eliminate duplicates` ` ` `vc ` `=` `dict` `()` ` ` ` ` `# Find the remainders` ` ` `for` `i ` `in` `range` `(` `1` `, ceil(sqrt(n)) ` `+` `1` `):` ` ` `vc[n ` `/` `/` `i] ` `=` `1` ` ` `for` `i ` `in` `range` `(n ` `/` `/` `ceil(sqrt(n)) ` `-` `1` `, ` `-` `1` `, ` `-` `1` `):` ` ` `vc[i] ` `=` `1` ` ` ` ` `# Print the contents of the set` ` ` `for` `it ` `in` `sorted` `(vc):` ` ` `print` `(it, end ` `=` `" "` `)` ` ` `# Driver code` `n ` `=` `5` ` ` `findRemainders(n)` ` ` `# This code is contributed by Mohit Kumar` |

## C#

`// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;` ` ` `class` `GFG` `{` ` ` `// Function to find all the distinct` `// remainders when n is divided by` `// all the elements from` `// the range [1, n + 1]` `static` `void` `findRemainders(` `long` `n)` `{` ` ` ` ` `// Set will be used to store` ` ` `// the remainders in order` ` ` `// to eliminate duplicates` ` ` `List<` `long` `> vc = ` `new` `List<` `long` `>();` ` ` ` ` `// Find the remainders` ` ` ` ` `for` `(` `long` `i = 1; i <= Math.Ceiling(Math.Sqrt(n)); i++)` ` ` `vc.Add(n / i);` ` ` `for` `(` `long` `i = (` `long` `) (n / Math.Ceiling(Math.Sqrt(n)) - 1); ` ` ` `i >= 0; i--)` ` ` `vc.Add(i);` ` ` `vc.Reverse();` ` ` ` ` `// Print the contents of the set` ` ` `foreach` `(` `long` `it ` `in` `vc)` ` ` `Console.Write(it + ` `" "` `);` `}` ` ` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `long` `n = 5;` ` ` ` ` `findRemainders(n);` `}` `}` ` ` `// This code is contributed by PrinciRaj1992` |

**Output:**

0 1 2 5

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