Given a positive integer **N**, the task is to divide it into **K** unique parts such that the sum of these parts is equal to the original number and the **gcd** of all the parts is maximum. Print the **maximum gcd** if such a division exists else print **-1**.

**Examples:**

Input:N = 6, K = 3Output:1

Only possible division with unique

elements is (1, 2, 3) and gcd(1, 2, 3) = 1.

Input:N = 18, K = 3Output:3

**Naive approach:** Find all the possible divisions of **N** and compute the maximum gcd of them. But this approach will take exponential time and space.

**Efficient approach:** Let the divisions of **N** be A_{1}, A_{2}……..A_{K}

Now, it is known that **gcd(a, b) = gcd(a, b, a + b)** and hence, **gcd(A _{1}, A_{2}….A_{K}) = gcd(A_{1}, A_{2}….A_{K}, A_{1} + A_{2}…. + A_{K})**

But

**A**and hence, the gcd of the divisions will be one of the factors of

_{1}+ A_{2}…. + A_{K}= N**N**.

Let

**a**be the factor of

**N**which can be the possible answer:

Since

**a**is the gcd, all the division will be the multiples of

**a**and hence, for

**K**unique

**B**,

_{i}**a * B**

_{1}+ a * B_{2}……. + a * B_{K}= N**a * (B**

_{1}+ B_{2}…….+ B_{K}) = NSince all the

**B**are unique,

_{i}**B**

_{1}+ B_{2}…….+ B_{K}≥ 1 + 2 + 3 ….. + K**B**

_{1}+ B_{2}…….+ B_{K}≥ K * (K + 1) / 2Hence, all the factors of

**N**whose complementary factor is greater than or equal to

**K * (K + 1) / 2**can be one of the possible answers and we have take to the maximum of all the possible answers.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; `
` ` `// Function to calculate maximum GCD ` `int` `maxGCD(` `int` `N, ` `int` `K) `
`{ ` ` ` ` ` `// Minimum possible sum for `
` ` `// K unique positive integers `
` ` `int` `minSum = (K * (K + 1)) / 2; `
` ` ` ` `// It is not possible to divide `
` ` `// N into K unique parts `
` ` `if` `(N < minSum) `
` ` `return` `-1; `
` ` ` ` `// All the factors greater than sqrt(N) `
` ` `// are complementary of the factors less `
` ` `// than sqrt(N) `
` ` `int` `i = ` `sqrt` `(N); `
` ` `int` `res = 1; `
` ` `while` `(i >= 1) { `
` ` ` ` `// If i is a factor of N `
` ` `if` `(N % i == 0) { `
` ` `if` `(i >= minSum) `
` ` `res = max(res, N / i); `
` ` ` ` `if` `(N / i >= minSum) `
` ` `res = max(res, i); `
` ` `} `
` ` `i--; `
` ` `} `
` ` ` ` `return` `res; `
`} ` ` ` `// Driver code ` `int` `main() `
`{ ` ` ` `int` `N = 18, K = 3; `
` ` ` ` `cout << maxGCD(N, K); `
` ` ` ` `return` `0; `
`} ` |

*chevron_right*

*filter_none*

`// Java implementation of the approach ` `import` `java.io.*; `
`import` `java.lang.Math; `
` ` `class` `GFG `
`{ ` ` ` `// Function to calculate maximum GCD `
` ` `static` `int` `maxGCD(` `int` `N, ` `int` `K) `
` ` `{ `
` ` ` ` `// Minimum possible sum for `
` ` `// K unique positive integers `
` ` `int` `minSum = (K * (K + ` `1` `)) / ` `2` `; `
` ` ` ` `// It is not possible to divide `
` ` `// N into K unique parts `
` ` `if` `(N < minSum) `
` ` `return` `-` `1` `; `
` ` ` ` `// All the factors greater than sqrt(N) `
` ` `// are complementary of the factors less `
` ` `// than sqrt(N) `
` ` `int` `i = (` `int` `) Math.sqrt(N); `
` ` `int` `res = ` `1` `; `
` ` `while` `(i >= ` `1` `) `
` ` `{ `
` ` ` ` `// If i is a factor of N `
` ` `if` `(N % i == ` `0` `) `
` ` `{ `
` ` `if` `(i >= minSum) `
` ` `res = Math.max(res, N / i); `
` ` ` ` `if` `(N / i >= minSum) `
` ` `res = Math.max(res, i); `
` ` `} `
` ` `i--; `
` ` `} `
` ` ` ` `return` `res; `
` ` `} `
` ` ` ` `// Driver code `
` ` `public` `static` `void` `main (String[] args) `
` ` `{ `
` ` `int` `N = ` `18` `, K = ` `3` `; `
` ` ` ` `System.out.println(maxGCD(N, K)); `
` ` `} `
`} ` ` ` `// This code is contributed by ApurvaRaj ` |

*chevron_right*

*filter_none*

`# Python3 implementation of the approach ` `from` `math ` `import` `sqrt,ceil,floor `
` ` `# Function to calculate maximum GCD ` `def` `maxGCD(N, K): `
` ` ` ` `# Minimum possible sum for `
` ` `# K unique positive integers `
` ` `minSum ` `=` `(K ` `*` `(K ` `+` `1` `)) ` `/` `2`
` ` ` ` `# It is not possible to divide `
` ` `# N into K unique parts `
` ` `if` `(N < minSum): `
` ` `return` `-` `1`
` ` ` ` `# All the factors greater than sqrt(N) `
` ` `# are complementary of the factors less `
` ` `# than sqrt(N) `
` ` `i ` `=` `ceil(sqrt(N)) `
` ` `res ` `=` `1`
` ` `while` `(i >` `=` `1` `): `
` ` ` ` `# If i is a factor of N `
` ` `if` `(N ` `%` `i ` `=` `=` `0` `): `
` ` `if` `(i >` `=` `minSum): `
` ` `res ` `=` `max` `(res, N ` `/` `i) `
` ` ` ` `if` `(N ` `/` `i >` `=` `minSum): `
` ` `res ` `=` `max` `(res, i) `
` ` ` ` `i` `-` `=` `1`
` ` ` ` `return` `res `
` ` `# Driver code ` ` ` `N ` `=` `18`
`K ` `=` `3`
` ` `print` `(maxGCD(N, K)) `
` ` `# This code is contributed by mohit kumar 29 ` |

*chevron_right*

*filter_none*

`// C# implementation of the approach ` `using` `System; `
` ` `class` `GFG `
`{ ` ` ` `// Function to calculate maximum GCD `
` ` `static` `int` `maxGCD(` `int` `N, ` `int` `K) `
` ` `{ `
` ` ` ` `// Minimum possible sum for `
` ` `// K unique positive integers `
` ` `int` `minSum = (K * (K + 1)) / 2; `
` ` ` ` `// It is not possible to divide `
` ` `// N into K unique parts `
` ` `if` `(N < minSum) `
` ` `return` `-1; `
` ` ` ` `// All the factors greater than sqrt(N) `
` ` `// are complementary of the factors less `
` ` `// than sqrt(N) `
` ` `int` `i = (` `int` `) Math.Sqrt(N); `
` ` `int` `res = 1; `
` ` `while` `(i >= 1) `
` ` `{ `
` ` ` ` `// If i is a factor of N `
` ` `if` `(N % i == 0) `
` ` `{ `
` ` `if` `(i >= minSum) `
` ` `res = Math.Max(res, N / i); `
` ` ` ` `if` `(N / i >= minSum) `
` ` `res = Math.Max(res, i); `
` ` `} `
` ` `i--; `
` ` `} `
` ` `return` `res; `
` ` `} `
` ` ` ` `// Driver code `
` ` `public` `static` `void` `Main() `
` ` `{ `
` ` `int` `N = 18, K = 3; `
` ` ` ` `Console.WriteLine(maxGCD(N, K)); `
` ` `} `
`} ` ` ` `// This code is contributed by AnkitRai01 ` |

*chevron_right*

*filter_none*

**Output:**

3

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