Divide N into K unique parts such that gcd of those parts is maximum

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 = 3
Output: 1
Only possible division with unique
elements is (1, 2, 3) and gcd(1, 2, 3) = 1.

Input: N = 18, K = 3
Output: 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 A1, A2……..AK
Now, it is known that gcd(a, b) = gcd(a, b, a + b) and hence, gcd(A1, A2….AK) = gcd(A1, A2….AK, A1 + A2…. + AK)
But A1 + A2…. + AK = N and hence, the gcd of the divisions will be one of the factors of 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 Bi,
a * B1 + a * B2……. + a * BK = N
a * (B1 + B2…….+ BK) = N
Since all the Bi are unique,
B1 + B2…….+ BK ≥ 1 + 2 + 3 ….. + K
B1 + B2…….+ BK ≥ K * (K + 1) / 2
Hence, 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:

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// 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;
}
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// 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
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# 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
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// 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
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Output:
3





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