将N分成K个唯一的部分，以使这些部分的gcd最大

给定正整数N ，任务是将其划分为K个唯一的部分，以使这些部分的总和等于原始数，并且所有部分的gcd最大。如果存在这样的除法，则打印最大的gcd ，否则打印-1 。

例子：

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

天真的方法：找到N的所有可能除法，并计算它们的最大gcd。但是这种方法将花费指数时间和空间。

高效的方法：让N的除法为A ₁ ，A ₂ ……..A _K
现在，已知gcd(a，b)= gcd(a，b，a + b) ，因此gcd(A ₁ ，A ₂ ….A _K )= gcd(A ₁ ，A ₂ ….A _K ，A ₁ + A ₂ …。+ A _K )
但是A ₁ + A ₂ …。 + A _K = N ，因此，除法的gcd将成为N的因素之一。
设a为N的因子，这可能是答案：
由于是GCD，所有的分裂将是的倍数，因此，对于K独特B _I，
a * B ₁ + a * B ₂ ……。 + a * B _K = N
a *(B ₁ + B ₂ ……. + B _K )= N
由于所有的B _i都是唯一的，
B ₁ + B ₂ ……。+ B _K≥1 + 2 + 3….. + K
乙₁ + B ₂ ……。+ b _k的序列≥K *(K + 1)/ 2
因此，互补因子大于或等于K *(K +1)/ 2的所有N个因子都可以作为可能的答案之一，而我们已考虑了所有可能的答案中的最大值。

下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
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;
}

Java

// 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

Python

# 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

C#

// 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

输出：

3