N的最小除数D，使得gcd(D，M)大于1

给定两个正整数N和M。 ，任务是找到N的最小除数D，使gcd(D，M)> 1 。如果没有这样的除数，则打印-1。

例子：

Input: N = 8, M = 10
Output: 2

Input: N = 8, M = 1
Output: -1

一个天真的方法是对每个因子进行迭代，并计算该因子和M的gcd。如果它超过M，那么我们就有了答案。

时间复杂度： O(N * log max(N，M))

一种有效的方法是迭代直到sqrt(n)并检查gcd(i，m)。如果gcd(i，m)> 1，则打印并破坏它，否则我们检查gcd(n / i，m)并存储其中的最小值。

下面是上述方法的实现。

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to find the minimum divisor
int findMinimum(int n, int m)
{
    int mini = m;
  
    // Iterate for all factors of N
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            int sec = n / i;
  
            // Check for gcd > 1
            if (__gcd(m, i) > 1) {
                return i;
            }
  
            // Check for gcd > 1
            else if (__gcd(sec, m) > 1) {
                mini = min(sec, mini);
            }
        }
    }
  
    // If gcd is m itself
    if (mini == m)
        return -1;
    else
        return mini;
}
// Drivers code
int main()
{
    int n = 8, m = 10;
    cout << findMinimum(n, m);
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
  
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
      
} 
  
// Function to find the minimum divisor
static int findMinimum(int n, int m)
{
    int mini = m;
  
    // Iterate for all factors of N
    for (int i = 1; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            int sec = n / i;
  
            // Check for gcd > 1
            if (__gcd(m, i) > 1) 
            {
                return i;
            }
  
            // Check for gcd > 1
            else if (__gcd(sec, m) > 1)
            {
                mini = Math.min(sec, mini);
            }
        }
    }
  
    // If gcd is m itself
    if (mini == m)
        return -1;
    else
        return mini;
}
  
// Driver code
public static void main (String[] args) 
{
    int n = 8, m = 10;
    System.out.println(findMinimum(n, m));
}
}
  
// This code is contributed by chandan_jnu

Python3

# Python3 implementation of the above approach
import math
  
# Function to find the minimum divisor 
def findMinimum(n, m): 
  
    mini, i = m, 1
      
    # Iterate for all factors of N 
    while i * i <= n: 
        if n % i == 0: 
            sec = n // i 
  
            # Check for gcd > 1 
            if math.gcd(m, i) > 1: 
                return i 
  
            # Check for gcd > 1 
            elif math.gcd(sec, m) > 1: 
                mini = min(sec, mini) 
              
        i += 1
  
    # If gcd is m itself 
    if mini == m:
        return -1
    else:
        return mini 
  
# Drivers code 
if __name__ == "__main__": 
  
    n, m = 8, 10
    print(findMinimum(n, m)) 
  
# This code is contributed by Rituraj Jain

C#

// C# implementation of the above approach
using System;
  
class GFG
{
  
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
      
} 
  
// Function to find the minimum divisor
static int findMinimum(int n, int m)
{
    int mini = m;
  
    // Iterate for all factors of N
    for (int i = 1; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            int sec = n / i;
  
            // Check for gcd > 1
            if (__gcd(m, i) > 1) 
            {
                return i;
            }
  
            // Check for gcd > 1
            else if (__gcd(sec, m) > 1)
            {
                mini = Math.Min(sec, mini);
            }
        }
    }
  
    // If gcd is m itself
    if (mini == m)
        return -1;
    else
        return mini;
}
  
// Driver code
static void Main()
{
    int n = 8, m = 10;
    Console.WriteLine(findMinimum(n, m));
}
}
  
// This code is contributed by chandan_jnu

PHP

 1
            if (__gcd($m, $i) > 1) 
            {
                return $i;
            }
  
            // Check for gcd > 1
            else if (__gcd($sec, $m) > 1)
            {
                $mini = min($sec, $mini);
            }
        }
    }
  
    // If gcd is m itself
    if ($mini == $m)
        return -1;
    else
        return $mini;
}
  
// Driver code
$n = 8; $m = 10;
echo(findMinimum($n, $m));
  
// This code is contributed by Code_Mech.

输出：

2

时间复杂度： O(sqrt N * log max(N，M))

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