数的最大除数不能被一个完美的平方整除

给定一个正整数N。查找给定数字的最大除数，该除数不能被大于1的完美平方除以。

例子：

Input : 12
Output : 6
Explanation : Divisors of 12 are 1, 2, 3, 4, 6 and 12. 
Since 12 is divisible by 4 (a perfect square), 
it can't be required divisor. 6 is not divisible 
by any perfect square.
 
Input :97
Output :97

一种简单的方法是通过迭代到N的平方根来查找给定数字N的所有除数，并将它们按排序顺序(降序)保留在列表中。在这里，我们将它们按降序插入到集合中以保持它们的排序。另外，通过从1到10 ⁵进行迭代，列出直至10 ¹⁰的所有理想平方。
现在，对于从最大的除数开始的每个除数，请检查其是否可以被列表中的任何理想平方整除。如果除数不能被任何完美数整除，只需将其返回为答案即可。

下面是上述方法的实现。

C++

// C++ Program to find the largest
// divisor not divisible by any
// perfect square greater than 1
#include 
using namespace std;
 
const int MAX = 1e5;
 
// Function to find the largest
// divisor not divisible by any
// perfect square greater than 1
int findLargestDivisor(int n)
{
    // set to store divisors in
    // descending order
    int m = n;
    set > s;
    s.insert(1);
    s.insert(n);
 
    for (int i = 2; i < sqrt(n) + 1; i++) {
        // If the number is divisible
        // by i, then insert it
        if (n % i == 0) {
            s.insert(n / i);
            s.insert(i);
            while (m % i == 0)
                m /= i;
        }
    }
 
    if (m > 1)
        s.insert(m);
 
    // Vector to store perfect squares
    vector vec;
    for (int i = 2; i <= MAX; i++)
        vec.push_back(i * i);
 
    // Check for each divisor, if it is not
    // divisible by any perfect square,
    // simply return it as the answer.
    for (auto d : s) {
        int divi = 0;
        for (int j = 0; j < vec.size()
                        && vec[j] <= d;
             j++) {
            if (d % vec[j] == 0) {
                divi = 1;
                break;
            }
        }
        if (!divi)
            return d;
    }
}
 
// Driver Code
int main()
{
    int n = 12;
    cout << findLargestDivisor(n) << endl;
 
    n = 97;
    cout << findLargestDivisor(n) << endl;
    return 0;
}

Java

// Java Program to find the largest
// divisor not divisible by any
// perfect square greater than 1
import java.util.*;
class Main{
     
static int MAX = (int)1e5;
     
// Function to find the largest
// divisor not divisible by any
// perfect square greater than 1
public static int findLargestDivisor(int n)
{
  // set to store divisors in
  // descending order
  int m = n;
  Set s =
      new HashSet();
  s.add(1);
  s.add(n);
 
  for (int i = 2;
           i < (int)Math.sqrt(n) + 1; i++)
  {
    // If the number is divisible
    // by i, then insert it
    if (n % i == 0)
    {
      s.add(n / i);
      s.add(i);
      while (m % i == 0)
        m /= i;
    }
  }
 
  if (m > 1)
    s.add(m);
 
  List l =
       new ArrayList(s);
 
  Collections.sort(l);
  Collections.reverse(l);
 
  // Vector to store
  // perfect squares
  Vector vec =
         new Vector();
 
  for (int i = 2; i <= MAX; i++)
    vec.add(i * i);
 
  // Check for each divisor, if
  // it is not divisible by any
  // perfect square, simply return
  // it as the answer.
  for (int d : l)
  {
    int divi = 0;
     
    for (int j = 0;
             j < vec.size() &&
             vec.get(j) <= d; j++)
    {
      if (d % vec.get(j) == 0)
      {
        divi = 1;
        break;
      }
    }
    if (divi == 0)
      return d;
  }
  return 0;
}
 
// Driver code   
public static void main(String[] args)
{
  int n = 12;
  System.out.println(findLargestDivisor(n));
 
  n = 97;
  System.out.println(findLargestDivisor(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 Program to find the largest
# divisor not divisible by any
# perfect square greater than 1
 
MAX = 10 ** 5
 
# Function to find the largest
# divisor not divisible by any
# perfect square greater than 1
def findLargestDivisor(n):
 
    # set to store divisors in
    # descending order
    m = n
    s = set()
    s.add(1)
    s.add(n)
 
    for i in range(2, int(n ** (0.5)) + 1):
         
        # If the number is divisible
        # by i, then insert it
        if n % i == 0:
            s.add(n // i)
            s.add(i)
            while m % i == 0:
                m //= i
         
    if m > 1:
        s.add(m)
 
    # Vector to store perfect squares
    vec = [i**2 for i in range(2, MAX + 1)]
 
    # Check for each divisor, if it is not
    # divisible by any perfect square,
    # simply return it as the answer.
    for d in sorted(s, reverse = True):
         
        divi, j = 0, 0
        while j < len(vec) and vec[j] <= d:
 
            if d % vec[j] == 0:
                divi = 1
                break
            j += 1       
         
        if not divi:
            return d
 
# Driver Code
if __name__ == "__main__":
 
    n = 12
    print(findLargestDivisor(n))
 
    n = 97
    print(findLargestDivisor(n))
     
# This code is contributed by Rituraj Jain

C#

// C# program to find the largest
// divisor not divisible by any
// perfect square greater than 1
using System;
using System.Collections.Generic;
 
class GFG{
     
static int MAX = (int)1e5;
 
// Function to find the largest
// divisor not divisible by any
// perfect square greater than 1
static int findLargestDivisor(int n)
{
     
    // Set to store divisors in
    // descending order
    int m = n;
     
    HashSet s = new HashSet();
    s.Add(1);
    s.Add(n);
     
    for(int i = 2;
            i < (int)Math.Sqrt(n) + 1;
            i++)
    {
         
        // If the number is divisible
        // by i, then insert it
        if (n % i == 0)
        {
            s.Add(n / i);
            s.Add(i);
             
            while (m % i == 0)
                m /= i;
        }
    }
     
    if (m > 1)
        s.Add(m);
     
    List l = new List(s);
    l.Sort();
    l.Reverse();
     
    // Vector to store
    // perfect squares
    List vec = new List();
     
    for(int i = 2; i <= MAX; i++)
        vec.Add(i * i);
     
    // Check for each divisor, if
    // it is not divisible by any
    // perfect square, simply return
    // it as the answer.
    foreach(int d in l)
    {
        int divi = 0;
         
        for(int j = 0;
                j < vec.Count && vec[j] <= d;
                j++)
        {
            if (d % vec[j] == 0)
            {
                divi = 1;
                break;
            }
        }
        if (divi == 0)
            return d;
    }
    return 0;
} 
 
// Driver code
static void Main()
{
    int n = 12;
    Console.WriteLine(findLargestDivisor(n));
     
    n = 97;
    Console.WriteLine(findLargestDivisor(n));
}
}
 
// This code is contributed by divyesh072019

输出：

6
97

一种有效的方法是为每个i除以n，以便(i * i)除n。

C++

// Efficient C++ Program to find the
// largest divisor not divisible by any
// perfect square greater than 1
#include 
using namespace std;
  
// Function to find the largest
// divisor not divisible by any
// perfect square greater than 1
int findLargestDivisor(int n)
{
    for (int i = 2; i < sqrt(n) + 1; i++) {
         
        // If the number is divisible
        // by i*i, then remove one i
        while (n % (i * i) == 0) {
            n = n / i;
        }
    }
   
    // Now all squares are removed from n
    return n;   
}
  
// Driver Code
int main()
{
    int n = 12;
    cout << findLargestDivisor(n) << endl;
  
    n = 97;
    cout << findLargestDivisor(n) << endl;
    return 0;
}

Java

// Efficient Java Program to find the
// largest divisor not divisible by any
// perfect square greater than 1
 
public class GFG
{
 
    // Function to find the largest
    // divisor not divisible by any
    // perfect square greater than 1
    static int findLargestDivisor(int n)
    {
        for (int i = 2; i < Math.sqrt(n) + 1; i++) {
             
            // If the number is divisible
            // by i*i, then remove one i
            while (n % (i * i) == 0) {
                n = n / i;
            }
        }
         
        // Now all squares are removed from n
        return n;    
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int n = 12;
        System.out.println(findLargestDivisor(n)) ;
     
        n = 97;
        System.out.println(findLargestDivisor(n)) ;
     
    }
    // This code is contributed
    // by Ryuga
}

Python3

# Efficient Python3 Program to find the
# largest divisor not divisible by any
# perfect square greater than 1
import math
 
# Function to find the largest
# divisor not divisible by any
# perfect square greater than 1
def findLargestDivisor( n):
 
    for i in range (2, int(math.sqrt(n)) + 1) :
         
        # If the number is divisible
        # by i*i, then remove one i
        while (n % (i * i) == 0) :
            n = n // i
     
    # Now all squares are removed from n
    return n
 
# Driver Code
if __name__ == "__main__":
 
    n = 12
    print (findLargestDivisor(n))
 
    n = 97
    print (findLargestDivisor(n))
 
# This code is contributed by ita_c

C＃

// Efficient C# Program to find the
// largest divisor not divisible by any
// perfect square greater than 1
using System;
public class GFG
{
 
    // Function to find the largest
    // divisor not divisible by any
    // perfect square greater than 1
    static int findLargestDivisor(int n)
    {
        for (int i = 2; i < Math.Sqrt(n) + 1; i++) {
             
            // If the number is divisible
            // by i*i, then remove one i
            while (n % (i * i) == 0) {
                n = n / i;
            }
        }
         
        // Now all squares are removed from n
        return n;    
    }
     
    // Driver Code
    public static void Main()
    {
        int n = 12;
        Console.WriteLine(findLargestDivisor(n)) ;
     
        n = 97;
        Console.WriteLine(findLargestDivisor(n)) ;
     
    }
}
    // This code is contributed
    // by Mukul Singh

Java脚本

    // Efficient Javascript Program to find the
    // largest divisor not divisible by any
    // perfect square greater than 1
     
    // Function to find the largest
    // divisor not divisible by any
    // perfect square greater than 1
    function findLargestDivisor(n)
    {
        for (let i = 2; i < Math.sqrt(n) + 1; i++)
        {
 
            // If the number is divisible
            // by i*i, then remove one i
            while (n % (i * i) == 0) {
                n = n / i;
            }
        }
 
        // Now all squares are removed from n
        return n;  
    }
     
    let n = 12;
    document.write(findLargestDivisor(n) + "</br>");
   
    n = 97;
    document.write(findLargestDivisor(n) + "</br>");
     
    // This code is contributed by suresh07.

输出：

6
97