通过应用给定的操作最小化N的值

给定整数N ，以下操作可以对N执行任意次：

1. 将N乘以任何正整数X，即N = N * X。
2. 随N平方根替换N(N必须是一个整数)，即N = SQRT(N)。

任务是找到可以通过上述操作将N减少为的最小整数。

例子：

Input: N = 20
Output: 10
We can multiply 20 by 5, then take sqrt(20*5) = 10, this is the minimum number that 20 can be reduced to with the given operations.

Input: N = 36
Output: 6
Take sqrt(36). Number 6 can’t be reduced further.

方法：

1. 首先分解数字N。
2. 假设12具有因子2、2和5 。使用sqrt(n)只能减少重复的因子，即sqrt(2 * 2)= 2 。
3. 在这些因素中仅出现一次的数字无法进一步减少。
4. 因此，最终答案将是数字N的所有不同素数的乘积

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
#define ll long long int
using namespace std;
  
// function to return the product of
// distinct prime factors of a number
ll minimum(ll n)
{
    ll product = 1;
  
    // find distinct prime
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            while (n % i == 0)
                n = n / i;
            product = product * i;
        }
    }
    if (n >= 2)
        product = product * n;
  
    return product;
}
  
// Driver code
int main()
{
    ll n = 20;
    cout << minimum(n) << endl;
  
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
  
class solution
{
  
 // function to return the product of
 // distinct prime factors of a number
static int minimum(int n)
{
    int product = 1;
  
    // find distinct prime
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            while (n % i == 0)
                n = n / i;
            product = product * i;
        }
    }
    if (n >= 2)
        product = product * n;
  
    return product;
}
  
// Driver code
public static void main(String arr[])
{
    int n = 20;
    System.out.println(minimum(n));
  
}
}
//This code is contributed by
//Surendra_Gangwar

Python3

# Python3 implementation of above approach
  
# function to return the product 
# of distinct prime factors of a 
# numberdef minSteps(str):
def minimum(n):
      
    product = 1
      
    # find distinct prime
    i = 2
    while i * i <= n:
        if n % i == 0:
            while n % i == 0:
                n = n / i
            product = product * i
        i = i + 1 
    if n >= 2:
        product = product * n
    return product
  
# Driver code
  
# Get the binary string
n = 20
print(minimum(n))
          
# This code is contributed
# by Shashank_Sharma

C#

// C# implementation of the above approach
class GFG
{
  
// function to return the product of
// distinct prime factors of a number
static int minimum(int n)
{
    int product = 1;
  
    // find distinct prime
    for (int i = 2; i * i <= n; i++) 
    {
        if (n % i == 0) 
        {
            while (n % i == 0)
                n = n / i;
            product = product * i;
        }
    }
    if (n >= 2)
        product = product * n;
  
    return product;
}
  
// Driver code
static void Main()
{
    int n = 20;
    System.Console.WriteLine(minimum(n));
}
}
  
// This code is contributed by mits

PHP

= 2)
        $product = $product * $n;
  
    return $product;
}
  
// Driver code
$n = 20;
echo minimum($n),"\n";
  
// This code is contributed by ANKITRAI1
?>

输出：

10