通过乘以任意数或取平方根来减少N的最小运算

给定数字N ，任务是通过多次应用以下运算来找到N的最小值：

• N乘以任何正整数
• 仅当N是一个完美的正方形时，才用sqrt(N)替换N。

例子：

Input: N = 20 
Output: 10
Explanation: 
Multiply -> 20 * 5 = 100 
sqrt(100) = 10, which is the minimum value obtainable.

Input: N = 5184 
Output: 6 
Explanation:
sqrt(5184) = 72. 
Multiply -> 72*18 = 1296 
sqrt(1296) = 6, which is the minimum value obtainable.
 

方法：可以使用贪婪方法解决此问题。步骤如下：

1. 继续将N替换为sqrt(N)，直到N是一个完美的正方形为止。
2. 完成上述步骤后，从sqrt(N)迭代到2 ，并且对于每一次，如果N被i ²整除，我将继续用N / i替换N。
3. 在上述步骤之后的N值将是最小可能值。

下面是上述方法的实现：

C++

// C++ program for above approach 
#include  
using namespace std; 
  
// Function to reduce N to its minimum 
// possible value by the given operations 
void minValue(int n) 
{ 
    // Keep replacing n until is 
    // an integer 
    while (int(sqrt(n)) == sqrt(n) 
        && n > 1) { 
        n = sqrt(n); 
    } 
  
    // Keep replacing n until n 
    // is divisible by i * i 
    for (int i = sqrt(n); 
        i > 1; i--) { 
  
        while (n % (i * i) == 0) 
            n /= i; 
    } 
  
    // Print the answer 
    cout << n; 
} 
  
// Driver Code 
int main() 
{ 
    // Given N 
    int N = 20; 
  
    // Function Call 
    minValue(N); 
}

Java

// Java implementation of the above approach 
import java.lang.Math; 
  
class GFG{ 
  
// Function to reduce N to its minimum 
// possible value by the given operations 
static void minValue(int n) 
{ 
      
    // Keep replacing n until is 
    // an integer 
    while ((int)Math.sqrt(n) == 
                Math.sqrt(n) && n > 1) 
    { 
        n = (int)(Math.sqrt(n)); 
    } 
  
    // Keep replacing n until n 
    // is divisible by i * i 
    for(int i = (int)(Math.sqrt(n)); 
            i > 1; i--)
    { 
        while (n % (i * i) == 0) 
            n /= i;
    } 
      
    // Print the answer 
    System.out.println(n); 
}
  
// Driver code 
public static void main(String args[])
{
      
    // Given N 
    int N = 20; 
      
    // Function call 
    minValue(N); 
} 
}
  
// This code is contributed by vikas_g

Python3

# Python3 program for the above approach 
import math 
  
# Function to reduce N to its minimum 
# possible value by the given operations 
def MinValue(n):
      
    # Keep replacing n until is 
    # an integer 
    while(int(math.sqrt(n)) == 
              math.sqrt(n) and n > 1):
        n = math.sqrt(n)
          
    # Keep replacing n until n 
    # is divisible by i * i 
    for i in range(int(math.sqrt(n)), 1, -1):
        while (n % (i * i) == 0):
            n /= i
              
    # Print the answer 
    print(n)
  
# Driver code
n = 20
  
# Function call
MinValue(n)
  
# This code is contributed by virusbuddah_

C#

// C# implementation of the approach 
using System; 
  
class GFG{ 
      
// Function to reduce N to its minimum 
// possible value by the given operations 
static void minValue(int n) 
{ 
      
    // Keep replacing n until is 
    // an integer 
    while ((int)Math.Sqrt(n) == 
                Math.Sqrt(n) && n > 1)
    { 
        n = (int)(Math.Sqrt(n)); 
    } 
      
    // Keep replacing n until n 
    // is divisible by i * i 
    for (int i = (int)(Math.Sqrt(n));
             i > 1; i--)
    { 
        while (n % (i * i) == 0) 
            n /= i; 
    } 
      
    // Print the answer 
    Console.Write(n); 
}
  
// Driver code 
public static void Main() 
{
      
    // Given N 
    int N = 20; 
      
    // Function call 
    minValue(N);
}
}
  
// This code is contributed by vikas_g

输出：

10

时间复杂度： O(N)
辅助空间： O(1)