通过乘以任意数或取平方根来减少 N 的最小操作

给定一个数字N ，任务是通过应用以下操作任意次数来找到N的最小值：

• 将N乘以任意正整数
• 将N替换为sqrt(N) ，仅当 N 是一个完美的平方时。

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：这个问题可以使用贪心方法解决。以下是步骤：

1. 继续将N替换为sqrt(N)直到N是一个完美的平方。
2. 在上述步骤之后，从sqrt(N)迭代到2 ，对于每一个，如果N 可以被 i ²整除，我会继续用N / i替换N 。
3. 上述步骤后的N值将是可能的最小值。

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

Javascript

 
// Javascript program for above approach
 
// Function to reduce N to its minimum
// possible value by the given operations
function minValue(n)
{
 
    // Keep replacing n until is
    // an integer
    while (parseInt(Math.sqrt(n)) == Math.sqrt(n)
        && n > 1)
    {
        n = parseInt(Math.sqrt(n));
    }
 
    // Keep replacing n until n
    // is divisible by i * i
    for (var i = parseInt(Math.sqrt(n));
        i > 1; i--) {
 
        while (n % (i * i) == 0)
            n /= i;
    }
 
    // Print the answer
    document.write(n);
}
 
// Driver Code
 
// Given N
var N = 20;
 
// Function Call
minValue(N);
 
// This code is contributed by rutvik_56.

输出：

10

时间复杂度： O(N)
辅助空间： O(1)