计数直到N的数字都是完美的平方和完美的立方

给定数字N。任务是对N下的总数进行计数，这些总数既是完美的平方，又是一些整数的立方。
例子：

Input: N = 100
Output: 2
They are 1 and 64.

Input: N = 100000
Output: 6

方法：对于给定的正数N是一个完美的平方，它必须满足P ² = N类似地，对于一个理想的立方体，其中P和Q是一些正整数，Q ³ =N。
N = P ² = Q ³
因此，如果N是6的幂，那么这肯定会起作用。说N = A ⁶ ，可以写为(A ³ ) ²或(A ² ) ³ 。
因此，请选择小于N的每个正整数的6幂。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to return required count
int SquareCube(long long int N)
{
 
    int cnt = 0, i = 1;
 
    while (int(pow(i, 6)) <= N) {
        ++cnt;
        ++i;
    }
 
    return cnt;
}
 
int main()
{
    long long int N = 100000;
 
    // function call to print required answer
    cout << SquareCube(N);
    return 0;
}

Java

// Java implementation of the above approach
 
public class GFG{
 
    // Function to return required count
    static int SquareCube(long N)
    {
     
        int cnt = 0, i = 1;
     
        while ((int)(Math.pow(i, 6)) <= N) {
            ++cnt;
            ++i;
        }
     
        return cnt;
    }
     
    public static void main(String []args)
    {
        long N = 100000;
     
        // function call to print required answer
        System.out.println(SquareCube(N)) ;
    }
    // This code is contributed by Ryuga
}

Python3

# Python3 implementation of the
# above approach
 
# Function to return required count
def SquareCube( N):
 
    cnt, i = 0, 1
 
    while (i ** 6 <= N):
        cnt += 1
        i += 1
 
    return cnt
 
# Driver code
N = 100000
 
# function call to print required
# answer
print(SquareCube(N))
 
# This code is contributed
# by saurabh_shukla

C#

// C# implementation of the above approach
using System;
 
public class GFG{
  
    // Function to return required count
    static int SquareCube(long N)
    {
      
        int cnt = 0, i = 1;
      
        while ((int)(Math.Pow(i, 6)) <= N) {
            ++cnt;
            ++i;
        }
      
        return cnt;
    }
      
    public static void Main()
    {
        long N = 100000;
      
        // function call to print required answer
        Console.WriteLine(SquareCube(N)) ;
    }
}
 
/*This code is contributed by 29AjayKumar*/

Javascript

// JavaScript implementation of the above approach
 
// Function to return required count
function SquareCube(N)
{
 
    let cnt = 0, i = 1;
    while (Math.floor(Math.pow(i, 6)) <= N)
    {
        ++cnt;
        ++i;
    }
    return cnt;
}
 
    let N = 100000;
 
    // function call to print required answer
    document.write(SquareCube(N));
 
// This code is contributed by Surbhi Tyagi.

输出：

6

时间复杂度： O(N ^1/6 )

辅助空间： O(1)