对，它们的五次幂之差为X

给定一个整数X ，任务是找到对A和B ，使它们的五次幂之差为X，即A ⁵ – B ⁵ =X 。如果没有这样的配对，则打印“不可能”。

Input: X = 33 
Output: 1 -2 
Explanation: 
$1^{5} - (-2)^{5} = 1 - (-32) = 33 == N$
Input: N = 211 
Output: -2 -3 
Explanation: 
$(-2)^{5} - (-3)^{5} = -32 - (-243) = 211 == N$
 

天真的方法：一个简单的解决方案是使用两个for循环，一个用于A，一个用于B，范围为-10 ⁹到10 ⁹ 。
高效方法：想法是使用数学技术缩小A和B的范围。

由于A ⁵ – B ⁵ = X => A ⁵ = X + B ⁵ 。从不等式可以明显看出，要使A尽可能高，B也必须尽可能高。

Consider A = N and B = N – 1 
=> N⁵ – (N – 1)⁵ = X. 
 

通过二项式展开，我们知道

(p + 1)y^p <= (y + 1)^p+1 – y^p+1 <= (p+1)(y+1)^p 
 

因此，可以说LHS的最大值为4N ⁴ 。

Hence 4N⁵ <= X 
=> N <= (X/5)^1/5. 
=> This gives us N ~ 120. 
 

由于A和B也可以为负，因此我们简单地推断范围，最终得到的范围是[-120，120]。

下面是上述方法的实现：

C++

// C++ implementation to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
 
#include 
using namespace std;
 
// Function to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
void findPair(int x)
{
    int lim = 120;
 
    // Loop to choose every possible
    // pair with in the range
    for (int i = -lim; i <= lim; i++) {
        for (int j = -lim; j <= lim; j++) {
 
            // Check if equation holds
            if (pow(i, 5) - pow(j, 5) == x) {
                cout << i << ' ' << j << endl;
                return;
            }
        }
    }
    cout << "-1";
}
 
// Driver Code
signed main()
{
    int X = 33;
 
    // Function Call
    findPair(X);
    return 0;
}

Java

// Java implementation to find a
// pair of integers A & B such
// that difference of fifth power
// is equal to the given number X
class GFG{
 
// Function to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
static void findPair(int x)
{
    int lim = 120;
 
    // Loop to choose every possible
    // pair with in the range
    for(int i = -lim; i <= lim; i++)
    {
       for(int j = -lim; j <= lim; j++)
       {
            
          // Check if equation holds
          if (Math.pow(i, 5) -
              Math.pow(j, 5) == x)
          {
              System.out.print(i + " " +
                               j + "\n");
              return;
          }
       }
    }
    System.out.print("-1");
}
 
// Driver Code
public static void main(String[] args)
{
    int X = 33;
 
    // Function Call
    findPair(X);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to find 
# a pair of integers A & B such 
# that difference of fifth power
# is equal to the given number X
import math
 
# Function to find a pair
# of integers A & B such that
# difference of fifth power is
# equal to the given number X
def findPair(x):
 
    lim = 120
 
    # Loop to choose every possible
    # pair with in the range
    for i in range(-lim, lim + 1):
        for j in range(-lim, lim + 1):
             
            # Check if equation holds
            if (math.pow(i, 5) -
                math.pow(j, 5) == x):
                print (i, end = ' ')
                print (j, end = '\n')
                return
     
    print ("-1")
 
# Driver Code
X = 33
 
# Function Call
findPair(X)
 
# This code is contributed by PratikBasu

C#

// C# implementation to find a
// pair of integers A & B such
// that difference of fifth power
// is equal to the given number X
using System;
 
class GFG{
 
// Function to find a pair of
// integers A & B such that
// difference of fifth power is
// equal to the given number X
static void findPair(int x)
{
    int lim = 120;
 
    // Loop to choose every possible
    // pair with in the range
    for(int i = -lim; i <= lim; i++)
    {
       for(int j = -lim; j <= lim; j++)
       {
           
          // Check if equation holds
          if (Math.Pow(i, 5) -
              Math.Pow(j, 5) == x)
          {
              Console.Write(i + " " +
                            j + "\n");
              return;
          }
       }
    }
    Console.Write("-1");
}
 
// Driver code
public static void Main(String[] args)
{
    int X = 33;
 
    // Function call
    findPair(X);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript implementation to find a
// pair of integers A & B such
// that difference of fifth power
// is equal to the given number X
 
 
// Function to find a pair
// of integers A & B such that
// difference of fifth power is
// equal to the given number X
function findPair(x)
{
    let lim = 120;
 
    // Loop to choose every possible
    // pair with in the range
    for(let i = -lim; i <= lim; i++)
     
    for(let j = -lim; j <= lim; j++)
     
             
        // Check if equation holds
        if (Math.pow(i, 5) -Math.pow(j, 5) == x)
        {
            document.write(i + " "+ j);
            return;
        }
     
    document.write("-1");
}
 
// Driver Code
 
    let X = 33;
 
    // Function Call
    findPair(X);
 
// This code is contributed by mohan
 

输出：

1 -2

时间复杂度： O(240 * 240)