一对整数，其五次幂之差为 X
给定一个整数 X ，任务是找到一对 A 和 B 使得它们的五次方差为 X，即A5–B5= X。如果没有这样的双印“不可能”。
输入: X = 33 输出: 1 -2 解释:
输入: N = 211 输出: -2 -3 解释:
输入: N = 211
输出: -2 -3
解释:
天真方法: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，范围从-10 9 到 10 9 。 高效途径:思路是利用数学技术缩小 A 和 B 的范围。
自 A5–B5= X =>A5= X+B5。要使 A 尽可能高，B 也必须尽可能高，这从不等式中可以明显看出。
考虑 A = N 和 B = N–1 =>N5–(N–1)5= x .
通过二项式展开，我们知道
(p+1)ypT9】=(y+1)p+1–yp+1T10】=(p+1)(y+1)pT8】
所以我们可以说 LHS 的最大值是 4N 4 。
因此 4N5<= X =>N<=(X/5)1/5。 = >这就给了我们 N ~ 120。
由于 A 和 B 也可以是负数，我们简单的外推一下范围，最终得到的范围是 [-120，120]。
以下是上述方法的实现:
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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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
</script>
Output:
1 -2
时间复杂度: O(240*240)
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