差为K的整数对，其元素为另一个的第K倍

给定整数K ，任务是找到一对数字(A，B) ，使A – B = K并且A / B =K。如果无法生成这样的对，则打印“ No” 。

例子：

Input: K = 6 
Output: 7.2 1.2 
Explanation:  
Since 7.2 – 1.2 = 6 and 7.2 / 1.2 = 6, The pair {7.2, 1.2} satisfy the necessary conditions.

Input: K = 1 
Output: No 
 

方法：
需要解决以下问题：

• 给定的条件可以用等式的形式写成：
  1. 公式(1)： A – B = K => A – B – K = 0
  2. 公式(2)： A / B = K => A –(K * B)= 0
• 求解这两个方程，我们得到：

( K * Equation (1) ) – Equation(2) = 0 
=> K*A – K*B – K² – (A – K*B) = 0 
=> K*A – K² – A – K*B + K*B = 0 
=> A*(K-1) – K² = 0 
=> A*(K-1) = K² 
=> A = K² / (K -1)
 

• 替换公式(1)中的A值，可以得出B的值为K /(K – 1)
• 可以观察到，如果K的值为1 ，则找不到这样的对，因为A和B的分母都变为0。

请按照以下步骤解决问题：

• 如果K等于1 ，则打印“ NO” 。
• 否则，如果K等于0以外的任何其他值，则计算(K * K)/(K -1)和K /(K – 1)的值并打印出来。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above problem
#include 
using namespace std;
 
// Function to find the
// required pair
void computePair(double K)
{
    // No pair possible
    if (K == 1) {
        cout << "No";
        return;
    }
    else {
 
        cout << K * K / (K - 1) << " ";
        cout << K / (K - 1) << endl;
    }
}
 
// Driver Code
int main()
{
    double K = 6;
    computePair(K);
 
    return 0;
}

Java

// Java program to implement
// the above problem
class GFG{
 
// Function to find the
// required pair
static void computePair(double K)
{
     
    // No pair possible
    if (K == 1)
    {
        System.out.print("No");
        return;
    }
    else
    {
        System.out.print(K * K / (K - 1) + " ");
        System.out.print(K / (K - 1) + "\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    double K = 6;
     
    computePair(K);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above problem
 
# Function to find the
# required pair
def computePair(K):
     
    # No pair possible
    if (K == 1):
        print("No")
        return
 
    else:
        print(K * K / (K - 1), end = " ")
        print(K / (K - 1))
 
# Driver Code
if __name__ == "__main__":
 
    K = 6
     
    computePair(K)
 
# This code is contributed by chitranayal

C#

// C# program to implement
// the above problem
using System;
 
class GFG{
 
// Function to find the
// required pair
static void computePair(double K)
{
     
    // No pair possible
    if (K == 1)
    {
        Console.Write("No");
        return;
    }
    else
    {
        Console.Write(K * K / (K - 1) + " ");
        Console.Write(K / (K - 1) + "\n");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    double K = 6;
     
    computePair(K);
}
}
 
// This code is contributed by gauravrajput1

输出：

7.2 1.2

时间复杂度： O(1)
辅助空间； O(1)