找到两个与 N 相同的差和除法的数字

给定一个整数 N ，任务是找到两个数字 a 和 b ，使得a / b = N和a – b = N 。如果不可能有这样的数字，请打印“否” 。
例子：

Input: N = 6 
Output: 
a = 7.2
b = 1.2
Explanation:
For the given two numbers a and b, a/b = 6 = N and a-b = 6 = N

Input: N = 1  
Output: No
Explanation:
There are no values of a and b that satisfy the condition.

方法：要解决这个问题，请观察下面推导出的方程：

$\begin{cases} a - b &= N \\ a - Nb& = 0 \end{cases}$

同时求解上述方程，我们得到：

$A=\dfrac{N^2}{N-1}$
$B=\dfrac{N}{N-1}$

由于分母是N – 1 ，所以当N = 1时，答案是不可能的。对于所有其他情况，答案是可能的。因此，分别求出 a 和 b 的值。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find two numbers with
// difference and division both as N
void findAandB(double N)
{
    // Condition if the answer
    // is not possible
 
    if (N == 1) {
        cout << "No";
        return;
    }
 
    // Calculate a and b
    double a = N * N / (N - 1);
    double b = a / N;
 
    // Print the values
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
}
 
// Driver Code
int main()
{
    // Given N
    double N = 6;
 
    // Function Call
    findAandB(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to find two numbers with
// difference and division both as N
static void findAandB(double N)
{
     
    // Condition if the answer
    // is not possible
    if (N == 1)
    {
        System.out.print("No");
        return;
    }
 
    // Calculate a and b
    double a = N * N / (N - 1);
    double b = a / N;
 
    // Print the values
    System.out.print("a = " + a + "\n");
    System.out.print("b = " + b + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given N
    double N = 6;
 
    // Function call
    findAandB(N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
 
# Function to find two numbers with
# difference and division both as N
def findAandB(N):
     
    # Condition if the answer
    # is not possible
    if (N == 1):
        print("No")
        return
     
    # Calculate a and b
    a = N * N / (N - 1)
    b = a / N
 
    # Print the values
    print("a = ", a)
    print("b = ", b)
 
# Driver Code
 
# Given N
N = 6
 
# Function call
findAandB(N)
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find two numbers with
// difference and division both as N
static void findAandB(double N)
{
 
    // Condition if the answer
    // is not possible
    if (N == 1)
    {
        Console.Write("No");
        return;
    }
 
    // Calculate a and b
    double a = N * N / (N - 1);
    double b = a / N;
 
    // Print the values
    Console.Write("a = " + a + "\n");
    Console.Write("b = " + b + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Given N
    double N = 6;
 
    // Function call
    findAandB(N);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

// Javascript program for the above approach
 
 
    // Function to find two numbers with
    // difference and division both as N
    function findAandB( N) {
 
        // Condition if the answer
        // is not possible
        if (N == 1) {
            document.write("No");
            return;
        }
 
        // Calculate a and b
        let a = N * N / (N - 1);
        let b = a / N;
 
        // Prlet the values
        document.write("a = " + a + "<br/>");
        document.write("b = " + b + "<br/>");
    }
 
    // Driver Code
      
 
        // Given N
        let N = 6;
 
        // Function call
        findAandB(N);
 
 
 
// This code contributed by aashish1995

输出：

a = 7.2
b = 1.2

时间复杂度： O(1)
辅助空间： O(1)