查找两个数字，总和与乘积都与N相同

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。如果没有这样的数字，请打印“ NO”。
例子：

Input: N = 69 
Output: a = 67.9851 
b = 1.01493
Input: N = 1 
Output: NO 
 

方法：如果仔细观察，我们将得到二次方程根的和与乘积。
如果N ² – 4 * N <0，则方程只有虚数根，因此“ NO”将是答案。其他a和b将是：

a = ( N + sqrt( N² – 4*N ) ) / 2 
b = ( N – sqrt( N² – 4*N ) ) / 2 
 

下面是上述方法的实现：

C++

// C++ program to find a and b
// such that a*b=N and a+b=N
#include 
using namespace std;
 
// Function to return the smallest string
void findAandB(double N)
{
    double val = N * N - 4.0 * N;
 
    // Not possible
    if (val < 0) {
        cout << "NO";
        return;
    }
 
    // find a and b
    double a = (N + sqrt(val)) / 2.0;
    double b = (N - sqrt(val)) / 2.0;
 
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
}
 
// Driver Code
int main()
{
    double N = 69.0;
    findAandB(N);
    return 0;
}

Java

// Java program to find a and b
// such that a*b=N and a+b=N
 
class GFG{
// Function to return the smallest string
static void findAandB(double N)
{
    double val = N * N - 4.0 * N;
 
    // Not possible
    if (val < 0) {
        System.out.println("NO");
        return;
    }
 
    // find a and b
    double a = (N + Math.sqrt(val)) / 2.0;
    double b = (N - Math.sqrt(val)) / 2.0;
 
    System.out.println("a = "+a);
    System.out.println("b = "+b);
}
 
// Driver Code
public static void main(String[] args)
{
    double N = 69.0;
    findAandB(N);
}
}
// This Code is contributed by mits

Python3

# Python 3 program to find a and b
# such that a*b=N and a+b=N
from math import sqrt
 
# Function to return the
# smallest string
def findAandB(N):
    val = N * N - 4.0 * N
 
    # Not possible
    if (val < 0):
        print("NO")
        return
     
    # find a and b
    a = (N + sqrt(val)) / 2.0
    b = (N - sqrt(val)) / 2.0
 
    print("a =", '{0:.6}' . format(a))
    print("b =", '{0:.6}' . format(b))
 
# Driver Code
if __name__ == '__main__':
    N = 69.0
    findAandB(N)
     
# This code is contributed
# by SURENDRA_GANGWAR

C#

// C# program to find a and b
// such that a*b=N and a+b=N
 
using System;
class GFG
{
// Function to return the smallest string
static void findAandB(double N)
{
double val = N * N - 4.0 * N;
 
// Not possible
if (val < 0) {
Console.WriteLine("NO");
return;
}
 
// find a and b
double a = (N + Math.Sqrt(val)) / 2.0;
double b = (N - Math.Sqrt(val)) / 2.0;
 
Console.WriteLine("a = "+a);
Console.WriteLine("b = "+b);
}
 
// Driver Code
static void Main()
{
double N = 69.0;
findAandB(N);
}
// This code is contributed by ANKITRAI1
}

Javascript

    // Javascript program to find a and b
    // such that a*b=N and a+b=N
     
    // Function to return the smallest string
    function findAandB(N)
    {
      let val = N * N - 4.0 * N;
 
      // Not possible
      if (val < 0) {
        document.write("NO");
        return;
      }
 
      // find a and b
      let a = (N + Math.sqrt(val)) / 2.0;
      let b = (N - Math.sqrt(val)) / 2.0;
 
      document.write("a = "+a.toFixed(4) + "</br>");
      document.write("b = "+b.toFixed(5));
    }
     
    let N = 69.0;
    findAandB(N);
         

输出：

a = 67.9851
b = 1.01493

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。