Program to find the value of P(N + r) for a polynomial of a degree N such that P(i) = 1 for 1 ≤ i ≤ N and P(N + 1) = a

Given three positive integers N, R, and A and a polynomial P(X) of degree N, P(i) = 1 for 1 ? i ? N and the value of P(N + 1) is A, the task is to find the value of the P(N + R).

Examples:

Input: N = 1, R = 3, A = 2
Output: 4
Explanation:
The equation of the given polynomial is P(x) = x. Therefore, the value of P(N + R) = P(4) = 4.

Input: N = 2, R = 3, A = 1
Output: 1

Approach: The given problem can be solved by finding the expression of the given polynomial P(X) and then calculate the value of P(N + R). The general form of a polynomial with degree N is:

P(x) = k * (x – x₁) * (x – x₂) * (x – x₃) * … *(x – x_n) + c
where x₁, x₂, x₃, …, x_n are the roots P(x) where k and c are arbitrary constants.

Consider F(x) be another polynomial of degree N such that
F(x) = P(x) + c — (1)

Now, if c = -1, F(x) = 0 since P(x) = 1 for x ? [1, N].
? F(x) has N roots which are equal to 1, 2, 3, …, N.
Therefore, F(x) = k * (x – 1) * (x – 2) * … * (x – N), for all c = -1.

Rearranging terms in equation (1),  
P(x) = F(x) – c 
P(x) = k * (x – 1) * (x – 2) * … *(x – N) + 1 — (2)

Putting x = N + 1 in equation (2),  
k = (a – 1) / N!

Therefore, P(x) = ((a – 1)/N!) * (x – 1) * (x – 2) * … *(x – N) + 1 — (3)

Therefore, the idea is to substitute the value of (N + R) in equation (3) and print the value of the expression as the result. Follow the steps below to solve the problem:

• Initialize a variable, say ans, as the value of (A – 1) / N!.
• Iterate over the range [1, N] using the variable i and update the value of ans as ans * (N + R – i).
• After completing the above steps, print the value of (ans + 1) as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// factorial of N
int fact(int n)
{

    // Base Case
    if (n == 1 || n == 0)
        return 1;

    // Otherwise, recursively
    // calculate the factorial
    else
        return n * fact(n - 1);
}

// Function to find the value of
// P(n + r) for polynomial P(X)
int findValue(int n, int r, int a)
{

    // Stores the value of k
    int k = (a - 1) / fact(n);

    // Store the required answer
    int answer = k;

    // Iterate in the range [1, N] and
    // multiply (n + r - i) with answer
    for(int i = 1; i < n + 1; i++)
        answer = answer * (n + r - i);

    // Add the constant value C as 1
    answer = answer + 1;

    // Return the result
    return answer;
}

// Driver Code
int main()
{
    int N = 1;
    int A = 2;
    int R = 3;

    cout << (findValue(N, R, A));

    return 0;
}

// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to calculate
// factorial of N
static int fact(int n)
{

    // Base Case
    if (n == 1 || n == 0)
        return 1;

    // Otherwise, recursively
    // calculate the factorial
    else
        return n * fact(n - 1);
}

// Function to find the value of
// P(n + r) for polynomial P(X)
static int findValue(int n, int r, int a)
{

    // Stores the value of k
    int k = (a - 1) / fact(n);

    // Store the required answer
    int answer = k;

    // Iterate in the range [1, N] and
    // multiply (n + r - i) with answer
    for(int i = 1; i < n + 1; i++)
        answer = answer * (n + r - i);

    // Add the constant value C as 1
    answer = answer + 1;

    // Return the result
    return answer;
}

// Driver Code
public static void main(String args[])
{
    int N = 1;
    int A = 2;
    int R = 3;
    System.out.print(findValue(N, R, A));
}

}

// This code is contributed by SURENDRA_GANGWAR.

Python3

# Python program for the above approach

# Function to calculate
# factorial of N
def fact(n):

    # Base Case
    if n == 1 or n == 0:
        return 1

    # Otherwise, recursively
    # calculate the factorial
    else:
        return n * fact(n-1)

# Function to find the value of
# P(n + r) for polynomial P(X)
def findValue(n, r, a):

    # Stores the value of k
    k = (a-1) // fact(n)

    # Store the required answer
    answer = k

    # Iterate in the range [1, N] and
    # multiply (n + r - i) with answer
    for i in range(1, n + 1):
        answer = answer*(n + r-i)

    # Add the constant value C as 1
    answer = answer + 1

    # Return the result
    return answer


# Driver Code

N = 1
A = 2
R = 3

print(findValue(N, R, A))

C#

// C# program for the above approach
using System;

class GFG{

// Function to calculate
// factorial of N
static int fact(int n)
{

    // Base Case
    if (n == 1 || n == 0)
        return 1;

    // Otherwise, recursively
    // calculate the factorial
    else
        return n * fact(n - 1);
}

// Function to find the value of
// P(n + r) for polynomial P(X)
static int findValue(int n, int r, int a)
{

    // Stores the value of k
    int k = (a - 1) / fact(n);

    // Store the required answer
    int answer = k;

    // Iterate in the range [1, N] and
    // multiply (n + r - i) with answer
    for(int i = 1; i < n + 1; i++)
        answer = answer * (n + r - i);

    // Add the constant value C as 1
    answer = answer + 1;

    // Return the result
    return answer;
}

// Driver Code
static public void Main()
{
    int N = 1;
    int A = 2;
    int R = 3;

    Console.Write((findValue(N, R, A)));
}
}

// This code is contributed by code_hunt

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to calculate
// factorial of N
function fact(n)
{

    // Base Case
    if (n == 1 || n == 0)
        return 1;

    // Otherwise, recursively
    // calculate the factorial
    else
        return n * fact(n - 1);
}

// Function to find the value of
// P(n + r) for polynomial P(X)
function findValue(n, r, a)
{

    // Stores the value of k
    let k = (a - 1) / fact(n);

    // Store the required answer
    let answer = k;

    // Iterate in the range [1, N] and
    // multiply (n + r - i) with answer
    for(let i = 1; i < n + 1; i++)
        answer = answer * (n + r - i);

    // Add the constant value C as 1
    answer = answer + 1;

    // Return the result
    return answer;
}

// Driver code
    let N = 1;
    let A = 2;
    let R = 3;
    document.write(findValue(N, R, A));

 // This code is contributed by code_hunt.
</script>

Output:

4

Time Complexity: O(N)
Auxiliary Space: O(1)