程序找到一个次数为N的多项式的P(N + r)的值，使得对于1≤i≤N且P(N + 1)= a的P(i)= 1

给出三个正整数N，R，A和多项式P的程度N，P(i)用于1 = 1(X)≤I≤N和P的(N + 1)是A的值，则任务是找出P(N + R)的值。

例子：

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

方法：可以通过找到给定多项式P(X)的表达式，然后计算P(N + R)的值来解决给定的问题。次数为N的多项式的一般形式为：

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)

因此，其想法是用等式(3)中的(N + R)值代替并打印表达式的值作为结果。请按照以下步骤解决问题：

• 将变量ans初始化为(A – 1)/ N的值！ 。
• 使用变量i在范围[1，N]上进行迭代，并将ans的值更新为ans *(N + R – i) 。
• 完成上述步骤后，打印(ans + 1)的值作为结果。

下面是上述方法的实现：

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))

输出：

4

时间复杂度： O(N)
辅助空间： O(1)