查找系列 0、1、1、2、5、29、841 的第 N 项...

给定一个正整数N ，任务是找到系列0、1、1、2、5、29、841中的第N项

例子：

Input: N = 6
Output: 29
Explanation: The 6th term of the given series is 29.

Input: N = 1
Output: 1

Input: N = 8
Output: 750797

方法：给定的问题是一个基于数学的基本问题，其中 A _i = A _i-1 ² + A _i-2 ² 。 因此，创建变量a = 0和b = 1 。使用[2, N]范围内的变量i进行迭代，对于每个i ，计算第ⁱ项并将a和b的值分别更新为第 i – ¹项和ⁱ – 2项。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find N-th term
// of the given series
int getNthTerm(int N)
{
    if (N < 3)
        return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    long int a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {
 
        // pow((i - 2)th term, 2) +
        // pow((i - 1)th term, 2)
        long int c = a * a + b * b;
 
        // Update a and b
        a = b;
        b = c;
    }
 
    // Return Answer
    return b;
}
 
// Driver Code
int main()
{
    int N = 8;
    cout << getNthTerm(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
  // Function to find N-th term
  // of the given series
  static int getNthTerm(int N)
  {
    if (N < 3)
      return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    int a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {
 
      // Math.pow((i - 2)th term, 2) +
      // Math.pow((i - 1)th term, 2)
      int c = a * a + b * b;
 
      // Update a and b
      a = b;
      b = c;
    }
 
    // Return Answer
    return b;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 8;
    System.out.print(getNthTerm(N));
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
 
# Function to find N-th term
# of the given series
def getNthTerm(N):
    if (N < 3):
        return N - 1
       
    # Initialize Variables repre-
    # senting 1st and 2nd term
    a = 0
    b = 1
     
    # Loop to iterate through the
    # range [3, N] using variable i
    for i in range(3, N + 1):
       
        # pow((i - 2)th term, 2) +
        # pow((i - 1)th term, 2)
        c = a * a + b * b
         
        # Update a and b
        a = b
        b = c
         
    # Return Answer
    return b
 
# Driver Code
N = 8
print(getNthTerm(N))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{
   
// Function to find N-th term
// of the given series
static int getNthTerm(int N)
{
    if (N < 3)
        return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    long a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (int i = 3; i <= N; i++) {
 
        // pow((i - 2)th term, 2) +
        // pow((i - 1)th term, 2)
        long c = a * a + b * b;
 
        // Update a and b
        a = b;
        b = c;
    }
 
    // Return Answer
    return (int)b;
}
 
// Driver Code
public static void Main()
{
    int N = 8;
    Console.Write(getNthTerm(N));
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

  // JavaScript code for the above approach
 
 
  // Function to find N-th term
  // of the given series
  function getNthTerm(N) {
    if (N < 3)
      return N - 1;
 
    // Initialize Variables repre-
    // senting 1st and 2nd term
    let a = 0, b = 1;
 
    // Loop to iterate through the
    // range [3, N] using variable i
    for (let i = 3; i <= N; i++) {
 
      // pow((i - 2)th term, 2) +
      // pow((i - 1)th term, 2)
      let c = a * a + b * b;
 
      // Update a and b
      a = b;
      b = c;
    }
 
    // Return Answer
    return b;
  }
 
  // Driver Code
 
  let N = 8;
  document.write(getNthTerm(N));
 
 
// This code is contributed by Potta Lokesh

输出

750797

时间复杂度： O(N)
辅助空间： O(1)