📜  K-斐波那契系列

📅  最后修改于: 2021-06-26 01:21:41             🧑  作者: Mango

给定整数“ K”和“ N”,任务是找到K-斐波那契数列的第N个项。

例子:

Input: N = 4, K = 2
Output: 3
The K-Fibonacci series for K=2 is 1, 1, 2, 3, ...
And, the 4th element is 3.

Input: N = 5, K = 6
Output: 1
The K-Fibonacci series for K=6 is 1, 1, 1, 1, 1, 1, 6, 11, ...

一种简单的方法:

  • 首先,将第一个“ K”元素初始化为“ 1”。
  • 然后,通过运行从“ ik”到“ i-1”的循环来计算先前的“ K”个元素的总和。
  • 将第i个值设置为总和。

时间复杂度: O(N * K)
一种有效的方法:

  • 首先,将第一个“ K”元素初始化为“ 1”。
  • 创建一个名为“ sum”的变量,该变量将以“ K”初始化。
  • 将第(K + 1)个元素的值设置为sum。
  • 将下一个值设置为Array [i] = sum – Array [ik-1] + Array [i-1],然后更新sum = Array [i]。
  • 最后,显示数组的第N个项。

下面是上述方法的实现:

C++
// C++ implementation of above approach
#include 
using namespace std;
 
// Function that finds the Nth
// element of K-Fibonacci series
void solve(int N, int K)
{
    vector Array(N + 1, 0);
 
    // If N is less than K
    // then the element is '1'
    if (N <= K) {
        cout << "1" << endl;
        return;
    }
 
    long long int i = 0, sum = K;
 
    // first k elements are 1
    for (i = 1; i <= K; ++i) {
        Array[i] = 1;
    }
 
    // (K+1)th element is K
    Array[i] = sum;
 
    // find the elements of the
    // K-Fibonacci series
    for (int i = K + 2; i <= N; ++i) {
 
        // subtract the element at index i-k-1
        // and add the element at index i-i
        // from the sum (sum contains the sum
        // of previous 'K' elements )
        Array[i] = sum - Array[i - K - 1] + Array[i - 1];
 
        // set the new sum
        sum = Array[i];
    }
    cout << Array[N] << endl;
}
 
// Driver code
int main()
{
    long long int N = 4, K = 2;
 
    // get the Nth value
    // of K-Fibonacci series
    solve(N, K);
 
    return 0;
}


Java
// Java implementation of above approach
 
public class GFG {
 
    // Function that finds the Nth
    // element of K-Fibonacci series
    static void solve(int N, int K)
    {
        int Array[] = new int[N + 1];
         
 
        // If N is less than K
        // then the element is '1'
        if (N <= K) {
            System.out.println("1") ;
            return;
        }
 
        int i = 0 ;
        int sum = K;
 
        // first k elements are 1
        for (i = 1; i <= K; ++i) {
            Array[i] = 1;
        }
 
        // (K+1)th element is K
        Array[i] = sum;
 
        // find the elements of the
        // K-Fibonacci series
        for (i = K + 2; i <= N; ++i) {
 
            // subtract the element at index i-k-1
            // and add the element at index i-i
            // from the sum (sum contains the sum
            // of previous 'K' elements )
            Array[i] = sum - Array[i - K - 1] + Array[i - 1];
 
            // set the new sum
            sum = Array[i];
        }
        System.out.println(Array[N]);
    }
 
    public static void main(String args[])
    {
          int N = 4, K = 2;
 
            // get the Nth value
            // of K-Fibonacci series
            solve(N, K);
 
    }
    // This code is contributed by ANKITRAI1
}


Python3
# Python3 implementation of above approach
 
# Function that finds the Nth
# element of K-Fibonacci series
def solve(N, K) :
    Array = [0] * (N + 1)
     
    # If N is less than K
    # then the element is '1'
    if (N <= K) :
        print("1")
        return
     
    i = 0
    sm = K
     
    # first k elements are 1
    for i in range(1, K + 1) :
        Array[i] = 1
         
    # (K+1)th element is K
    Array[i + 1] = sm
     
    # find the elements of the
    # K-Fibonacci series
    for i in range(K + 2, N + 1) :
         
        # subtract the element at index i-k-1
        # and add the element at index i-i
        # from the sum (sum contains the sum
        # of previous 'K' elements )
        Array[i] = sm - Array[i - K - 1] + Array[i - 1]
 
        # set the new sum
        sm = Array[i]
 
    print(Array[N])
     
     
# Driver code
N = 4
K = 2
 
# get the Nth value
# of K-Fibonacci series
solve(N, K)
 
# This code is contributed by Nikita Tiwari.


C#
// C# implementation of above approach
using System;
 
class GFG {
 
    // Function that finds the Nth
    // element of K-Fibonacci series
    public static void solve(int N, int K)
    {
        int[] Array = new int[N + 1];
 
 
        // If N is less than K
        // then the element is '1'
        if (N <= K)
        {
            Console.WriteLine("1");
            return;
        }
 
        int i = 0;
        int sum = K;
 
        // first k elements are 1
        for (i = 1; i <= K; ++i)
        {
            Array[i] = 1;
        }
 
        // (K+1)th element is K
        Array[i] = sum;
 
        // find the elements of the
        // K-Fibonacci series
        for (i = K + 2; i <= N; ++i)
        {
 
            // subtract the element at index i-k-1
            // and add the element at index i-i
            // from the sum (sum contains the sum
            // of previous 'K' elements )
            Array[i] = sum - Array[i - K - 1] +
                                 Array[i - 1];
 
            // set the new sum
            sum = Array[i];
        }
        Console.WriteLine(Array[N]);
    }
 
    // Main Method
    public static void Main(string[] args)
    {
        int N = 4, K = 2;
 
            // get the Nth value
            // of K-Fibonacci series
            solve(N, K);
 
    }
     
}
 
// This code is contributed
// by Shrikant13


PHP


Javascript


输出:
3

时间复杂度: O(N)

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