给定整数“ K”和“ N”，任务是找到K-斐波那契数列的第N个项。
In K – Fibonacci series, the first ‘K’ terms will be ‘1’ and after that every ith term of the series will be the sum of previous ‘K’ elements in the same series.
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 Shrikant13PHP
Javascript
输出：
3时间复杂度： O(N)
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