备用斐波那契数

给出一个数字N，打印替代的斐波那契数字，直到第n个斐波那契。

例子：

Input :  N = 7
Output : 0 1 3 8 

Input  : N = 15
Output : 0 1 3 8 21 55 144 377 

阅读下面的文章中的方法2 ：斐波那契数

方法：使用动态编程方法。继续存储先前计算出的斐波那契数，并使用前两个存储下一个斐波那契数。

C++

// Alternate Fibonacci Series using Dynamic Programming
#include 
using namespace std;
  
void alternateFib(int n)
{
    if (n < 0)
      return;
  
    /* 0th and 1st number of the series are 0 and 1*/
    int f1 = 0;
    int f2 = 1;
  
    cout << f1 << " ";
    for (int i = 2; i <= n; i++) {
         int f3 = f2 + f1;
            
         if (i % 2 == 0)
            cout << f3 << " ";
  
         f1 = f2;
         f2 = f3;
    }
}
  
int main()
{
    int N = 15;
    alternateFib(N);
    return 0;
}

Java

// Alternate Fibonacci Series 
// using Dynamic Programming
import java.io.*;
  
class GFG 
{
static void alternateFib(int n)
{
    if (n < 0)
    return;
  
    /* 0th and 1st number of the
       series are 0 and 1*/
    int f1 = 0;
    int f2 = 1;
  
    System.out.print(f1 + " ");
    for (int i = 2; i <= n; i++) 
    {
        int f3 = f2 + f1;
          
        if (i % 2 == 0)
            System.out.print(f3 + " ");
  
        f1 = f2;
        f2 = f3;
    }
}
  
// Driver Code
public static void main (String[] args) 
{
    int N = 15;
    alternateFib(N);
}
}
  
// This code is contributed
// by chandan_jnu.

Python3

# Alternate Fibonacci Series
# using Dynamic Programming
def alternateFib(n):
    if (n < 0):
        return -1;
  
    # 0th and 1st number of
    # the series are 0 and 1
    f1 = 0;
    f2 = 1;
  
    print(f1, end = " ");
    for i in range(2, n + 1):
        f3 = f2 + f1;
          
        if (i % 2 == 0):
            print(f3, end = " ");
  
        f1 = f2;
        f2 = f3;
  
# Driver Code
N = 15;
alternateFib(N);
  
# This code is contributed by mits

C#

// Alternate Fibonacci Series 
// using Dynamic Programming
using System;
  
class GFG 
{
static void alternateFib(int n)
{
    if (n < 0)
    return;
  
    /* 0th and 1st number of 
    the series are 0 and 1*/
    int f1 = 0;
    int f2 = 1;
  
    Console.Write(f1 + " ");
    for (int i = 2; i <= n; i++) 
    {
        int f3 = f2 + f1;
          
        if (i % 2 == 0)
            Console.Write(f3 + " ");
  
        f1 = f2;
        f2 = f3;
    }
}
  
// Driver Code
public static void Main () 
{
    int N = 15;
    alternateFib(N);
}
}
  
// This code is contributed
// by chandan_jnu.

输出：

0 1 3 8 21 55 144 377

时间复杂度： O(n)
辅助空间： O(1)