# "C Language" for Fibonacci (Fibonacci) sequence entries (recursive method, non-recursive method)

Source: Internet
Author: User

The Italian mathematician, Leon, discovered this sequence in 1202 when he studied the problem of rabbits producing cubs. Set up a pair of rabbits each month to give birth to a rabbit, each newborn rabbit one months after birth and then the Cubs, if the rabbit is not dead. Q: A pair of rabbits, how many pairs of rabbits can be bred in a year? In nature there are two types of rabbits: one is the rabbit that can reproduce, referred to as the Big Rabbit, the newborn rabbit can not reproduce, referred to as a small rabbit, rabbit one months into a big rabbit. It is the sum of the big rabbits and the little rabbits.

Month    Part Ⅰⅱ Ⅲ Ⅳ ⅤⅥ Ⅶ ⅧⅨⅩ Ⅺ Ⅻ
Large Rabbit logarithm1 1  2  3  5 8  -  + the - the144
Small Rabbit logarithm0 1  1  2  3 5  8  - + the -  the by December There were big rabbits 144 pairs, bunny 89 pair, total rabbit 144+89=233 from the table above:

① monthly small rabbit logarithm = The logarithm of the big Rabbit last month.
② each month the logarithm of the large rabbit is equal to the logarithm of the large rabbit and the small rabbit.
Comprehensive ①② Two points, we have: the monthly large rabbit logarithm is equal to the first two months the sum of the large rabbit logarithm.

If the N-month large rabbit logarithm is expressed in un, there is
UN = un-1 + un-2, n >2 each month the number of large rabbits in the series is: 1,1,2,3,5,8,13,21,34,55,89,144, /c5> this sequence is called the Fibonacci sequence.

Recursive method:

Using the formula F[n]=f[n-1]+f[n-2], recursive calculation in turn, the recursive end condition is f[1]=1,f[2]=1.

code example:

`#include <iostream>using namespace std;long long fib (int n) {     if  (n == 0)     {         return 0;    }    else if  (n == 1)      {        return 1;    }     else if (n > 1)     {         return fib (n - 1)  + fib (n - 2);     }    //return n > 1 ? fib (n - 1)  +  Fib (N&NBSP;-&NBSP;2)  : n; //conditional operator is simple, one line of code can be}void test () {    int n &NBSP;=&NBSP;0;&NBSP;&NBSP;&NBSP;&NBSP;SCANF ("%d",  &n);     int ret = &NBSP;FIB (N);     printf ("%d\n",  ret);} Int main () {    test ();     system ("Pause");     return 0;}`

However, the recursive approach to solving this problem is not efficient, let's look at the non-recursive method.

Non-recursive method:

Iterative implementations are the most efficient, with a time complexity of n*1 = 0 (n) and a spatial complexity of 0 (1).

`#include <iostream>using namespace std;long long fib (int n) {     if  (n == 0)     {         return 0;    }    else if  (n == 1)      {        return 1;    }     else if  (n > 1)     {         int a = 1;        int  b = 1;        int c = 1;         for  (int i = 2; i < n; i++)         {             c = a + b;            a = b;             b = c;         }        return c;    }}void  test () {    int n = 0;    scanf ("%d",  &N) ;     int ret = fib (N);     printf ("%d\n",  ret);} Int main () {    test ();     system ("Pause");     return 0;}`

"C Language" for Fibonacci (Fibonacci) sequence entries (recursive method, non-recursive method)

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