11-01 Exercise 3: Fibonacci numbers

Question

In mathematics, the Fibonacci numbers or Fibonacci sequence are the numbers in the following integer sequence: 0,1,1,2,3,5,8,13,21,34,55,89,144

前13的斐波那契數列：0,1,1,2,3,5,8,13,21,34,55,89,144

By definition, the first two numbers in the Fibonacci sequence is 0 and 1, and each subsequent number is the sum of the previous two. In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the relation Fn = Fn-1 + Fn-2 with seed values F0 = 0, F1= 1

斐波那契數由0和1開始，其後的數就由之前的兩個數相加。用數學表示 Fn = (Fn-1) + (Fn-2)，F0 = 0, F1 = 1

Write a program that asks the user to type an integer n, then display Fn on the screen.

寫一個輸入一個整數n，輸出第n個斐波那契數的程式

Example input 1:

5

Example output 1:

5

 

Example input 2:

12

Example output 2:

144

 

Answer 1

Hidden content!
#include <stdio.h>
int main()
{
    int a,i=0,j=1,k,c=2;
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    }
    return 0;
}

Comments

Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output

Answer 2

Hidden content!
#include<stdio.h>

int Fibonacci(int i);

int main(){
*   ** *** * * ** * *****  * a;
 * **  ***  **    *  *  ***   *  *  **   ***  ****   *  
*** *    ** *    **  *  *** *     *** *   *  ***** * **** **    *

*         * * ** **  ****** *     0;
}

int Fibonacci(int i){
  *** * *** * **  ******** == 0){
* *  ** ***     ***  *     ** * ******     *** * *  ***  ******* 0;
**   *    **  *  ***  * *   *  * 
 ****   **    * * **     * *  if(i==1){
  *******    ** **  *   *   ****  ** * * * ****    *   1;
**  *** * * *** * * ** ** **
     *  * ** ***   * *  ****
***   *       **  **** * **  *    * ****        * *  ** ** Fibonacci(i-1)+Fibonacci(i-2);
   ** *    *** ****    *  * 
}

Comments

Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output

Answer 3

Hidden content!
#include *   * * ***  * 
#include  * *      *  

int main()
{
  * ******   *     *** **  ** num[100] = { 0 }, n;
** *  * *  ** *    * *      ** i;

 **  ******   ** *  **  ****           ****   *   *  **
*  *****  ***  ***  * ******** * = 1;

***      *       ***   ** *  ( i = 2; i *  n; i++)
***  *   **** *  **  **
 *  *  *   *   *  *  *** ***  *****   **        ** ***  **   = num[i - 1] + num[i - 2];
* * **    *    *      *  ****
  **    * *  ** ** *****    * **  ** * *  *  *  num[i - 1]);

***  *  **   * ** ***     *  * ** 0;
}

Comments

Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output