Thread: Fibbonacci Code

Hey all, Programming newbie here.

I'm programming the Fibbonacci code, but I can't figure out how to set the condition

"if NOT EQUAL to any numerical value"

This is what I have so far:

Code:

#include <stdio.h>
#include <conio.h>

long fib(int);

int main()
{
  int n;

  printf("Enter the Fibbonacci number you would like to view: ");
  scanf("%i", &n);

  if (WHAT GOES HERE?)
  {
    printf("Input error.\n");
    getch();
  }
  else if (n<0) {
    printf("Fibonacci is undefined for negative integers.\n");
    getch();
  }
  else {
    printf("The %i-th Fibonacci number is %ld\n", n, fib(n));
  }
  getch();
  return 0;
}

long fib(int n)
{
  if (n == 1 || n == 2) 
  {
    return 1;
  } 
  else 
  {
    return (fib(n-1) + fib(n-2));
  }
}

Thanks for the help in advance!

Thanks Richie, I was trying your code, but every return was "the 2th number is 1". I think it just keeps refering to the base value for some reason

But, I was playing with it, and this one worked somehow :S

Code:

if (scanf("%i", &n) != 1) {
    printf("Input error.\n");
    getch();

Could someone explain Why it worked? From what I understand, it means "If the inputted integer is not equal to 1, print "input error" (which makes no sence to me, since 1 is an integer...)

Thank you VERY Much. ALl's working fine, now just to inprove the GUI. Thanks to all who've helped me!

Final Code

Code:

#include <stdio.h>
#include <conio.h>

long fib(int);

int main()
{
  int n;
  int num;
  
  num = 0;
  
  printf("Enter the Fibbonacci number you would like to view: ");

  if (scanf("%i", &n) != 1) 
  {
    printf("Input error.\n");
    getch();
  }
  else if (n<0) {
    printf("Fibonacci is undefined for negative integers.\n");
    getch();
  }
  else {
    printf("The %i-th Fibonacci number is %ld\n", n, fib(n));
  }
  getch();
  return 0;
}

long fib(int n)
{
  if (n == 1 || n == 2) 
  {
    return 1;
  } 
  else 
  {
    return (fib(n-1) + fib(n-2));
  }
}

A deadly more efficient fibo function using dynamic programming. it may help you later.

Code:

 
long long int fibo( int n )
{
 long long int* mem = new long long int[n +1];
 long long int ret;
 
 mem[1] = mem[2] = 1;

 for (int i = 3 ; i <= n ; i ++)
 {
  mem[i] = mem[i -1] + mem[i -2];
 }
 
 ret = mem[n];
 
 delete[] mem;
 return ret;
}

Yeah, also windows calculator gives me warnings whenever I try to do 1,000,000!. It says that it will take a long time. I've not out waited it yet :/ I hope to do so some time.

Originally Posted by Prelude

There are only 46 Fibonacci numbers that can fit within a 32-bit unsigned type. Use a table.

Actually 48

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073