Question

I am using Dev-C++. It doesn’t show any code error, but fails to work.

It works when I try small numbers like 10 or 20

I am working on this problem :

Each new term in the Fibonacci sequence is generated by adding the
previous two terms. By starting with 1 and 2, the first 10 terms will
be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not
exceed four million, find the sum of the even-valued terms.

    #include<stdio.h>
#include<stdlib.h>
int main()
{
const int N=100;
int a=1,b=2,i,t[N],S=0,c,j;
t[0]=1;
t[1]=2;
for(i=2;i<N;i++){
t[i]=t[i-2]+t[i-1];
if(t[i]>4000000)
{
for(j=1;j<=i-1;j++){
                    c=t[j]%2;
                    if(c==0){
                             S=S+t[j];
                            }
                    else    {
                             continue;
                    }}
break;
}
}
printf("%d\n",S);
system("pause");
}

Answer 1

You don’t need an array to store all those numbers, you can get away with storing the last two terms in the sequence, since that’s all that’s need to calculate the next term.

Trying to allocate that much space on the stack is asking for trouble since the stack is a relatively limited resource.

In fact, that exact code entered into gcc on a Linux box gives me a segmentation violation when I try to run it, for precisely that reason.

On top of that, your code is not getting the even valued terms, it’s getting every term, and you’re getting the first four million values, rather than the values below four million which was specified.

The sort of code you’re after would look like this:

#include <stdio.h>

int main (void) {
    // Accumulator and terms (acc is zero because first two terms are odd).

    int acc = 0, t1 = 1, t2 = 1, t3;

    // Continue until next term is 4mil or more.

    while ((t3 = t1 + t2) < 4000000) {
        // printf ("DEBUG: %d %d %d %s\n", t1, t2, t3,
        //     ((t3 % 2) == 0) ? "<<" : "");

        // Accumulate only even terms.

        if ((t3 % 2) == 0) acc += t3;

        // Cycle through terms.

        t1 = t2; t2 = t3;
    }

    // Print the accumulated value.

    printf ("%d\n", acc);

    return 0;
}

And the output:

4613732

If you test that program by un-commenting the debug statement, you see:

DEBUG: 1 1 2 <<
DEBUG: 1 2 3 
DEBUG: 2 3 5 
DEBUG: 3 5 8 <<
DEBUG: 5 8 13 
DEBUG: 8 13 21 
DEBUG: 13 21 34 <<
DEBUG: 21 34 55 
DEBUG: 34 55 89 
DEBUG: 55 89 144 <<
DEBUG: 89 144 233 
DEBUG: 144 233 377 
DEBUG: 233 377 610 <<
DEBUG: 377 610 987 
DEBUG: 610 987 1597 
DEBUG: 987 1597 2584 <<
DEBUG: 1597 2584 4181 
DEBUG: 2584 4181 6765 
DEBUG: 4181 6765 10946 <<
DEBUG: 6765 10946 17711 
DEBUG: 10946 17711 28657 
DEBUG: 17711 28657 46368 <<
DEBUG: 28657 46368 75025 
DEBUG: 46368 75025 121393 
DEBUG: 75025 121393 196418 <<
DEBUG: 121393 196418 317811 
DEBUG: 196418 317811 514229 
DEBUG: 317811 514229 832040 <<
DEBUG: 514229 832040 1346269 
DEBUG: 832040 1346269 2178309 
DEBUG: 1346269 2178309 3524578 <<
4613732

and, if you add up all the even numbers at the end of those DEBUG lines, you do indeed get the given value.