Thread: Project Euler Problem 2 Solution

Hi, I have just started working on Project Euler and I have completed problem 2 .I was just wondering if there is a better implementation that is better than one I have implemented and what could be ideal or most efficient solution for this problem.

Here is my code implementation for this.

Code:

#include<stdio.h>
#include<stdlib.h>


main(){
	int i;
	int j=0;
	int numberOfItems=100;
	long sum[numberOfItems];
	int sumOfEvenNumbers;
	for(i=0;i<numberOfItems;i++){
		
		if(i==0|| i==1){
			
			if(i==0){
				sum[0]=1;
				
			}else{
				
				sum[1]=sum[0]+1;
				
				
			}
					
		}else{
			
			sum[i]=sum[i-1]+sum[i-2];
			
			
		}
		
		if(sum[i]%2==0){
				
				if(sum[i]>4000000){
					
					break;
					
				}		
				if(j==0){
						
					sumOfEvenNumbers=sum[i];
						
				}else{
						
						sumOfEvenNumbers=sumOfEvenNumbers+sum[i];
						
				}
				printf("%d\n",sum[i]);
				j++;
				
				
		}
		
		
	}
	
	printf("\nThe sum of even numbers is %d\n",sumOfEvenNumbers);
	
	
	return EXIT_SUCCESS;	
}

You could do it like this:

Code:

#include <stdio.h>

#define MAX_FIB 4000000

int main(void) {
	int a = 1, b = 1, c = 1, sum = 0;

	while (c <= MAX_FIB) {
		if (c % 2 == 0)
			sum += c;
		c = a + b;
		a = b;
		b = c;
	}

	printf("\nSum: %d\n", sum);

	return 0;
}