（Problem 2）Even Fibonacci numbers

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.

题目大意：

斐波那契数列中的每一项被定义为前两项之和。从1和2开始，斐波那契数列的前十项为：

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

考虑斐波那契数列中数值不超过4百万的项，找出这些项中值为偶数的项之和。

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
  
#define N 4000000
  
int a[1001];
  
void solve()
{
   int a,b,c,n,count=2;
   a=1,c=0,b=2;
   n=3;
   while(c<=N)
   {
     c=a+b;
     if(n%2!=0)
     {
        a=c;
     }
     else
     {
        b=c;
     }
     n++;
     if(c%2==0)
     {
       count+=c;
     }
   }
   printf("%d",count);
}
  
int main()
{
  solve();
  return 0;
}

（Problem 2）Even Fibonacci numbers

原文：http://www.cnblogs.com/acutus/p/3544290.html