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[];

void solve()

{

   int a,b,c,n,count=;

   a=,c=,b=;

   n=;

   while(c<=N)

   {

     c=a+b;

     if(n%!=)

     {

        a=c;

     }

     else

     {

        b=c;

     }

     n++;

     if(c%==)

     {

       count+=c;

     }

   }

   printf("%d",count);

}

int main()

{

  solve();

  return ;

}