Wu Yudong's Blog

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;
}

Answer:4613732
Completed on Tue, 2 Apr 2013, 06:36