【2014-11-19】斐波那契数列

1. 递归

f(n) = f(n-1) + f(n-2)

2. 从下向上计算

long long Fibonacci(int n)
{
    int result[2] = { 0, 1 };
    if (n < 2)
    {
        return result[n];
    }
    long long fibNminusOne = 1;
    long long fibNminusTwo = 0;
    long long fibN = 0;
    for (unsigned int i = 2; i <= n; i++)
    {
        fibN = fibNminusOne + fibNminusTwo;
        fibNminusTwo = fibNminusOne;
        fibNminusOne = fibN;
    }
    return fibN;
}

3. 数学公式

【2014-11-19】斐波那契数列

原文：http://www.cnblogs.com/sjtujoe/p/4108989.html