想法源于题目:一个人一次可以上一个台阶,也可以上两个台阶,问上到20级台阶有多少种走法?
我们也会发现:
f(3) = f(2) + f(1);
f(4) = 2*(f2)+1*f(1);
f(5) = 3*(f2) + 2*f(1);
f(6) = 5*f(2) + 3*f(1);
..........
f(n) = a*f(x) + b * f(y);
a,b同样是斐波那契数列中的数;同时发现当:
a+x == n && b+y ==n-1 && x == y+1时等式成立;
可以得到如下O(log(n))的算法:
/// 基本原理为:
/// n为偶数时f(n)=f(n/2)*f(n/2)+f(n-1)*f(n-1);
/// n为基数时f(n)=f(n/2+1)*f(n/2)+f(n/2)*f(n/2-1);
/// </summary>
/// <param name="n"></param>
/// <returns></returns>
public static long Fn2(int n)
{
if (1 < n)
{
var steps = new Stack<int>();
while (n > 2)
{
steps.Push(n);
n /= 2;
}
long r1 = 2, r2 = 3;
while (steps.Count > 0)
{
int tmp = steps.Pop();
if (3 < tmp)
{
long tr1;
long tr2;
if (0 == tmp%2)
{
tr1 = 2*r1*r1 + r2*r2 - 2*r1*r2;
tr2 = 2*r1*r2 - r1*r1;
r1 = tr1;
r2 = tr2;
}
else
{
tr1 = 2*r1*r2 - r1*r1;
tr2 = r1*r1 + r2*r2;
r1 = tr1;
r2 = tr2;
}
}
else
{
r1 = 3;
r2 = 5;
}
}
return r1;
}
if (1 == n) return 1;
return -1;
}