F(0) = 0
F(1) = 1
F(n) = F(n-1)+F(n-2) (n>=2)
![](https://img.haomeiwen.com/i8346323/a1e5133e5357ead2.png)
递推下去变成
![](https://img.haomeiwen.com/i8346323/56ce44655148fe70.png)
![](https://img.haomeiwen.com/i8346323/f54d52ecfd1388fe.png)
的来由
![](https://img.haomeiwen.com/i8346323/157a84b976d69e9b.png)
问题已经描述完,接下来就是如何求解矩阵幂。
reference:https://blog.csdn.net/wjlwangluo/article/details/105252473
F(0) = 0
F(1) = 1
F(n) = F(n-1)+F(n-2) (n>=2)
的来由
问题已经描述完,接下来就是如何求解矩阵幂。
reference:https://blog.csdn.net/wjlwangluo/article/details/105252473
本文标题:斐波那契数列矩阵求法
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