编程之美 斐波拉契数列 log(n) Fibonacci

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

f(n-1)=f(n-2)+f(n-3)

So write like a matrix operation, ignore f(n-3), we could get

[f(n),f(n-1)]^T = [[1,1],[1,0]]*[f(n-1),f(n-2)]^T

void multiply(int A[], int B[], int C[]) {
  C[0] = A[0]*B[0] + A[1]*B[2];
  C[1] = A[0]*B[1] + A[1]*B[3];
  C[2] = A[2]*B[0] + A[3]*B[2];
  C[3] = A[2]*B[1] + A[3]*B[3];
}
void power(int A[], int n, int C[]) {
  if (n==1) { 
    memcpy(C, A, 4*sizeof(int)); 
    return; 
  }
  int tmp[4];
  power(A, n>>1, C);
  multiply(C, C, tmp);
  if (n & 1 == 1)
    multiply(tmp, A, C);
  else
    memcpy(C, tmp, 4*sizeof(int));
}
int f(int n) {
  int A[4] = {1,1,1,0};
  int result[4];
  power(A, n, result);
  return result[0];
}

再贴一版Pytorch代码：

import torch

def matrix_power(matrix, n):
    """计算矩阵的n次幂"""
    result = torch.eye(matrix.size(0), dtype=torch.int64)
    base = matrix.clone()
    while n > 0:
        if n % 2 == 1:
            result = torch.mm(result, base)
        base = torch.mm(base, base)
        n //= 2
    return result

def fibonacci(n):
    """使用矩阵幂计算第n个斐波那契数"""
    if n == 0:
        return 0
    elif n == 1:
        return 1
    F = torch.tensor([[1, 1],
                      [1, 0]], dtype=torch.int64)
    # 计算F^(n-1)
    result = matrix_power(F, n-1)
    # F(n)是矩阵第一行第一列的元素
    return result[0, 0].item()

# 测试函数
print(fibonacci(4))  # 输出第10个斐波那契数