找到斐波纳契数的总和

def fibonacci_sum(n):
    if n < 0:
        return 0
    a, b = 0, 1
    total = a + b  # 初始化为F(0)+F(1)
    for _ in range(2, n + 1):
        a, b = b, a + b
        total += b
    return total

# 示例：计算前10项的和（F(0)到F(9)）
print(fibonacci_sum(9))  # 输出：88

def fibonacci(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
    return a

def fibonacci_sum_math(n):
    return fibonacci(n + 2) - 1

print(fibonacci_sum_math(9))  # 输出：88

def fibonacci_recursive(n):
    if n <= 1:
        return n
    return fibonacci_recursive(n - 1) + fibonacci_recursive(n - 2)

def fibonacci_sum_recursive(n):
    return sum(fibonacci_recursive(i) for i in range(n + 1))

print(fibonacci_sum_recursive(9))  # 输出：88

def matrix_mult(a, b):
    return [
        [a[0][0]*b[0][0] + a[0][1]*b[1][0], a[0][0]*b[0][1] + a[0][1]*b[1][1]],
        [a[1][0]*b[0][0] + a[1][1]*b[1][0], a[1][0]*b[0][1] + a[1][1]*b[1][1]]
    ]

def matrix_pow(mat, power):
    result = [[1, 0], [0, 1]]  # 单位矩阵
    while power > 0:
        if power % 2 == 1:
            result = matrix_mult(result, mat)
        mat = matrix_mult(mat, mat)
        power //= 2
    return result

def fibonacci_fast(n):
    if n == 0:
        return 0
    mat = [[1, 1], [1, 0]]
    result = matrix_pow(mat, n - 1)
    return result[0][0]

def fibonacci_sum_fast(n):
    return fibonacci_fast(n + 2) - 1

print(fibonacci_sum_fast(100))  # 快速计算大n的和