#     The Fibonacci numbers are the numbers in the following integer sequence (Fn):
#           0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
# such as
# F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

# Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying
#           F(n) * F(n+1) = prod.

# Your function productFib takes an integer (prod) and returns an array:
#           [F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True)
# depending on the language if F(n) * F(n+1) = prod.

# If you don't find two consecutive F(m) verifying F(m) * F(m+1) = prod you will return

# [F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False)
# F(m) being the smallest one such as F(m) * F(m+1) > prod.

# Examples
# productFib(714) # should return [21, 34, true], 
#                 # since F(8) = 21, F(9) = 34 and 714 = 21 * 34

# productFib(800) # should return [34, 55, false], 
#                 # since F(8) = 21, F(9) = 34, F(10) = 55 and 21 * 34 < 800 < 34 * 55
# Notes: Not useful here but we can tell how to choose the number n up to which to go: we can use the "golden ratio" phi
#   which is (1 + sqrt(5))/2 knowing that F(n) is asymptotic to: phi^n / sqrt(5). That gives a possible upper bound to n.

# You can see examples in "Example test".

# References
# http://en.wikipedia.org/wiki/Fibonacci_number
# http://oeis.org/A000045

def productFib(n):
    fib = 0
    a = 0
    b = 1
    while a * b < n:
        aux = b
        b += a
        a = aux
        fib += 1
    return [a, b, a * b == n]

print(productFib(1870))
print(productFib(800))
print(productFib(714))
#     The Fibonacci numbers are the numbers in the following integer sequence (Fn):
#           0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
# such as
# F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

# Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying
#           F(n) * F(n+1) = prod.

# Your function productFib takes an integer (prod) and returns an array:
#           [F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True)
# depending on the language if F(n) * F(n+1) = prod.

# If you don't find two consecutive F(m) verifying F(m) * F(m+1) = prod you will return

# [F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False)
# F(m) being the smallest one such as F(m) * F(m+1) > prod.

# Examples
# productFib(714) # should return [21, 34, true], 
#                 # since F(8) = 21, F(9) = 34 and 714 = 21 * 34

# productFib(800) # should return [34, 55, false], 
#                 # since F(8) = 21, F(9) = 34, F(10) = 55 and 21 * 34 < 800 < 34 * 55
# Notes: Not useful here but we can tell how to choose the number n up to which to go: we can use the "golden ratio" phi
#   which is (1 + sqrt(5))/2 knowing that F(n) is asymptotic to: phi^n / sqrt(5). That gives a possible upper bound to n.

# You can see examples in "Example test".

# References
# http://en.wikipedia.org/wiki/Fibonacci_number
# http://oeis.org/A000045

def productFib(n):
    fib = 0
    a = 0
    b = 1
    while a * b < n:
        aux = b
        b += a
        a = aux
        fib += 1
    return [a, b, a * b == n]

print(productFib(1870))
print(productFib(800))
print(productFib(714))
#     The Fibonacci numbers are the numbers in the following integer sequence (Fn): #           0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ... # such as # F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1. # Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying #           F(n) * F(n+1) = prod. # Your function productFib takes an integer (prod) and returns an array: #           [F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True) # depending on the language if F(n) * F(n+1) = prod. # If you don't find two consecutive F(m) verifying F(m) * F(m+1) = prod you will return # [F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False) # F(m) being the smallest one such as F(m) * F(m+1) > prod. # Examples # productFib(714) # should return [21, 34, true], #                 # since F(8) = 21, F(9) = 34 and 714 = 21 * 34 # productFib(800) # should return [34, 55, false], #                 # since F(8) = 21, F(9) = 34, F(10) = 55 and 21 * 34 < 800 < 34 * 55 # Notes: Not useful here but we can tell how to choose the number n up to which to go: we can use the "golden ratio" phi #   which is (1 + sqrt(5))/2 knowing that F(n) is asymptotic to: phi^n / sqrt(5). That gives a possible upper bound to n. # You can see examples in "Example test". # References # http://en.wikipedia.org/wiki/Fibonacci_number # http://oeis.org/A000045