def fibo(n):
a = 1
b = 1
if n <= 2:
return 1
else:
while n > 2:
if a <= b:
a = a + b
else:
b = a + b
return a + b
n = int(input())
x = fibo(n)
print(x)
I want to return the nth Fibonacci number and print it out. It is solved only by variables and while statements without a recursive function A and b are variables to store Fibonacci numbers, and I know how to increase the value of a and b in the when statement, but as n increases, I'm not sure I get the value of those two plus the value.
python
I read the Fibonacci sequence properties again and worked on a function, and there was one tricky part of the Fibonacci code without recursion. Compare it to your own code and try to figure out why it works.
def fibo(n) :
# Defines the default state. Paragraph 0 is 0 and paragraph 1 is 1.
# Here, b is the "last obtained term" and a is the "second and last obtained term".
a = 0
b = 1
# We are currently in the state of seeking paragraph 1.
SequenceNo = 1
# Until b becomes paragraph n:
while SequenceNo < n :
# The value before changing b is stored in a separate variable.
lastB = b
# Add a to b. Now b is the changed b.
b += a
# Change a to b before the change.
# This way, the b + = a operation works as intended when the next while is executed.
a = lastB
# Prepare to deal with the next section.
SequenceNo += 1
# Now that we've come this far, b is the value of paragraph n. Print it out and finish.
return b
# Testing
print(fibo(7)) # 13
print(fibo(12)) # 144
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