Python/project_euler/problem_29/solution.py
Bruno Simas Hadlich 267b5eff40 Added doctest and more explanation about Dijkstra execution. (#1014)
* Added doctest and more explanation about Dijkstra execution.

* tests were not passing with python2 due to missing __init__.py file at number_theory folder

* Removed the dot at the beginning of the imported modules names because 'python3 -m doctest -v data_structures/hashing/*.py' and 'python3 -m doctest -v data_structures/stacks/*.py' were failing not finding hash_table.py and stack.py modules.

* Moved global code to main scope and added doctest for project euler problems 1 to 14.

* Added test case for negative input.

* Changed N variable to do not use end of line scape because in case there is a space after it the script will break making it much more error prone.

* Added problems description and doctests to the ones that were missing. Limited line length to 79 and executed python black over all scripts.

* Changed the way files are loaded to support pytest call.

* Added __init__.py to problems to make them modules and allow pytest execution.

* Added project_euler folder to test units execution

* Changed 'os.path.split(os.path.realpath(__file__))' to 'os.path.dirname()'
2019-07-17 01:09:53 +02:00

52 lines
1.2 KiB
Python

"""
Consider all integer combinations of ab for 2 <= a <= 5 and 2 <= b <= 5:
2^2=4, 2^3=8, 2^4=16, 2^5=32
3^2=9, 3^3=27, 3^4=81, 3^5=243
4^2=16, 4^3=64, 4^4=256, 4^5=1024
5^2=25, 5^3=125, 5^4=625, 5^5=3125
If they are then placed in numerical order, with any repeats removed, we get
the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab
for 2 <= a <= 100 and 2 <= b <= 100?
"""
from __future__ import print_function
def solution(n):
"""Returns the number of distinct terms in the sequence generated by a^b
for 2 <= a <= 100 and 2 <= b <= 100.
>>> solution(100)
9183
>>> solution(50)
2184
>>> solution(20)
324
>>> solution(5)
15
>>> solution(2)
1
>>> solution(1)
0
"""
collectPowers = set()
currentPow = 0
N = n + 1 # maximum limit
for a in range(2, N):
for b in range(2, N):
currentPow = a ** b # calculates the current power
collectPowers.add(currentPow) # adds the result to the set
return len(collectPowers)
if __name__ == "__main__":
print("Number of terms ", solution(int(str(input()).strip())))