-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathPE_Q26.py
More file actions
42 lines (32 loc) · 1.05 KB
/
PE_Q26.py
File metadata and controls
42 lines (32 loc) · 1.05 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
"""
A unit fraction contains 1 in the numerator.
The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
"""
from decimal import getcontext, Decimal
# Increase this number till your max repeating length doesnt change
decimal_depth = 5000
getcontext().prec = decimal_depth
def find_lrf(x):
for i in range(2, len(x) // 2):
seqs = [x[k : k + i] for k in range(0, len(x), i)][:-1]
if seqs.count(seqs[0]) == len(seqs):
return i
return 0
repeating_lengths = []
for i in range(1, 1001):
fraction = str(Decimal(1) / Decimal(i))[2:]
repeating_lengths.append((i, find_lrf(fraction)))
repeating_lengths.sort(key=lambda x: x[1])
print(repeating_lengths)