Not that hard, just a loopty loop and keep track of which ones you did already
This commit is contained in:
83
55/main.py
Normal file
83
55/main.py
Normal file
@@ -0,0 +1,83 @@
|
||||
'''
|
||||
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
|
||||
|
||||
Not all numbers produce palindromes so quickly. For example,
|
||||
|
||||
349 + 943 = 1292,
|
||||
1292 + 2921 = 4213
|
||||
4213 + 3124 = 7337
|
||||
|
||||
That is, 349 took three iterations to arrive at a palindrome.
|
||||
|
||||
Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).
|
||||
|
||||
Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.
|
||||
|
||||
How many Lychrel numbers are there below ten-thousand?
|
||||
|
||||
NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.
|
||||
'''
|
||||
|
||||
import time
|
||||
import math
|
||||
|
||||
def reverseInt(n):
|
||||
res = 0
|
||||
while n > 0:
|
||||
res *= 10
|
||||
res += n % 10
|
||||
n = n // 10
|
||||
|
||||
return res
|
||||
|
||||
def palint(n):
|
||||
if n == 0:
|
||||
return True
|
||||
if int(math.log10(n)) + 1 <= 1:
|
||||
return True
|
||||
s = str(n)
|
||||
if s[0] != s[-1]:
|
||||
return False
|
||||
if len(s) == 2:
|
||||
return True
|
||||
return palint(int(s[1:-1]))
|
||||
|
||||
def lychrel(n, iterations):
|
||||
res = n
|
||||
visited = set()
|
||||
|
||||
for _ in range(iterations):
|
||||
visited.add(res)
|
||||
res += reverseInt(res)
|
||||
if palint(res):
|
||||
return res, visited
|
||||
|
||||
return False, visited
|
||||
|
||||
|
||||
def main():
|
||||
print("Hello, this is Patrick")
|
||||
t0 = time.time()
|
||||
|
||||
counter = 0
|
||||
lychrels = set()
|
||||
nonLychrels = set()
|
||||
|
||||
for i in range(1, 10000):
|
||||
if i in lychrels:
|
||||
counter += 1
|
||||
elif i in nonLychrels:
|
||||
continue
|
||||
else:
|
||||
ans, visited = lychrel(i, 50)
|
||||
if ans:
|
||||
nonLychrels = nonLychrels.union(visited)
|
||||
else:
|
||||
lychrels = lychrels.union(visited)
|
||||
counter += 1
|
||||
|
||||
print(counter, time.time() - t0)
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
main()
|
||||
Reference in New Issue
Block a user