mirror of
https://github.com/TheAlgorithms/Python.git
synced 2025-07-05 01:09:40 +08:00
Simplify code by dropping support for legacy Python (#1143)
* Simplify code by dropping support for legacy Python * sort() --> sorted()
This commit is contained in:
@ -1,27 +1,15 @@
|
||||
from __future__ import print_function
|
||||
|
||||
try:
|
||||
xrange #Python 2
|
||||
except NameError:
|
||||
xrange = range #Python 3
|
||||
|
||||
try:
|
||||
raw_input #Python 2
|
||||
except NameError:
|
||||
raw_input = input #Python 3
|
||||
|
||||
'''
|
||||
The number of partitions of a number n into at least k parts equals the number of partitions into exactly k parts
|
||||
plus the number of partitions into at least k-1 parts. Subtracting 1 from each part of a partition of n into k parts
|
||||
gives a partition of n-k into k parts. These two facts together are used for this algorithm.
|
||||
'''
|
||||
def partition(m):
|
||||
memo = [[0 for _ in xrange(m)] for _ in xrange(m+1)]
|
||||
for i in xrange(m+1):
|
||||
memo = [[0 for _ in range(m)] for _ in range(m+1)]
|
||||
for i in range(m+1):
|
||||
memo[i][0] = 1
|
||||
|
||||
for n in xrange(m+1):
|
||||
for k in xrange(1, m):
|
||||
for n in range(m+1):
|
||||
for k in range(1, m):
|
||||
memo[n][k] += memo[n][k-1]
|
||||
if n-k > 0:
|
||||
memo[n][k] += memo[n-k-1][k]
|
||||
@ -33,7 +21,7 @@ if __name__ == '__main__':
|
||||
|
||||
if len(sys.argv) == 1:
|
||||
try:
|
||||
n = int(raw_input('Enter a number: '))
|
||||
n = int(input('Enter a number: ').strip())
|
||||
print(partition(n))
|
||||
except ValueError:
|
||||
print('Please enter a number.')
|
||||
|
Reference in New Issue
Block a user