獲取通過打破一個數字形成的完美正方形列表的所有可能排列。例如：如果 N = 14，則列表為 [1,1,4,4,4], [1,4,9] , [1,1,1,1,1,9] , [1,1,1, 1,1,1,1,1,1,1,1,1,1,1]輸出列表可以是任何順序我得到了這個代碼，但它只能按順序給出完美的正方形。l = []b = int(input())for i in range(1,b): k = i*i l.append(k) if sum(l)>b: l.pop() break else: passprint(l)
2 回答
Qyouu
TA貢獻1786條經驗 獲得超11個贊
以下代碼導致 N = 14 的 6 種可能性，而不是發布的 4。
代碼
from itertools import chain, combinations
from pprint import pprint
# flatten and powerset from
# https://docs.python.org/3/library/itertools.html#itertools-recipes
def flatten(list_of_lists):
"Flatten one level of nesting"
return chain.from_iterable(list_of_lists)
def powerset(iterable):
"powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
s = list(iterable)
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
def solve(n):
" Get all possible permutations of list of perfect squares formed by breaking a number "
squares = (i*i for i in range(1, int(b**0.5)+1)) # squares that can be used
multiples = ([i]*int(b//i) for i in squares) # repetition of squares that can be used
numbers = flatten(multiples) # flatten to single list
# Compute set of powerset, and take results which sum to b
return [x for x in set(powerset(numbers)) if sum(x) == b]
測試
b = int(input('input number: ')) # Enter 14
result = solve(b)
pprint(result)
輸出
input number: 14
[(1, 1, 1, 1, 1, 1, 4, 4),
(1, 1, 1, 1, 1, 9),
(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4),
(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1),
(1, 4, 9),
(1, 1, 4, 4, 4)]
限制最大長度
def solve_maxlen(n, maxlen):
" Get all possible permutations of list of perfect squares formed by breaking a number "
squares = (i*i for i in range(1, int(b**0.5)+1)) # squares that can be used
multiples = ([i]*int(b//i) for i in squares) # repetition of squares that can be used
numbers = flatten(multiples) # flatten to single list
# Compute set of powerset, and take results which sum to b
return [x for x in set(powerset(numbers)) if sum(x) == b and len(x) <= maxlen]
pprint(solve_maxlen(14, 6))
輸出
[(1, 1, 1, 1, 1, 9), (1, 4, 9), (1, 1, 4, 4, 4)]
慕尼黑5688855
TA貢獻1848條經驗 獲得超2個贊
import itertools
up_to = int(input())
def is_perfect_square(number):
squared = pow(number, 0.5)
return int(squared) == squared
perfect_squares = filter(is_perfect_square, range(1, up_to))
permutations = list(itertools.permutations(perfect_squares))
print(permutations)
輸出是：
[(1, 4, 9), (1, 9, 4), (4, 1, 9), (4, 9, 1), (9, 1, 4), (9, 4, 1)]
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