Question

如果我有这个简单的代码，我该如何显示所有可用的变体？

#Display all possible variants until reach input sum with coins of 2 and 5
#Not fixed: Output only first result !

#Input
summ = int(input("Enter sum: "))
two = 2
five = 5

#Calculations/Checks/Output
for i in range(0, two+1):
    for j in range(0, five+1):
        if (i*2 + j*5 == summ):
            print(i, "*2 + ", j, "*5 = ", summ)

#End

例如：

Input: 17

Output:
1*2 + 3*5 = 17
6*2 + 1*5 = 17

但代码只显示第一个结果？
我该如何解决这个问题？

Answer 1

更改range的输入：

summ = int(input("Enter sum: "))
two = 2
five = 5

for i in range(summ//two+1):
    for j in range(summ//five+1):
        if (i*2 + j*5 == summ):
            print(i, "* 2 +", j, "* 5 =", summ)

Answer 2

您可以使用itertools和生成器：

In [17]: def correct_sum_i(i, val):
    ...:     sum_perm = [comb for comb in itertools.combinations_with_replacement([2,5],i) if sum(comb) == val]
    ...:     return sum_perm[0] if sum_perm else None
    ...: 

In [18]: val = 14

In [19]: next(correct_sum_i(i,val) for i in range(1000) if correct_sum_i(i, val))
Out[19]: (2, 2, 5, 5)

编辑：您可能还想检查无法做到的病态情况，例如1,3等......

编辑：所有金额

In [50]: def all_sums(char_set, sum_value):
    ...:     def correct_sum_i(i):
    ...:         
    ...:         sum_perm = [comb for comb in itertools.combinations_with_replacement(char_set,i) if sum(comb) == sum_value]
    ...:         return sum_perm if sum_perm else None
    ...:     
    ...:     assert min(char_set)  > 0
    ...:     
    ...:     return list(correct_sum_i(i) for i in range(int(sum_value / min(char_set))+ 1) if correct_sum_i(i))
    ...: 

In [51]: print (all_sums([2,5], 20))
[[(5, 5, 5, 5)], [(2, 2, 2, 2, 2, 5, 5)], [(2, 2, 2, 2, 2, 2, 2, 2, 2, 2)]]

这也为您提供了所需的行为：

In [53]: print (all_sums([2,5], 17))
[[(2, 5, 5, 5)], [(2, 2, 2, 2, 2, 2, 5)]]

如何在python中显示所有可用的变体？

2 个答案: