Question

我正在尝试制作一个程序，输出所有可能在数字1,2，…，9之间加+或-或不加任何运算，使得结果为100

可以理解，有些人已经在互联网上上传了解决方案，但是我想提出自己的解决方案。这是无效的代码：

"""
This program outputs all possibilities to put + or - or nothing between the numbers 1,2,…,9 (in this order) such that the result is 100
"""

class solver:

def __init__(self):
        """
        self.possibilities stores Arrays of type : [0]- always the sum of all operations 
        [1:9] all operations in int, where 0 equals plus, 1 equals minus, 2 equals nothing
        """
        self.possibilities = []

        self.possibilities.append([100])

        for i in range(7):
            self.possibilities.extend(self.makeNewIteration(i+1))

        print(self.possibilities)

        for obj in self.possibilities:
            if 100 is obj[0]:
                print(obj)

def makeNewIteration(self, i):
        for obj in self.possibilities:
            if(len(obj)<9):
                if(obj[-1] is 3):#if case 3
                    #recalculate the result
                    currentResult = int(obj[0] + self.conceal(i-1, i))
                else: currentResult = int(obj[0])
                #print(obj)
                possibilitiesNew = []
                possibilitiesNew.append([currentResult + i] + obj[1:] + [1])#case 1
                possibilitiesNew.append([currentResult - i] + obj[1:] + [2])#case 2
                possibilitiesNew.append([currentResult] + obj[1:] + [3])#case 3

                print("Iteration: "+str(i)+" : "+str(possibilitiesNew))

                self.possibilities.remove(obj)#remove the old object
            else:
                print("finished.")
        return possibilitiesNew

def conceal(self, x, y):
    # makes 12 out of x=1 and y=2
    return int(f'{x}{y}')

solve = solver()

我思考得越多，我遇到的问题就越多。我曾经以OOP的心态来学习编程，而事实上这是很久以前的事了，而且程序操作流程使这个问题变得容易得多，这让我陷入了困境。例如，如果连续两次“什么都没有”会发生什么？ 1，2，3，4变成12,23？ ...我希望有人可以修复该代码，并找出我做错了什么

Answer 1

我的第一个方法是

import itertools as it
import re

all_those_terms = (''.join([(sgn + str(i+1)) for i, sgn in enumerate(tpl)]) for tpl in it.product(['', '-', '+'], repeat=9) if tpl[0]!='')
for s in all_those_terms:
    r = re.findall('[+-]\d+', '+'+s)
    calc = sum(map(int, r))
    if calc == 100:
        print(s, '=', 100)

-1+2-3+4+5+6+78+9 = 100
+123-45-67+89 = 100
+123-4-5-6-7+8-9 = 100
+123+45-67+8-9 = 100
+123+4-5+67-89 = 100
+12-3-4+5-6+7+89 = 100
+12+3-4+5+67+8+9 = 100
+12+3+4+5-6-7+89 = 100
+1+23-4+56+7+8+9 = 100
+1+23-4+5+6+78-9 = 100
+1+2+34-5+67-8+9 = 100
+1+2+3-4+5+6+78+9 = 100

但是，eval应该替换，因为这很危险...

我会再编辑一遍...

编辑：替换为eval ...

Answer 2

Although eval should be avoided, here's another solution that uses it:

from itertools import product

[''.join(i) for i in product(*[[str(j)+'+', str(j) + '-', str(j)] for j in range(1, 9)] + ['9']) if eval(''.join(i))==100]
#['1+2+3-4+5+6+78+9', '1+2+34-5+67-8+9', '1+23-4+5+6+78-9', '1+23-4+56+7+8+9', '12+3+4+5-6-7+89', '12+3-4+5+67+8+9', '12-3-4+5-6+7+89', '123+4-5+67-89', '123+45-67+8-9', '123-4-5-6-7+8-9', '123-45-67+89']

如何解决我几天无法解决的Array任务？

2 个答案: