# 枚举思想-24点游戏

## 24点游戏

规则：4个数字的运算结果等于24；

```python
import itertools,random

def twentyfour(card):
    # itertools.permutations(card) ： 对card对4个数字进行全排列，穷举所有组合
    for nums in itertools.permutations(card):
        # itertools.product：对+-*/符合进行笛卡尔积组合，3个一对
        for ops in itertools.product('+-*/',repeat=3):
            # 这里只构造3中形式的计算式，其它形式可能得不到24，或者可化成和下面的一致，总之，枚举尽可能所有可能得到解的情况
            # *nums：0，1，2，3；*ops：4，5，6
            bds1='({0}{4}{1}){5}({2}{6}{3})'.format(*nums,*ops) #(a+b)*(c-d)
            bds2='(({0}{4}{1}){5}{2}){6}{3}'.format(*nums,*ops) #(a+b)*c-d
            bds3='{0}{4}({1}{5}({2}{6}{3}))'.format(*nums,*ops) #a/(b-(c/d)
            for bds in [bds1,bds2,bds3]:
                try:
                    if abs(eval(bds)-24.0)<1e-10:
                        return 1,bds
                except ZeroDivisionError:#除数是0的错误
                    continue
    return 0,card

cards=[i+1 for i in range(10)]
print(cards)

for _ in range(10):#随机抽10次
    card=random.sample(cards, 4) #从cards中抽4张数字
    succ,out=twentyfour(card)
    if succ ==0:
        print('%s 不能计算到 24'%out)
    else:
        print('%s =24'%out)
```

```
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
(5+9)+(10*1) =24
((10-8)*9)+6 =24
6*(5-(9-8)) =24
(7+9)*(6/4) =24
(3*10)-(6*1) =24
(7*5)-(1+10) =24
8*(6-(10-7)) =24
((10-9)+2)*8 =24
(9+8)+(3+4) =24
(4+9)+(1+10) =24
```

## itertools

是python的迭代器模块，itertools提供的工具相当高效且节省内存。

```python
# permutations 全排列
for item in itertools.permutations(['a', 'b', 'c']):
    print(item)
print('-'*50)

for item in itertools.permutations(['a', 'b', 'c'],2):
    print(item)
print('-'*50)

# itertools.product 笛卡尔积的元组
for item in itertools.product([1,2,3],[100,200]):
    print(item)

print('-'*50)
# product(A, repeat=4) means the same as product(A, A, A, A).
# for ops in itertools.product('+-*/','+-*/','+-*/',):
for ops in itertools.product('+-*/',repeat=3):
    print(ops)

for i in itertools.count(10): #从10开始无限循环
    print(i)
    break

for i in itertools.islice(itertools.count(10), 5): #从10开始，输出5个元素后结束
    print(i)

for item in itertools.cycle('XYZ'):#X,Y,Z轮流输出，无限循环
    print(item)
    break

import operator
# accumulate 迭代器将返回累计求和结果
print(list(itertools.accumulate(range(1,10)))) #默认，累加
print(list(itertools.accumulate(range(1,10),operator.mul))) #operator.mul：累乘
print(list(itertools.accumulate(range(1,10),operator.add))) #operator.add：累加
```

## random随机操作

```python
import random
values = [1, 2, 3, 4, 5, 6]

print(random.choice(values))

print(random.sample(values, 4))

print(values)
random.shuffle(values)
print(values)
```

```
4
[2, 3, 1, 5]
[1, 2, 3, 4, 5, 6]
[2, 4, 5, 3, 6, 1]
```


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