题解 | #24点游戏算法#

# 将4个数排列组合有4！=24种情况；4个数中间的3个计算符号每个都有加减乘除4种可能，共有4**3 =64种可能
# 所以最多循环4！*4**3 = 24*64 次遍历全部情况，其中任意情况得到24即可退出循环。

from itertools import permutations

l_num = list(map(str,input().split(' ')))
l_signal = ['+','-','*','/']

l_possible_con =list(permutations(l_num))
new_l = []

is_24 =0 
for x1 in l_signal:
    for x2 in l_signal:
        for x3 in l_signal:
            for new_l in l_possible_con:
                new_l = list(new_l)
                y = '((' + new_l[0] + x1 + new_l[1] + ')' + x2 + new_l[2] + ')' + x3 + new_l[3]
                y1 =eval(y)
                if y1 ==  24:
                    is_24 = 1
                    break

if is_24 == 1:
    print('true')
else :
    print('false')