题解 | #24点# 直接遍历须全排列关键字 优化后舒服了

import itertools
def compute(s1_list,s2):
    out = []
    for cmp in ['+','-','*','/']:
        for s1 in s1_list:
            out.append('('+s1+cmp+s2+')')


    return out



a,b,c,d = input().split()

#取两个运算
compute_s = []
for x1,x2,x3,x4 in itertools.permutations([a,b,c,d]):
    temp = compute([x1],x2)
    compute_s.extend(compute(compute(temp,x3),x4))
    for res in compute([x3],x4):
        compute_s.extend(compute(temp,res))


flag = "false"
for cs in compute_s:
    try:
        if eval(cs) == 24:
            flag = "true"
            break
			#有除零得排除
    except:
        pass

print(flag)
        





  
  

优化了一下 既然全排列了可以不考虑 小数情况 比如 7/5*10+10 这种会在7*10/5+10 这里遍历到 然后把除零一起提前去掉

import itertools
def compute(n1_list, n2_list):
    out = []
    for i in n1_list:
        for j in n2_list:
            out.extend([i+j, i-j, i*j])
            if j != 0 and i % j == 0:
                out.append(i/j)

    return out


a, b, c, d = map(int, input().split())


compute_s = []
for x1, x2, x3, x4 in itertools.permutations([a, b, c, d]):
    # 取两个运算
    temp1 = compute([x1], [x2])
    # 再一个一个算
    temp2 = compute(temp1, [x3])
    compute_s.extend(compute(temp2, [x4]))

    # 再两个两个算
    temp2 = compute([x3], [x4])
    compute_s.extend(compute(temp1, temp2))


flag = "false"
for res in compute_s:
    if res == 24:
        flag = "true"
        break

print(flag)