题解 | #24点游戏算法#

# 基本上已考虑所有情况，包括各种运算顺序和带括号的情况

arr = input().split()

def check_24(arr, exp=""):

    if len(arr) == 1:

        exp += arr[0]

        # 找出3个运算符所在位置
        flag = 0
        for i in range(len(exp)):
            if not exp[i].isdigit() and flag == 0:
                flag = 1
                s0 = i
                continue
            if not exp[i].isdigit() and flag == 1:
                flag = 2
                s1 = i
                continue
            if not exp[i].isdigit() and flag == 2:
                s2 = i
                break

        for exp_ in [  # 考虑各种带括号的形式，这里有很大优化空间，但是踩坑已经踩到吐血不想深究了
            exp,  # a + b + c + d
            f"({exp[:s1]}){exp[s1:]}",  # (a + b) * c + d
            f"{exp[:s1+1]}({exp[s1+1:]})",  # a + b * (c + d)
            f"({exp[:s1]}){exp[s1]}({exp[s1+1:]})",  # (a + b) * (c + d)
            f"({exp[:s2]}){exp[s2:]}",  # (a + b + c) * d
            f"{exp[:s0+1]}({exp[s0+1:]})",  # a * (b + c + d)
        ]:

            # 执行表达式，用try跳过/0的情况
            try:
                locals()["res"] = 0
                exec(f"res={exp_}")
            except:
                pass

            if round(locals()["res"], 5) == 24:  #保留5位小数，使类似23.99999...这种结果可以通过
                # print(exp_)
                raise  # 结果等于24就raise直接跳出

    else:

        # 生成所有表达式
        for i in range(len(arr)):
            check_24(arr[:i] + arr[i + 1 :], exp + arr[i] + "+")
            check_24(arr[:i] + arr[i + 1 :], exp + arr[i] + "-")
            check_24(arr[:i] + arr[i + 1 :], exp + arr[i] + "*")
            check_24(arr[:i] + arr[i + 1 :], exp + arr[i] + "/")


try:
    check_24(arr)
    print("false")
except:
    print("true")