题解 | #24点运算#

b = []
def all_order (a,start,end): #全排列，即将输入的四个扑克进行全排列，输出4*3*2*1个组合
    if(start == end):
        b.append([a[0],a[1],a[2],a[3]])
    else:
        for i in range(start,end + 1):
            d = a[i]
            a[i] = a[start]
            a[start] = d
            all_order(a, start + 1, end)
            d = a[i]
            a[i] = a[start]
            a[start] = d
while True:
    try:
        poker = input().split()
        if('joker' in poker) or ('Joker' in poker): #有大小王则无法计算
            print('ERROR')
            continue
        S = {'2':'2', '3':'3', '4':'4', '5':'5', '6':'6', '7':'7', '8':'8', '9':'9', '10':'10', 'A':'1', 'J':'11', 'Q':'12', 'K':'13'} #将扑克牌和对应的值存入字典
        sign = ['+','-','*','/'] #四个符号
        all_order(poker, 0, 3)
        c = []
        for i in range(4):
            for j in range(4):
                for k in range(4): #穷举
                    for z in range(24):
                        if(int(eval('(' + '(' + S[b[z][0]] + sign[i] + S[b[z][1]] + ')' + sign[j] + S[b[z][2]] + ')' + sign[k] + S[b[z][3]])) == 24): #通过eval计算值，加括号的意义是使例如1+2*3 的值为9而不是7
                            c.append(b[z][0] + sign[i] + b[z][1] + sign[j] + b[z][2] + sign[k] + b[z][3]) #将所有得到24的组合算式放入c
        if(len(c) == 0): #没有等于24的
            print('NONE') 
        else: #若是有则输出第一个即可
            print(c[0])
    except:
        break