题解 | #24点运算#
24点运算
https://www.nowcoder.com/practice/7e124483271e4c979a82eb2956544f9d
题目明确表示了从前往后依次运算，所以其实从后往前看，始终是(ABC)+/-/*//D的形式，所以dfs采用倒序遍历的方式，最末尾的数直接影响前面几个数应该的输出值，直接倒序打印遍历，即可找到对应的计算流程了
a = input().split(' ')
dictory = {'J':11,'Q':12,'K':13,'A':1}
def dfs(wait,target,out):
if len(wait) == 1:
if wait[0] in dictory:
c = dictory[wait[0]]
else:
c = int(wait[0])
if target == c:
L.append(wait[0]+out)
else:
for i in range(len(wait)):
w = wait[:i]+wait[i+1::]
if wait[i] in dictory:
c = dictory[wait[i]]
else:
c = int(wait[i])
dfs(w,target-c,'+'+wait[i]+out)
dfs(w,target+c,'-'+wait[i]+out)
dfs(w,target*c,'/'+wait[i]+out)
dfs(w,target/c,'*'+wait[i]+out)
L = []
if 'joker' in a or 'JOKER' in a:
print('ERROR')
else:
dfs(a,24,'')
if not L:
print('NONE')
else:
print(L[0])
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。