Python算法笔试题目，破解Hash值，回溯法

Find the string which has this hash: 25267566250558

The string has length8.

Characters can befrom: c,e,i,a,r,w,u,s,p

The hash functionworks like this:

hash(str):

1.LETTERS = c, e, i, a, r, w, u, s, p

2. h =7

3. forc in str:

1.i = index of c in LETTERS

2.h = 37 * h + i

4.return h

Please send us thestring you found, and the code you used to find it.

上边是题目。

下边是使用回溯法的枚举解法，运行代码几分钟后出现结果：

# calculate the hash number of the str_hash
str_hash_length = 8
hash_result = 25267566250558
LETTERS = ['c', 'e', 'i', 'a', 'r', 'w', 'u', 's', 'p']

def calculate_hash(str_hash):
h = 7
for c in str_hash:
# print c
i = LETTERS.index(c) #+ 1 ###
# print i,
h = 37*h + i
return h

## for test of calculate_hash()
# str_hash = 'ec'
# print calculate_hash(str_hash)

## calculate the str_hash using n-nary and recursive
# instialize a list to store the characters
str_hash_list = ['' for i in xrange(str_hash_length)]
# print 'the type is ',type(str_hash_list)
def find_str(k):
if str_hash_length == k:
# this conversion can move to just before it needs to print
str_hash = ''.join(str_hash_list)
str_hash_value = calculate_hash(str_hash)
# print str_hash_value
if str_hash_value == hash_result:
print "yes! the string '%s' is found ,its hash is %d" %(str_hash,str_hash_value)
#return   ##!neglected, will it be ok when negelected?, ans: yes

## can add

else:
for c in LETTERS:
#if str_hash_list[len(str_hash_list)-1] is not '':
#print str_hash_list
str_hash_list[k] = c
find_str(k+1)

def main():
find_str(0)

if __name__ == '__main__':
main()