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.
上邊是題目。
下邊是使用回溯法的枚舉解法，運行代碼幾分鐘後出現結果：
# 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()