HASHIT - Hash it!


Your task is to calculate the result of the hashing process in a table of 101 elements, containing keys that are strings of length at most 15 letters (ASCII codes 'A',...,'z'). Implement the following operations:

  • find the index of the element defined by the key (ignore, if no such element),
  • insert a new key into the table (ignore insertion of the key that already exists),
  • delete a key from the table (without moving the others), by marking the position in table as empty (ignore non-existing keys in the table)

When performing find, insert and delete operations define the following function:
integer Hash(string key),
which for a string key=a1...an returns the value:
Hash(key)=h(key) mod 101, where
h(key)=19 *(ASCII(a1)*1+...+ASCII(an)*n).
Resolve collisions using the open addressing method, i.e. try to insert the key into the table at the first free position: (Hash(key)+j2+23*j) mod 101, for j=1,...,19. After examining of at least 20 table entries, we assume that the insert operation cannot be performed.

Input

t [the number of test cases <= 100]
n1 [the number of operations (one per line)[<= 1000]
ADD:string
[or]
DEL:string [other test cases, without empty lines betwee series]

Output

For every test case you have to create a new table, insert or delete keys, and write to the output:
the number of keys in the table [first line]
index:key [sorted by indices]

Example

Input:
1
11
ADD:marsz
ADD:marsz
ADD:Dabrowski
ADD:z
ADD:ziemii
ADD:wloskiej
ADD:do
ADD:Polski
DEL:od
DEL:do
DEL:wloskiej

Output:
5
34:Dabrowski
46:Polski
63:marsz
76:ziemii
96:z

hide comments
cake_is_a_lie: 2017-03-06 14:46:25

Not double inserting existing items can become problematic after deletions, so be careful! Cost me a lot of WAs.

vengatesh15: 2017-02-13 08:47:28

There should be a blank line after every test case .. that cost me 2WA

samnik: 2017-02-12 07:41:58

can anyone help me working for test cases but still getting WA
here is my code

<snip>

Last edit: 2022-08-09 22:39:06
Bartosz: 2016-10-23 13:15:46

Empty string is also considered to be a valid key, and it have to be put on the position 0. It cost me few WAs to realize that.

Last edit: 2016-10-23 13:43:45
gautam: 2016-08-06 15:40:10

yeah....ac.....!!!!

souravjaiswal: 2016-07-05 20:58:17

guys just use http://spojtoolkit.com/ to get extra test cases.

KD : 2016-06-07 09:06:09

AC after 5 WA ....be careful about the deletion and insertion of same elements multiple time:

root8950: 2016-03-22 12:38:39

SHIT is a substring of HASHIT :/ :D

Vipul Srivastava: 2016-02-01 19:45:46

j cannot be 20..!!

proficient: 2016-01-30 17:44:22

Awesome problem. Lesson learned.


Added by:mima
Date:2004-06-01
Time limit:3s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS PERL6 VB.NET
Resource:-