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
harsh:
2014-08-29 08:02:15
I'm getting a WA answer could you please tell me what am i doing wrong my source code is <snip> Last edit: 2022-08-09 22:39:12 |
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Raman Shukla:
2014-07-02 14:49:05
Congrats Arpit Uppal... U deserve it...
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Arpit Uppal:
2014-07-02 14:17:57
just do as the problem says :) my 300th classical on spoj :) |
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Raj Mehta:
2013-12-21 13:49:34
@Syaorann i did as u said still WA
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Syaorann:
2013-01-24 11:35:44
just simply write the program like what it said.no other thing need to take into consideration,and u can AC it |
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Siddharth Bora:
2012-08-10 13:34:52
Important: To note that the open-addressing method calculates hashes j=1...19 on the original hash and not on the previous hash. |
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Aman Gupta:
2012-07-19 10:33:16
my solution works on ideone but here it is repeatedly giving NZEC
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Giovanni Botta:
2011-09-14 21:04:53
Is the open addressing performed by adding j^2+23*j to the previous hash code or to the original hash code h(.)?
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asqwzxdfercv:
2011-05-29 11:45:37
"After examining of at least 20 table entries"
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Jeevan K. S. Chalam:
2011-05-26 16:41:20
Done :) Last edit: 2011-05-28 09:00:22 |
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: | - |