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ICODER - Instruction Decoder |
Mathews uses a brand new 16-bit instruction processor. (Yeah i am being sarcastic!). It has one register (say R) and it supports two instructions:
- ADD X; Impact: R = (R + X) mod 65536
- MUL X; Impact: R = (R * X) mod 65536
- [For both instructions 0 <= X <= 65535]
Input Format:
The input file consists of multiple testcases.
The first line of each testcase contains one integer, N. (1 <= N <= 100,000).
The following N lines contain one instructions each.
Input terminates with a line containing N=0, which must not be processed.
Output Format:
For each testcase print one integer in a single line, denoting the number of different values the register can take after code execution.
Sample Input:
1 ADD 3 1 MUL 0 5 MUL 3 ADD 4 MUL 5 ADD 3 MUL 2 8 ADD 32 MUL 5312 ADD 7 MUL 7 ADD 32 MUL 5312 ADD 7 MUL 7 0Sample Output:
65536 1 32768 16
Added by: | Prasanna |
Date: | 2007-10-08 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | NITT ACM ICPC Local Contest 2007 [Self] |
hide comments
2022-06-13 19:47:31 David
For the final program, the 16 possible values are [1393, 5489, 9585, 13681, 17777, 21873, 25969, 30065, 34161, 38257, 42353, 46449, 50545, 54641, 58737, 62833]. |
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2016-03-18 11:46:59 minhthai
can someone explain the result "16". I thought the equation in that test is not solvable ??? |
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2014-04-20 05:00:37 Kishlay Raj
Last edit: 2014-04-22 16:45:24 |
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2013-03-09 13:42:40 napster
can u tell me how the 3rd test case getting 32768?? |