ANARC05B - The Double HeLiX
Two finite, strictly increasing, integer sequences are given. Any common integer between the two sequences constitute an intersection point. Take for example the following two sequences where intersection points are
printed in bold:
- First= 3 5 7 9 20 25 30 40 55 56 57 60 62
- Second= 1 4 7 11 14 25 44 47 55 57 100
You can ‘walk” over these two sequences in the following way:
- You may start at the beginning of any of the two sequences. Now start moving forward.
- At each intersection point, you have the choice of either continuing with the same sequence you’re currently on, or switching to the other sequence.
The objective is finding a path that produces the maximum sum of data you walked over. In the above example, the largest possible sum is 450, which is the result of adding 3, 5, 7, 9, 20, 25, 44, 47, 55, 56, 57, 60, and 62
Input
Your program will be tested on a number of test cases. Each test case will be specified on two separate lines. Each line denotes a sequence and is specified using the following format:
n v1 v2 ... vn
Where n is the length of the sequence and vi is the ith element in that sequence. Each sequence will have at least one element but no more than 10,000. All elements are between -10,000 and 10,000 (inclusive).
The last line of the input includes a single zero, which is not part of the test cases.
Output
For each test case, write on a separate line, the largest possible sum that can be produced.
Sample
Input: 13 3 5 7 9 20 25 30 40 55 56 57 60 62 11 1 4 7 11 14 25 44 47 55 57 100 4 -5 100 1000 1005 3 -12 1000 1001 0 Output: 450 2100
hide comments
sieunhanbom04:
2016-08-03 11:53:48
if someone uses "merge-sort" like approach with sentinel, maybe this test can help you (in case you got WA):
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baadshah_:
2016-07-15 19:38:19
simple approach
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Karre Nagaraju:
2016-07-13 11:50:23
No binary search, no dp, simple logic. Thats it! Last edit: 2016-07-13 11:51:20 |
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hash7:
2016-06-25 13:21:06
Simple implementation ..Only take care of some corner cases . No need of Dp :) |
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mrinal_aich:
2016-06-21 21:01:41
No need for Dynamic Programming.... Just use simple logic, Hint: perform the algorithm you do mentally. |
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Jamil Siam:
2016-06-21 10:38:31
AC at 0.00 time. used O(n) algorithm. |
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atif_11:
2016-05-18 16:58:14
use Binary Index Tree......AC in 1st Go |
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ajay_5097:
2016-05-15 14:34:27
problem is easy ! Binary search + some tricks .. That's it |
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ALi Ibrahim:
2016-05-15 14:10:55
No need for binary search, use map |
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avisheksanvas:
2016-05-13 08:55:06
Another Binary Search implementation with Greedy :)
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Added by: | psetter |
Date: | 2009-07-05 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | ANARC 2005 |