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The third problem is looking to partition the entire sequence of numbers into contiguous subsequences, all of which add up to the same sum. In other words, if t
by DanielStraight 15y ago
The third problem is looking to partition the entire sequence of numbers into contiguous subsequences, all of which add up to the same sum. In other words, if the list were 23415, you subsequences could be 23, 41 and 5, all equaling 5. But the subsequences have to use up the entire string and they can't overlap.
- hugh3 15y agoDid anyone else do that one without writing any code?
- genesiss 15y agoYou mean with smart guessing? That's no fun!
- hugh3 15y agoYeah. It took me three guesses to get it right. :)
- wbhart 15y agoYou didn't need 3 guesses. The sum of each partition has to be a factor of the number. In particular the sum of the first partition has to be a factor. Then the second. That already gives you the answer I think.
- hugh3 15y agoSure, but then I'd actually have to look at the numbers. My solution works without having to download the file! ;)
- deleted 15y ago[deleted]
- llimllib 15y agoThey can't overlap! I don't think the problem specifies that? You have to interpret "break up" to mean "non-overlapping" to claim that the problem means what you're saying it means; I found 341 "contiguous subsequences such that the sums of each of the subsequences are equal", but they overlap. (If you had said "partition" I suspect I would have got what you meant)
- DanielStraight 15y agoContiguous implies non-overlapping.