Rei's blog

Round 51, B
By Rei, 14 years ago, translation, In English
Problem B.
One just needs to calculate all possible answers and find the minimum. For example one may run on a set of numbers, for all pairs of numbers apply next operation to that pair and recursively run on a new set of numbers. When only one number remains, compare it to the already obtained minimum, and change that minimum if  it's needed.
  • Vote: I like it
  • +19
  • Vote: I do not like it

Comments Comments (3)
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14 years ago, # |
  Vote: I like it 0 Vote: I do not like it
Can you describe how the answer to 1st test is 3?
1 1 1 1
+ + *

i proceed this way
+
2 2 1 1

+
2 3 3 1

*
2 3 3 3

???
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  • 14 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it
    Read the problem statement carefully. The operations are:

    1 1 1 1

    +
    2 1 1

    +
    3 1

    *
    3
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  • 14 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it
    I think you didn't understand the statement correctly. "... replaced them with their sum or their product." means that he erases two numbers and writes only one (sum or product). I see ambiguity in mentioned phrase, but you can derive right interpretation from the next sentence: "In the end he got one number". So after 3 minutes there should be only one number at the blackboard.
    In 1st test we have:
    1 1 1 1
    +
    2 1 1
    +
    3 1
    *
    3
    We apologize for that ambiguity, but believe that it is not very hard to resolve it by reading statement carefully one more time.

    Meh, I am too slow.
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