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UVA 11762
By Upgrading....., history, 8 years ago, In English

UVA 11762

Suppose we want to calculate E(N) where N is a number,k is the number of it’s distinct prime factors and p is the number of primes < n

Normal rule, E(N)=(1/p)*(1 + E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p)*(1 + E(N))

What is the wrong about this logic?

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8 years ago, hide # |
 
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The problem statement also counts "blanks" as a move. Even if you whiff, get an invalid prime, and remain at D, it still adds 1 to the tally because it still counts as a move.

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    8 years ago, hide # ^ |
     
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    it still adds 1 to the tally because it still counts as a move. don't understand this part.....

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    8 years ago, hide # ^ |
     
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    the first 1 is for blank move? so, why not another 1 for invalid prime?

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      8 years ago, hide # ^ |
       
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      Let's look at a simpler problem. How many coins do you have to flip on average to get 2 Heads in a row?

      For this we define E(n) being "expected number of flips if I already have n heads". Obviously E(2) = 0 and we just want E(0).

      How do we compute E(0)? There is a 50% chance that we get tails, so that is still E(0) but  + 1 because we still flipped the coin. There is a 50% chance that we get heads, so that is E(1) and also  + 1 because we also flipped the coin. So that is , and notice if you distribute the fractions everything becomes +1 in the end.

      So to answer your question, yes you put the +1 for invalid prime AND for valid prime. But you put the +1 inside the parenthesis, and the gets multiplied and distributed. The +1 is outside the parenthesis.

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