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kingofnumbers's blog

help with spoj problem
By kingofnumbers, 13 years ago, In English

Hi, I have no idea how to solve this problem. can anyone help me?

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13 years ago, hide # |
 
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Every positive number can be represented as a form like this: 2^x * 3^y * z, where z is not divisible by 2 and 3.

Let's suppose that n = 2^x * 3^y * z. Then we cannot have 2n = 2^(x+1) * 3^y * z and 3n = 2^x * 3^(y+1) * z in set S.

Now, let's define a plane P[z]. For each number 2^x * 3^y * z, draw a point in P[z]. Then, the problem will be solvable :)

Sorry for the poor English.

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    13 years ago, hide # ^ |
     
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    We can have 2n and 3n in set S if we didn't include n in set S.

    can you code your idea please to make me understand your idea very well.

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      13 years ago, hide # ^ |
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      Sorry for my mistake..

      Let's suppose that n = 2^x * 3^y * z. (where z is not divisible by 2 and 3) If we include n in set S, we cannot have 2n = 2^(x+1) * 3^y * z and 3n = 2^x * 3^(y+1) * z in set S.

      Let's define a plane P[z] where z is not divisible by 2 and 3. For each positive integer i = 2^x * 3^y * z (<= N), draw a point (x, y) at P[z].

      If we select the point (x, y) in P[z], then we cannot select the point (x+1, y) and (x, y+1) in P[z]. So the problem is: how many ways can we choose points such that the condition above is satisfied?

      The value of x and y are very small, so we can do this in DP.

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        13 years ago, hide # ^ |
         
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        Good idea, but how to do this in DP effectively, notice that we have n/3 numbers less than n that not divisible by 2 and 3 and we have 500 test cases per input.

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          13 years ago, hide # ^ |
           
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          You can use bitmasks.. It needs more complex solution than just bitmasking. Think about it :)

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