Tintin's blog

How can I implement it O(n)?
By Tintin, 13 months ago, In English,

Your text to link here... My dp solution is O(n^2) but here n=50000 and time is 1 sec. Is there any mathematical formula which reduces time within 1 sec. Any hints...Thanks

 
 
 
 

Comments Comments (6)
 
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13 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Try this idea: calculate sums in all prefixes (from empty to the whole sequence) and then sum of elements in a some substring is difference of two numbers. Hope this helps. If not, I can give you detailed solution.

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    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I don't get it properly.

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      13 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Look: you have an array: 1 2 3 4 5. It has the following prefixes: "", "1", "1 2", and so on. You calculate sum of numbers in each such prefix and get 0, 1, 1+2=3, 1+2+3=6, 10 and 15 (let's call it s[]). Then, to calculate, for example, sum from 2 to 4, you just calculate s[4] - s[1]=9. Elements before 2 were reduced.

      This idea can help you solving this problem: you run down left border of a substring and then you can calc amount of required substrings in O(1) for each left border.

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        13 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I get your first ideal and I use it in my code.But still don't get your second idea. here is my code:http://pastebin.com/abRxS9kg

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          13 months ago, # ^ |
          ← Rev. 2   Vote: I like it 0 Vote: I do not like it 
          s[0] = 0;
for (int i = 0; i < a.size(); i++)
  s[i + 1] += s[i] + a[i];
for (int i = 0; i <= s.size(); i++) {
  res += cnt[s[i] % d];
  cnt[s[i] % d]++;
}
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            13 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            I get it.Thanks to yeputons and ALias both of you

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