UVa OJ Board

rio

> junrajz
I used the same algorithm as you described. It got accepted in 0.65 s without any optimization.
Maybe some flaw in your code.

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Rio

rio

There's no need to enumerate through all O(n^3) triangles.

Hint: Given one point, think about how many triangles includes this point.
If this could be done in O(nlogn), the whole calculation could be done with O(n^2logn).
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Rio

rio

There's a non-random O(n^2logn) algorithm.

ADD: I also tried random algorithm, but couldn't get AC.
Didn't have enough accuracy.
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Rio

rio

he can not be away from any project for more than G consecutive days. (of course unless a project is already complete.)

aabaabbb
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Rio

rio

I have tried both Pollard and Brent versions for factorization and got TLE. However, the major bottleneck in my case is calculating (a*b)%c, where a, b and c are 64-bit integers, without an overflow. I am using O(log n) algorithm similar to fast exponentiation. Is there a way to speed up this compu...

rio

You could also get AC using simple Backtrack with a little trick.

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Rio

rio

Your code doesn't even pass the sample io.
Try to output it correctly, and you'll understand wheres your mistake is.

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Rio

rio

There is no restriction in how many commandos you use.
If there is N buildings, think about using N commandos (one command for one building).

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Rio

rio

Got the same answer, and WA...

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Rio

rio

Search google with "lca dag".
I found paper about it in the first page.
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Rio

rio

This problem is about finding LCA in DAG.

I used an algorithm with O(nm) for precalculation and O(1) for query.
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Rio

rio

Sorry for wasting your time

I edited the test case.
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Rio

rio

Try this.

Code: Select all

Correct output is 5 (10(or 9)-8-5-4-2) but your code outputs 4.
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Rio

rio

Part that has bug;

Code: Select all

      int pot = 1;
      for(i=1;i<=n;i++)
      {
         scanf("%d",&a[i]);
         if(a[i]<pot)b=0;
         pot = a[i];
      }

Code: Select all

      for(i=1;i<=k;i++)
      {
         for(j=i+1;j<=k;j++)
         {

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Rio

rio

Actually, I'm having problems with understanding the problem description, should i cut the matrix vertically or horizontally to make the comparison to check whether it's symmetric or not ???? sorry but I'm little confused here :oops: and thanx in advance No, you don't have to. If the matrix is like...

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