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PO
primarykey
Id
3706321
data
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2010-09-14T05:31:51.307
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2010-09-14T07:23:47.057
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2010-09-14T07:23:47.057
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395626
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3696693
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2
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text
Body
Just some hints.. You could impose better restrictions on the values. The requirement leads to condition <code>N1*(k+1) == S*k</code>, where <code>N1</code> is number of ones in an area, and <code>S=dx*dy</code> is its surface. It can be rewritten in better form: <code>N1/k == S/(k+1)</code>. Because the greatest common divisor of numbers <code>n</code> and <code>n+1</code> is always 1, then <code>N1</code> have to be multiple of <code>k</code> and <code>dx*dy</code> to be multiple of <code>k+1</code>. It reduces greatly the possible space of solutions, the larger is <code>k</code>, the better (for <code>dx*dy</code> case you'll need to play with prime divisors of <code>k+1</code>). Now, because you need just the surface of the largest area with such property, it would be wise to start from largest areas and move to smaller ones. By trying <code>dx*dy</code> from <code>n^2</code> downto <code>k+1</code> that would satisfy the divisor and the bounding conditions, you'll find quite fast the solution, muuuch faster than O(n^4), because of a special reason: except cases when the array was specially constructed, if we assume a random input, the probability that there are <code>N1</code> ones out of <code>S</code> values in the <code>(n-dx+1)*(n-dy+1)</code> areas that have the surface <code>S</code> will constantly grow with decrease of <code>S</code>. (large values of <code>k</code> will make the probability smaller, but in the same time they will make the filter for <code>dx</code> and <code>dy</code> pairs stronger). Also, this problem: <a href="http://ioinformatics.org/locations/ioi99/contest/land/land.shtml" rel="nofollow noreferrer">http://ioinformatics.org/locations/ioi99/contest/land/land.shtml</a> , looks somehow similar, maybe you'll find some ideas in their solution.
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1. POFinding the maximum area in given binary data
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1. PTAnswer
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1. USruslik
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1. USruslik
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