Editorial for COCI '11 Contest 6 #4 Rez

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The first thing to notice is that, in order to obtain as many parts as possible, no three cuts will intersect in a common point. This is easy to see: if there is an intersection of three cuts, then we will get an extra part by moving slightly one of those cuts.

Now we can find a maximum number of parts as a function of number of used cuts:

$\displaystyle number\_of\_parts(r) = 1 + \sum_{i=1}^r i$ $n u m b e r_o f_p a r t s (r) = 1 + \sum_{i = 1}^{r} i$

So we must find the minimum $r$ $r$ for which $number\_of\_parts(r) \ge K$ $n u m b e r_o f_p a r t s (r) \geq K$ . Since our function is obviously monotonic, we can use binary search to find the minimum $r$ $r$ .

If we output only the expected $r$ $r$ , we get $50\%$ $50 %$ of the total number of points.

There are many ways to find the actual cuts, and here is the simplest one. Make the $i^\text{th}$ $i^{th}$ cut by connecting these two points: $(5000-i, -5000)$ $(5000 - i, - 5000)$ and $(-5000, -5000+i)$ $(- 5000, - 5000 + i)$

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