Oh really? I always save all programming contest events in my phone's calendar and it always reminds me one day before each contest. I don't usually check for new email so I don't know when they send email about upcoming contest.

Btw, sorry for any inconvenience.

Same problem. However, I managed to clean cache after rebooting computer. Didn't help.

I am an ubuntu newbie, but typing "ControlPanel" in terminal and deleting temporary Java files in "Temporary files setting" helped me.

You should try command "javaws -Xclearcache" to clear cache.

Key observation: you can delete the node after you visit it. Let's create the sequence from the end. Let DP[i][j] be the number of ways to create a pattern and end up with i nodes in first row and j nodes in second row, while you start from the first row. Transitions: you got here from the other row ((j + 1)·DP[j + 1][i]) or from one of two neighbors in the same row (2·DP[i + 1][j]). Answer is 2·DP[1][0].

First thing to notice is that all configurations of x unused dots in a row are equivalent, because you can only move to the next dot to the right or left anyway.

Let s(x, y) be the number of ways you can make a pattern, if you have x unused dots in first row, y unused dots in second row, and you can be anywhere in first row.

You can move to the left or right in the first row, either of this leads you to s(x-1, y). So sum 2*s(x-1, y) to answer.

You can also go to the second row, this brings you to s(y, x-1). Remember you can be anywhere on first row, so add x * s(y, x-1) to answer.

It is rather easy to see that you basically add edge to graph from end to start and you do not add edges that would make 0 and N — 1 connected, but add theit cost to the answer. That is the intended solution, you just make it linear while it was easier to do it in qudratic time

I think it is the intended solution. Only one thing I still don't get is why the cost is 3ⁱ instead of 2ⁱ. Is the power of 2 too mainstream?

DSU is what I intend exactly. Because max flow equal to min cut, we can try to find min cut instead to max flow. In a cut, points of a graph will be divide to two group. we will try to make the edge of larger costs connect the points of same group due to 3ⁱ > 3¹ + 3² + ... + 3^i - 1

I'm actually kinda surprised the bruteforce only times out for a single case.

But you can transform it into correct solution easily: note that if your current product is not a divisor of goodValue, it's impossible to make it equal to goodValue by doing multiplications. So you should only add something to buffer when it divides goodValue.

This code passes systests spending less than 10 ms per case.

I found my mistake , at each stage we have to ensure that the current product is a divisor of the GOODVALUE. Thanks ffao.

People used dynamic programming to solve the problem. state dp(i,p) // i is the ith element and p is current product. Problem is p can be up to 2000000000. So we can not use such a huge memory for memoization purpose. For that people used map to save the states.