It may be better if it is an ACM-style contest, it is completely an ACM style one.

the beauty of ACM-Style is ::

1 — problems are not sorted

2 — problems are equals in points ( result is the number of solved problems)

I think Educational rounds should be unrated, according to reasons you have mentioned.

Same buddy :) CF is LOVE

Me too!!

Well, at least the Ducks won!

And all specialists to become experts :)

Your text to link here...

Hopefully they will be because i don't want to read all the problems and not be sure which one is harder.

Probably partially sorted if I had to guess. Wouldn't be surprised if C is harder than D, or E is harder than F, etc.

So you are saying that students actually pay attention to lectures?

Some cells may be blocked by multiple obstacles.

Second line contains max volumes, third line contains temperatures. So temperatures are 50 and 150, not 10 and 150.

It seems that %Lf dosen't work before G++14 on MinGW. And %lf, %f also don't work...It seems that only cout works...

when ui > vi

You can copy at most once!!!

Answer is 5(copy is operation too).

Thanks got it

There is no need for a segment tree. It can be done with sliding window technique. That will save a logn factor

You need to find the differences b/w the values of two consecutive cells. There should be only two difference values : 1. '1' for any two consecutive elements in a row. 2. 'y'(no of columns) for any two consecutive elements in a column(one above other).

There is no constraint on number of rows(x). So you can output 1e9 for each case. Note : Also you need to check the border contraints.

Let's assume for a moment that there are no obstacles at all. If a vector (v_{1}, v_{2}, v_{3}) represents a number of paths ending in a cell (i, n) for some n, then (v_{1} + v_{2}, v_{1} + v_{2} + v_{3}, v_{3} + v_{2}) does the same for n+1. For a large n we can compute the vector efficiently using a matrix multiplication.

Let's get back to the original problem. The board can be separated with vertical lines into O(n) rectangles of two kinds:

• rectangles with no obstacles,

• rectangles with at least one row fully occupied by some obstacle.

The first case can be solved using the algorithm from section 1. The solution for the latest one requires a straightforward cases analysis and can be done in logarithmic or constant time depending on a case.

Viva Chtholly! :)

Viva Chtholly!

Viva Chtholly! :)

http://codeforces.net/contest/940/standings

【这题可能有很多方法】 考虑枚举lca的深度（k）以及两点的深度（i，j）。 写出式子后发现可以发现枚举了k和j的时候，i的可行取值是一个后缀。 于是先加再乘就好了。

[This question may have many methods] Consider enumerating the depth of lca (k) and the depth of two points (i, j). After writing the formula and finding after enumerates k and j, the feasible value of i is a suffix. So first plus and then multiply.

Binary search and greedy. Let's binary search for answer(let's call it x). Now start from beginning, and if sum in range [max(1, i — r), min(n, i + r)] is less than x, increase min(n, i + r) value by x — sum. Now just check if sum of increases <= k.

Have a nice hacks!

The first screenshot was before your solution on B and C get hacked

bitset is very fast. but i think fft is better.

Where do you see that judge answer is 5?

Not in Educational rounds

First split all taps into "cold" with t_i < T, "warm" with t_i > T, and "ideal" with t_i = T. Note that you should always turn all "ideal" taps on completely since they do not change the temperature but contribute to the entire volume.

Then you have to balance "cold" and "warm" taps so that they kind of cancel out. Note that if you have at least one "cold" and one "warm" tap unused, you can increase total volume keeping the temperature constant by adding some volumes from both of the taps. This means that you can either use all "cold" taps, and then compensate with some "warm" taps, or use all "warm" taps and compensate temperature with some "cold" taps — depending on total temperature coming from all taps on.

As for compensation, I guess you can use a greedy approach — sort all the "compensation" taps in the order of |t_i — T| difference, and then turn them on one by one until you reach the desired temperature. It is easy to show that it is always better to turn the "less temperature-influencing" tap on first.

Why using Floyd Warshall for an unweighted graph ? just do a BFS..

Codeforces ignores penalty from compilation error and WA on test 1 verdicts .

Thanks . Nice contest/editorial !

There is also a ranking on who is the first to solve each problem.

Sorry, we got really lazy :D Will be posted in no longer than a hour.