Solution of Codeforces Round 156

Link: http://codeforces.ru/blog/entry/6161

255A - Greg's Workout

It is not hard problem. We must calculate sums of numbers for each group and print group with maximum count.

255B - Code Parsing

Not hard to see that after few operations of first type string will become: x..xy..y. After fer operations of second type, there will be only letters of one type, count of this letters will be: |count(x) — count(y)|

256A - Almost Arithmetical Progression

Заметим, что ответ это длина последовательности: a, b, a, b, ... где a и b — некоторые целые числа. Зафиксируем одно число (допустим a), будем перебирать число b, и считать какой мы получим ответ, если это будет последнее число в последовательности. Заметим, что для фиксированных a, b — ответ считается жадно. Так же будем действовать и тут. Будем искать последнее вхождение числа b до зафиксированного, что между ними есть число a, и будем брать ответ как длина до найденного числа +2 (икасть будем с помощью метода двух указателей). Так же нужно рассмотреть случай, когда это будет 1е или 2е вхождение в последовательность.
Так же существует решение с помощью динамического программирования.
Асимптотика обоих решений O(n^2).
Буду очень рад, если кто то напишет решение с лучшей асимптотикой.

256B - Mr. Bender and Square

Solution — binary search for answer. Next we have to calculate the area of a truncated square set at 45 degrees. This can be done as follows: Calculate its total area. Subtract area that cuts off the top line. Similarly, for the lower, left and right line. Add parts that are cutted by corners. You can write a function that finds the length of the truncation desired area, for that would not write a lot of code.

256C - Furlo and Rublo and Game

Note that after the first move any pile turns into a pile no larger than 1000000. We assume Grundy function for numbers less than 1 million. Grundy function is very small, you can start on the partial sums for each type of function that would quickly tell what function is in the interval, and which are not present. Knowing the answer is not difficult to find small response for all piles.

256D - Liars and Serge

If person say number x, and at all x was said by x persons, then we cannot tell anything about fixed person.

Now we understand which sequence are good for us. We will calculate their count wuth dynamic programming dp[n][m][k], n — which persons answers we set to the sequence right now, m — how mant persons gived theis answers, k — how many persons from them are liers.
Transfer:
dp[n][m][k]*cnk[N-m][n] -> dp[n+1][m+n][k]
dp[n][m][k]*cnk[N-m][p] -> dp[n+1][m+p][k+p] p = 1 .. N, p != n.
We assume, that N — total number of the persons. This solution get TLE, becouse complexity if O(N^4). We need to use precalc. It will not be so big, as N is power of 2.

256E - Lucky Arrays

Solution is — interval tree. We will save dynamic programming f[i,j] in each vertex, this dp means: in how many ways we can change all 0 to some numbers on interval, such that it will be valid and first element will be i and last will be j.

With normal implementation its easy to pass system tests.