A. Maximum in Table

Simulation.

n = int(raw_input())
g = [[1 for i in xrange(n)] for j in xrange(n)]

for i in xrange(1, n):
    for j in xrange(1, n):
        g[i][j] = g[i - 1][j] + g[i][j - 1]

print g[n - 1][n - 1]

B …

A. Pasha and Pixels

Brute force.

There are multiple ways to form a 2*2 square at one single step.

So at every step, we have to check the neighbours of pixel that is colored black.

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>

using namespace std;

#define …

题意

给你一个n * m的矩阵，让你做K次操作，使得最后得到的值最大。

操作有两种：

一是在任意一行上操作，最终的结果值加上这一行数的 …

A. Squats

Trun x => X or X => x to make the number of 'x' is equal to the number of 'X'.

n = int(raw_input())
hamsters = [c for c in raw_input()]

sits = hamsters.count('x')
stands = hamsters.count('X')

if sits == stands:
    print 0
    print ''.join(hamsters)
else:
    if sits > stands …

Overview

It has been months that I didn't participate in the contest on CF, now I'm back. :)

This round of contest makes me confused that the problem B and C is a little bit too twisted, if you can't catch the vital point, you will get a lot of WAs …

由于大号已经进Div. 1了，所以接下来的几场Div. 2都是用小号做的。

等有实力切D题了，再去打一区。（弱

事情一直很多，所以题解落后了好久才发。

A. Sereja …

A. Lever

水题，杠杆原理。

用^把字符串分割开。然后分别计算两边的重量即可。

#Result: Dec 24, 2013 6:04:41 PM    Wizmann  A - Lever   Python 2   Accepted     312 ms  4200 KB
def calc(ss …

A. K-Periodic Array

将Array切片，然后按位统计某一位上1的个数C(1)和2的个数C(2)。然后在这一位上的操作数就为M = min(C(1), C(2))。

简单题

B. Fox Dividing Cheese

傻逼才错 …

A. Sereja and Coat Rack

傻逼才错的题。不幸中枪。

没什么可说的。直接看代码就好。

B. Sereja and Suffixes

关键思想在于统计A[i...n-1]中有多少互不相 …