本文使用了一种概率算法，不是正解，只求骗分。

题意

原题

给一个长度为n的数组A[1...n]。

现在有q个查询请求（q <= 1e5），每个请求给定一个长度为m的子数组A …

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 …

Introduction

There are a lot of interview problem based on the 1D-array, which is the one of the easiest "data structure".

But the problem about that simple data structure might not be that simple. Here is the summary of the problem about 1D-array.

Of course, most of them come from …

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 …

啥？

Tic-tac-toe是我很久之前在CF上做的一道题。非常考细心的模拟题。

最近有同学和我讨论过类似的问题。于是拿出来重新做一遍。练练手。

原题做 …

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 …