## A. 2月29日 (Feb. 29th)

### Description

Given a starting date and an ending date. Count how many Feb. 29th are between the given dates.

### Solution

The easiest way, of course, the brute force, which is quite simple with Python using the `datetime` lib.

However, it's not an effective way for the …

100组数据。

## 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]
```

## 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 …```

## 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 …

## A. Sereja and Dima

```n = int(raw_input())
pokers = map(int, raw_input().split())

v = [0, 0]
p = 0

for i in xrange(n):
if pokers > pokers[-1]:
v[p] += pokers
del pokers
else:
v[p] += pokers[-1 …```

## A. Lever

`^`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):
res = 0
p = 1
for item in ss:
if item != '=':
t = int(item)
res += t * p
p += 1
return res

s = raw_input …```

## Codeforces Round #218 (Div. 2)不完全不正确题解

### D. Vessels 