Implementation
(n, m) = map(int, raw_input().split())
res = []
for i in xrange(n):
if i % 2 == 0:
res.append('#' * m)
elif (i / 2) % 2 == 0:
res.append('.' * (m - 1) + '#')
else:
res.append('#' + '.' * (m - 1))
for line in res:
print line
DFS …
more ...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]
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 …Simple and easy, solved by two lines of python code.
ls = filter(lambda y: y, map(lambda x: x.strip(), raw_input()[1:-1].split(",")))
print len(set(ls))
Brute force. Just enumerate the beginning and the end of the substring, and …
more ...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 …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 …
more ...