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 …
在一个平面上,有n+m条蛇,其中n条蛇沿水平方向(y轴方向)移动,m条蛇沿竖直方向(x轴方向)移动。
现给出这些蛇头和尾所在的坐标点,求出这n+m条蛇在此时共有多少个交点。在同一个方向移动的 …
more ...对于一个字符串S,求S的循环的同构字符串S’中字典序最小的一个。
我们举例说明,字符串”abcd”的循环同构字符串有:["abcd", "bcda", "cdab", "dabc"]
。
题目的目标是 …
more ...