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 ...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 …
more ...