b*r
2 楼
比如推荐信?
谢谢!
谢谢!
l*a
3 楼
Tomorrow Friday (4/2/10), Newegg e-blast: "SSD super sale tomorrow over 20
huge deals on hot SSDs" with brands from Intel, OCZ, and more. Stay tuned
with us for more details.
huge deals on hot SSDs" with brands from Intel, OCZ, and more. Stay tuned
with us for more details.
t*e
4 楼
放松心态,祝成功!
d*0
6 楼
假的还是真的?
w*y
10 楼
多谢楼上了!
又看到一道不会做的题目//囧
"N*N 的正方形内有黑白两色,求四边都是黑色的最大的子正方形"
这个题怎么做? 版上有讨论么? 我好像没有看到过. 我只见过那种, 所有'1' 围成的最
大方形, 不知道这种只要求'四边'的怎么做呀
又看到一道不会做的题目//囧
"N*N 的正方形内有黑白两色,求四边都是黑色的最大的子正方形"
这个题怎么做? 版上有讨论么? 我好像没有看到过. 我只见过那种, 所有'1' 围成的最
大方形, 不知道这种只要求'四边'的怎么做呀
t*e
12 楼
想搞两块 80GB 的 X25M,把家里台式机换了,明天 $150/ea 就跳。
p*2
13 楼
bless。看你最近很努力。
x*w
14 楼
要是有offer letter的话,应该要原件吧!
x*w
17 楼
要是有offer letter的话,应该要原件吧!
z*u
19 楼
Be sure know how to talk in the interview.
Prepare technical questions;
Prepare behavior questions.
The most important, only say what they want to hear in the interview.
Listen how the experienced people answer interview questions.
我主持的 FREE 华人周三晚找工周会,就是为你这样的朋友们开的,Will answer any
of your interview questions. Come join the weekly meeting or listen the
meeting record.
Here is the website for Weekly meeting:
http://bbs.wenxuecity.com/career/428047.html
我就是想让你的面试能准备的更充分一些,听周会录音, 是很多朋友一致暂不绝口的
好方法.谁对此不满,随他去吧!
Prepare technical questions;
Prepare behavior questions.
The most important, only say what they want to hear in the interview.
Listen how the experienced people answer interview questions.
我主持的 FREE 华人周三晚找工周会,就是为你这样的朋友们开的,Will answer any
of your interview questions. Come join the weekly meeting or listen the
meeting record.
Here is the website for Weekly meeting:
http://bbs.wenxuecity.com/career/428047.html
我就是想让你的面试能准备的更充分一些,听周会录音, 是很多朋友一致暂不绝口的
好方法.谁对此不满,随他去吧!
c*o
22 楼
所有'1' 围成的最大方形,怎么做?
G*e
25 楼
跳!
t*e
26 楼
这是我当年考过的面试题,可以负责的告诉你这题面试中不会再考到了,不过拿来练
coding还是很好的,topcoder上面也考过类似的问题。解法如下:
Let the input matrix be x[n][n]. The idea is to calculate two auxiliary
matrices a[n][n], where a[i][j] records the length of the all 1 horizontal
edge to the right of a[i][j], and b[n][n], where b[i][j] records the length
of the all 1 vertical edge above a[i][j]. Then the size of the largest all 1
boundary sub-square whose left bottom corner is a[i][j] is t = max{0≤t≤
min(a[i][j], b[i][j])|a[i-t+1][j]≥t and b[i][j+t-1]≥t}.
Obviously, a[i][j] and b[i][j] can be calculated recursively:
a[i][j] = 0; if x[i][j] == 0
a[i][j] = 1 + a[i][j+1]; if x[i][j] == 1
b[i][j] = 0; if x[i][j] == 0
b[i][j] = 1 + b[i-1][j]; if x[i][j] == 1
This is the C++ implementation of the algorithm:
template
int maxAllOneSquare(int (&x)[m][n], pair& output)
{
int a[m][n], b[m][n];
int maxSquare = 0;
output.first = -1;
output.second = -1;
for(int row = 0; row < m; ++row){
for(int col = n - 1; col >= 0; --col){
if(col == n -1){
a[row][col] = x[row][col] == 1 ? 1 : 0;
}
else{
a[row][col] = 0;
if(x[row][col] == 1){
a[row][col] = 1 + a[row][col + 1];
}
}
}
}
for(int row = 0; row < m; ++row){
for(int col = 0; col < n; ++col){
if(row == 0){
b[row][col] = x[row][col] == 1 ? 1 : 0;
}
else{
b[row][col] = 0;
if(x[row][col] == 1){
b[row][col] = 1 + b[row - 1][col];
}
}
}
}
for(int row = 0; row < m; ++row){
for(int col = 0; col < n; ++col){
if(x[row][col] == 1){
int t = min(a[row][col], b[row][col]);
while(t > 0 &&
(a[row - t + 1][col] < t || b[row][col + t - 1] < t)){
t--;
}
if(maxSquare < t){
maxSquare = t;
output.first = row;
output.second = col;
}
//maxSquare = max(maxSquare, t);
}
}
}
return maxSquare;
}
coding还是很好的,topcoder上面也考过类似的问题。解法如下:
Let the input matrix be x[n][n]. The idea is to calculate two auxiliary
matrices a[n][n], where a[i][j] records the length of the all 1 horizontal
edge to the right of a[i][j], and b[n][n], where b[i][j] records the length
of the all 1 vertical edge above a[i][j]. Then the size of the largest all 1
boundary sub-square whose left bottom corner is a[i][j] is t = max{0≤t≤
min(a[i][j], b[i][j])|a[i-t+1][j]≥t and b[i][j+t-1]≥t}.
Obviously, a[i][j] and b[i][j] can be calculated recursively:
a[i][j] = 0; if x[i][j] == 0
a[i][j] = 1 + a[i][j+1]; if x[i][j] == 1
b[i][j] = 0; if x[i][j] == 0
b[i][j] = 1 + b[i-1][j]; if x[i][j] == 1
This is the C++ implementation of the algorithm:
template
int maxAllOneSquare(int (&x)[m][n], pair
{
int a[m][n], b[m][n];
int maxSquare = 0;
output.first = -1;
output.second = -1;
for(int row = 0; row < m; ++row){
for(int col = n - 1; col >= 0; --col){
if(col == n -1){
a[row][col] = x[row][col] == 1 ? 1 : 0;
}
else{
a[row][col] = 0;
if(x[row][col] == 1){
a[row][col] = 1 + a[row][col + 1];
}
}
}
}
for(int row = 0; row < m; ++row){
for(int col = 0; col < n; ++col){
if(row == 0){
b[row][col] = x[row][col] == 1 ? 1 : 0;
}
else{
b[row][col] = 0;
if(x[row][col] == 1){
b[row][col] = 1 + b[row - 1][col];
}
}
}
}
for(int row = 0; row < m; ++row){
for(int col = 0; col < n; ++col){
if(x[row][col] == 1){
int t = min(a[row][col], b[row][col]);
while(t > 0 &&
(a[row - t + 1][col] < t || b[row][col + t - 1] < t)){
t--;
}
if(maxSquare < t){
maxSquare = t;
output.first = row;
output.second = col;
}
//maxSquare = max(maxSquare, t);
}
}
}
return maxSquare;
}
w*z
27 楼
好奇问一下,这题是你们自己想出来的吗?
还是哪里看来的?
length
1
【在 t******e 的大作中提到】
: 这是我当年考过的面试题,可以负责的告诉你这题面试中不会再考到了,不过拿来练
: coding还是很好的,topcoder上面也考过类似的问题。解法如下:
: Let the input matrix be x[n][n]. The idea is to calculate two auxiliary
: matrices a[n][n], where a[i][j] records the length of the all 1 horizontal
: edge to the right of a[i][j], and b[n][n], where b[i][j] records the length
: of the all 1 vertical edge above a[i][j]. Then the size of the largest all 1
: boundary sub-square whose left bottom corner is a[i][j] is t = max{0≤t≤
: min(a[i][j], b[i][j])|a[i-t+1][j]≥t and b[i][j+t-1]≥t}.
: Obviously, a[i][j] and b[i][j] can be calculated recursively:
: a[i][j] = 0; if x[i][j] == 0
还是哪里看来的?
length
1
【在 t******e 的大作中提到】
: 这是我当年考过的面试题,可以负责的告诉你这题面试中不会再考到了,不过拿来练
: coding还是很好的,topcoder上面也考过类似的问题。解法如下:
: Let the input matrix be x[n][n]. The idea is to calculate two auxiliary
: matrices a[n][n], where a[i][j] records the length of the all 1 horizontal
: edge to the right of a[i][j], and b[n][n], where b[i][j] records the length
: of the all 1 vertical edge above a[i][j]. Then the size of the largest all 1
: boundary sub-square whose left bottom corner is a[i][j] is t = max{0≤t≤
: min(a[i][j], b[i][j])|a[i-t+1][j]≥t and b[i][j+t-1]≥t}.
: Obviously, a[i][j] and b[i][j] can be calculated recursively:
: a[i][j] = 0; if x[i][j] == 0
t*e
28 楼
抱歉没说清楚,我是被面试者,题目是面试我的人出的,应该是他原创的,呵呵。
l*8
33 楼
为什么说“这题面试中不会再考到了”? 因为题目太老了?
length
1
【在 t******e 的大作中提到】
: 这是我当年考过的面试题,可以负责的告诉你这题面试中不会再考到了,不过拿来练
: coding还是很好的,topcoder上面也考过类似的问题。解法如下:
: Let the input matrix be x[n][n]. The idea is to calculate two auxiliary
: matrices a[n][n], where a[i][j] records the length of the all 1 horizontal
: edge to the right of a[i][j], and b[n][n], where b[i][j] records the length
: of the all 1 vertical edge above a[i][j]. Then the size of the largest all 1
: boundary sub-square whose left bottom corner is a[i][j] is t = max{0≤t≤
: min(a[i][j], b[i][j])|a[i-t+1][j]≥t and b[i][j+t-1]≥t}.
: Obviously, a[i][j] and b[i][j] can be calculated recursively:
: a[i][j] = 0; if x[i][j] == 0
length
1
【在 t******e 的大作中提到】
: 这是我当年考过的面试题,可以负责的告诉你这题面试中不会再考到了,不过拿来练
: coding还是很好的,topcoder上面也考过类似的问题。解法如下:
: Let the input matrix be x[n][n]. The idea is to calculate two auxiliary
: matrices a[n][n], where a[i][j] records the length of the all 1 horizontal
: edge to the right of a[i][j], and b[n][n], where b[i][j] records the length
: of the all 1 vertical edge above a[i][j]. Then the size of the largest all 1
: boundary sub-square whose left bottom corner is a[i][j] is t = max{0≤t≤
: min(a[i][j], b[i][j])|a[i-t+1][j]≥t and b[i][j+t-1]≥t}.
: Obviously, a[i][j] and b[i][j] can be calculated recursively:
: a[i][j] = 0; if x[i][j] == 0
l*8
34 楼
要修改t的值
t一直表示当前测试的正方形边长。
t一直表示当前测试的正方形边长。
p*2
35 楼
p*2
40 楼
11
11
变成
11
12
111
111
111
变成
111
122
123
11
变成
11
12
111
111
111
变成
111
122
123
p*2
44 楼
对我说的另一题
★ Sent from iPhone App: iReader Mitbbs Lite 7.52
★ Sent from iPhone App: iReader Mitbbs Lite 7.52
m*0
45 楼
good luck!!
w*o
47 楼
h*0
55 楼
bless
d*s
56 楼
Bless
f*2
57 楼
bless
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