M******1
发帖数: 67 | 1 py上的不多，就常见那几种，说几个dest上见到过的大招。
dest其实就是一个大招图。 出生点就两个， 有矿后空间，有第二通道到主矿，有第二
路线（非指挥探路侦察不到的中间路线）。
先说常见的
5d 这个 是永远的大招，经过多年的总结以后此大招已经被开发成为了经济型打法了。
也就是说，只要咬掉几个农民吓唬对手拖农民补防御后，根本就没有想像中那么亏。
要点:血池下了后补农民到6， 6狗后房子。 操作稍微细腻点，对马虎和大意的p双开重
创乃至必杀。对水平不错的p, 可以逼其拖农民，补光炮，自己狂补农民开矿，其实一
点不亏。 对单兵营双开的T同理。
4d, 这个就纯大招了，比5d快几秒，更关键的是可以续狗，但是失败了基本上就输了，
不像5d还有的打。其他和5d相同。 找漏洞，逼对手拖农民造防御。
8bb地图侧中造，和7p 侧中兵营， 都是前置兵营的rush打法，看操作。 注意不要被农
民包家夹，然后尽量杀伤农民和关键建筑。 前5个 枪兵，和前三个lot的时期是兵力优
势期一定要好好用。 配以两三个农民更佳。 如果对手果断放弃外围，退而防守。 要
牺牲兵力注意侦察对手是速科技还是暴兵一波流。 |
|
e*******e
发帖数: 342 | 2 这几天老板让找一下有没有真核细胞用的、非融合表达GFP的DEST载体，只是用GFP来监测转化
效率，并不想作为标签蛋白。各位大侠，如有知道的，请提供信息，谢谢！ |
|
j*****y
发帖数: 1071 | 3 void * memmove(void * dest, void * src, unsigned int n)
{
if(src == dest)
{
return dest;
}
if(src + n <= dest || dest + n <= src)
{
memcpy(dest, src, n);
return dest;
}
if(src < dest)
{
memmove(dest + (dest - src), dest, n - (dest - src));
memcpy(dest, src, dest - src);
}
else
{
memcpy(dest, src, src - dest);
... 阅读全帖 |
|
t**********y
发帖数: 374 | 4 多谢了！！！
====
#import "VideoFrame.h"
#import "opencv2/opencv.hpp"
static void YpCbCr422toBGRA8(uint8_t *src, uint8_t *dest, uint32_t width,
uint32_t height);
@implementation VideoFrame
@synthesize image = _image;
@synthesize presentationTime = _presentationTime;
- (id)initWithIplImageTakingOwnership:(IplImage *)image presentationTime:(
NSTimeInterval)presentationTime
{
if ((self = [super init])) {
NSAssert(image, @"image is required");
_image = image;
_presentationTime = presentationTime;
}
return... 阅读全帖 |
|
z****e
发帖数: 54598 | 5 居然官方api里面没有找到
嗯，自己按照stackoverflow上的写了几个
其中一个是horizontal flip
还有一个是用最小内存消耗的方式读取图片到内存
default的8888格式吃内存太离谱了
还有就是android居然限制一个app只能16m
我插，让我想起小时候玩sfc，老板拿个磁盘一点一点读
也就只有16m
private static Map rectCache;
static {
rectCache = new HashMap();
}
public static void drawImage(Canvas canvas, Bitmap bitmap, Rect source,
Rect dest, Paint paint, boolean horizontalRotate) {
if (horizontalRotate) {
if (rectCache.get(dest) == null) {
re... 阅读全帖 |
|
S****G
发帖数: 3 | 6 关于输赢那个题目，根据版上大牛们的讨论，写了code大家可以一看，主要用了zemel
定理（也可以不用但code会比较messy）。此外，博弈树怎么做这个题，谁有相关资料
可以发来看看。
http://feisyr.blogspot.com/2015/04/zermelos-theorem-game-theory
bool canW(int val1, std::vector &available, int curSum, int & desT){
if (val1 + curSum >= desT) return true;
else{
bool theOtherWin = false;
for (int val2 = 1; val2 < available.size(); ++val2){
if (!available[val2]) continue;
available[val2] = false;
theOtherWin = the... 阅读全帖 |
|
m***i
发帖数: 2480 | 7 Some straight forward way:
Put all your new data into a temp table X.
delete all rows in X which are already in the Dest table
delete X from X Join Dest where X.A = Dest.A and X.B = Dest.B
insert all records in X to the Dest table
insert into Dest select * from X |
|
d****p
发帖数: 685 | 8 I doubt the memcpy implementation is wrong.
Dereferencing a void*?
I think the following is an OK implementation:
void* memcpy(void* dest, const void* src, size_t size)
{
assert(dest != 0 && src != 0);
char* dest2 = (char*)dest;
char* src2 = (char*)src;
while (size-- > 0)
*dest2++ = *src2++;
return dest;
}
Actually the following is also OK:
void* memcpy(void* dest, const void* src, size_t size)
{
assert(dest != 0 && src != 0);
|
|
t*********7
发帖数: 255 | 9 public static int getLeastStep(int dest, int start, int step){
if(start+step==dest || start-step==dest){
return step;
}
if(step > (10^9) || step<0){
return -1;
}
int rightStep = getLeastStep(dest, start+step, step+1);
int leftStep = getLeastStep(dest, start-step, step+1);
if(rightStep != -1 && leftStep != -1){
return Math.min(rightStep, leftStep);
}
if(rightStep != -1 || leftStep !... 阅读全帖 |
|
J****3
发帖数: 427 | 10 char *compress(char* str){
int rlen;
int len = strlen(str);
char* dest = (char*)malloc(sizeof(char)*(2*len+1));
int j = 0;
for(int i = 0; i < len; i++){
dest[j++] = str[i];
rlen = 1;
while(i+1 < len && str[i] == str[i+1]){
rlen++;
i++;
}
dest[j++] = rlen+'0';
}
dest[j] = '\0';
return dest;
}
O(n) 但是不是inplace的 你看看行不 |
|
h**o
发帖数: 548 | 11 char *compress(char* str){
...
char* dest = (char*)malloc(sizeof(char)*(2*len+1));
...
};
cc150说如果COMPRESSION比原str短，才用新的dest。否则还用原来的str。那caller怎
么知道你返回的是原来的str，所以他不用free dest, 还是你返回了dest,需要caller
free 掉dest哪？ |
|
f*****g
发帖数: 15860 | 12 起码应该可以这么干(SQL SERVER)
SELECT org.city, org.state, dest.city, dest.state
FROM
(
(SELECT org_zip, dest_zip, city, state FROM table_A, table_zip
WHERE org_zip = zipcode AS org)
UNION
(SELECT org_zip, dest_zip, city, state FROM table_A, table_zip
WHERE dest_zip = zipcode AS dest)
)
WHERE org.org_zip = dest.org_zip AND org.dest_zip = dest.dest_zip |
|
z****n
发帖数: 1379 | 13 boolean HasPath(Point source, Point dest)
{
boolean up = false;
boolean down = false;
boolean left = false;
boolean right = false;
source.visited=true;
if(source.equals(dest))return true;
if(IsBlock(source))return false;
if(Up(source)!=null && !Up(source).visited)
up = HasPath(Up(source), dest);
if(Down(source)!=null && !Down(source).visited)
down = HasPath(Down(source), dest);
if(Left(source)!=null && !Left(source).visited)
|
|
j*****u
发帖数: 1133 | 14 c# implementation，还有没有更好方法？
static void ArrayShift(T[] array, T seed) where T:IComparable
{
if (array == null)
throw new ArgumentNullException("array");
int dest = 0;
for (int i = 0; i < array.Length; ++i)
{
if (i != dest && !array[i].Equals(seed))
{
array[dest++] = array[i];
}
}
for (int i = dest; i < array.Length; ++i)
{
array[i] = seed;
}
} |
|
j*****y
发帖数: 1071 | 15 memmove(dest, dest - 1, 10)
需要 call memcpy 10 次吧， 这里假定内存的布局中 dest - 1 在 dest的
左边 |
|
l*********1
发帖数: 936 | 16 我不太写java code, 不过思路是这样
一个source 指针，每次加一，
一个dest 指针，只有non-negative 才加一，
从source copy 到 destination,
最后call removeRange one time.
code (c++ style) is:
source = 0;
dest = 0;
while (source < array.size()) {
if (array[source] >= 0 ){
array[dest++] = array[source];
}
++source;
}
array.removeRange(dest, source);
you should check boundary cases:
1) array is empty
2) every element is negative
3) every element s positive |
|
I*******g
发帖数: 7600 | 17 Imagine the following compare_and_swap (CAS) primitive function is available
to
you. Using this function, write an integer counter class providing increment,
decrement, set and get APIs in an atomic and thread-safe way in a context
subject to
periodic highly intensive concurrent activity .
/* In a critical section, if (*dest) == old_val, then updates dest to new_
val and returns true otherwise leaves dest unchanged and returns false.
*/
bool compare_and_swap(int* dest, int old_val, int new_val) |
|
t*****g
发帖数: 1275 | 18 oracle?
select date, origin, dest from table where type = 4 group by date, orgin,
dest
minus
select date, origin, dest from table where type <> 4 group by date, orgin,
dest |
|
l*******d
发帖数: 101 | 19 网上下的一段程序。需要分析复杂度。中间的循环完全把我绕糊涂了。该怎样想呢？
谢谢回帖！
void repmat(char *dest, const char *src, int ndim, int *destdimsize,
int *dimsize, const int *dims, int *rep)
{
int d = ndim-1;
int i, chunk;
/* copy the first repetition into dest */
if(d == 0) {
chunk = dimsize[0];
memcpy(dest,src,chunk);
}
else {
/* recursively repeat each slice of src */
for(i=0;i
repmat(dest + i*destdimsize[d-1], src + i*dimsize[d-1],
ndim-1, destdimsize, dimsize, |
|
|
w*******y
发帖数: 60932 | 21 KLM is having a fare sale. Wide open dates and availability
Sample fares are (out of Boston, but great fares all over the East coast)
BOS-London - $611
BOS-Manchester - $714
BOS - Stockholm - $748
BOS - Moscow - $743
BOS - Delhi - $1176
BOS- Mumbai - $1094
All of these fares are tax included - great deals!
Sale:
http://www.klm.com/travel/us_en/apps/ebt/ebt_home.htm?goToPage= flex=true&c[0].os=BOS&c[0].ds=LHR&Dest=LHR
#91;0].os=BOS&c.ds=LHR&c.os=LHR&c.ds=BOS&c.dd=2011-09-06&c.dd=2011-09-20&%
20f... 阅读全帖 |
|
i**w
发帖数: 883 | 22 回拨&DISA：
context answer-with-option {
s => {
Noop(CALLBACK flag: ${DB(OPTIONS/CALLBACK)});
if ("${DB(OPTIONS/CALLBACK)}" == "1") {
Set(caller=${CALLERID(num)});
Noop(${caller});
Set(callbackNum=${SHELL(/etc/asterisk/scripts/callback -get ${caller})});
Noop(${callbackNum});
if (${callbackNum} != -1) {
Noop(call back to ${callbackNum});
System(/etc/asterisk/scripts/callback -... 阅读全帖 |
|
w****f
发帖数: 1420 | 23 大家帮忙看下价格如何
2011 Grand Cherokee Overland 4x4
MSRP $41910 + 780dest = 42690
Inovoice: $39423 + 780dest = 40203
5.7 V8 MDS HEMI
MSRP $1995
Inovice: $1776
Off Road Pack II
MSRP $275
Invoice: $245
430N
included
Adaptive cruise control
MSRP: $1295
Invoice: $1153
Engine heater
MSRP: $50
Invoice: $45
TOTAL MSRP = $46305 (dest included)
TOTAL Invoice = 43422 (dest included)
Dealer offered 41,346 (dest included) + TTL
是否是deal？？？不是现车，需要order。4-8week提车 |
|
B********m
发帖数: 98 | 24 不带navi,不要rear recreation,主要是在edmund和piloteer上找的，跟几个dealer也
email quote了一些价,除了车钱比较大头的是800的destination fee,200左右的
processing fee, tax之类的就自己算吧。NY/NJ地区的dealer第一轮quote价都比较高
，比较低的报价多是是下面几个地区的
Philadelphia 32500 (w/ dest. fee)
Baltimore 30500 (w/o dest. fee)
CA, TX 30500 (w/o dest. fee)
感觉地区差异还是很大的，欢迎版上同学share自己的经验 |
|
j*****t
发帖数: 2225 | 25 还是搜索引擎好用……我拿的就是26700，贵了63块，忘记了，我加了一个55块的轮子
锁，所以贵了8块钱，还行……
2015 SUBARU FORESTER
Model/Trim----Trans--MSRP-----MSRP+Dest--Invoice---Invoice+Dest--VIP------
VIP+Dest
2.5i Limited--CVT----$28,095--$28,945----$26,313---$27,163-------$25,787--$
26,637 |
|
c*****n
发帖数: 96 | 26 should be DFS ?
PrintPath(string currentPath, node current, node dest)
{
if (current == dest)
{
printf (currentPath + node);
return;
}
MarkNodeAsVisited(node);
foreach (node n in node.AdjacentNodes)
{
if (n is not visited)
{
PrintPath(currentPath + n, n, dest);
}
}
MarkNodeAsNotVisited(node);
} |
|
S******t
发帖数: 151 | 27 我大概写一个吧，没编译，不知道靠谱不
#include
#include
#include
using namespace std;
string dict[MAXNUM];
vector graph[MAXNUM];
bool visited[MAXNUM];
bool diff_one(string a,string b) {
int len_a = (int)a.size();
int len_b = (int)b.size();
if (len_a != len_b) return false;
int cnt = 0;
for(int i=0;i
if(a[i]!=b[i]) cnt++;
return cnt == 1;
}
void preprocess() {
int dict_sz = (int)dict.size();
for(int i=0;i
for(int j=i+1;j
... 阅读全帖 |
|
S******t
发帖数: 151 | 28 我大概写一个吧，没编译，不知道靠谱不
#include
#include
#include
using namespace std;
string dict[MAXNUM];
vector graph[MAXNUM];
bool visited[MAXNUM];
bool diff_one(string a,string b) {
int len_a = (int)a.size();
int len_b = (int)b.size();
if (len_a != len_b) return false;
int cnt = 0;
for(int i=0;i
if(a[i]!=b[i]) cnt++;
return cnt == 1;
}
void preprocess() {
int dict_sz = (int)dict.size();
for(int i=0;i
for(int j=i+1;j
... 阅读全帖 |
|
h********g
发帖数: 496 | 29 我也来一个。发现这个题目其实现场写不容易，细节很容易忽略。算法很简单，可还是
写了半个小时，怎么办？
void f(char*s, char*dest){
int set[256]={0};
int i=0, len=strlen(s);
int count=0, maxWin=0;
int l=0, r=len; //local window edges
int ll, rr; //global window edges
while(i
if(count<2 || set[s[i]]){
r=i;
set[s[i]]++;
i++;
if(!set[s[i]]) count++;
if(maxWin
maxWin=r-l+1;
ll=l;
... 阅读全帖 |
|
h**o
发帖数: 548 | 30 1）请问为什么malloc 是 2*len + 1？
2）如何处理让caller知道你返回dest,他应该用后 删去,还是你返回原str， 他不用删
dest。（如果dest比str长） |
|
h**o
发帖数: 548 | 31 你的意思是说如果dest比原str还长，就把str拷给dest,依然返回dest? |
|
w*******e
发帖数: 395 | 32 如果，src和dest有不同的aligment怎么办呢？
比如src移了2个byte后就是4-byte aligned，但是dest本身就是4-byte aligned，不需
要移动，那你优化了src，但是dest写的时候又不是aligned。
这总情况怎么办？ |
|
F****n
发帖数: 3271 | 33 不用 line sweep 的简单解法：
import java.awt.geom.Rectangle2D;
import java.util.ArrayList;
import java.util.List;
public class Aggregation
{
protected List rects;
protected Aggregation overlap;
public Aggregation()
{
this.rects = new ArrayList<>();
this.overlap = new Aggregation();
}
public void add(Rectangle2D r)
{
List intersects = new ArrayList<>(this.rects.size());
for (Rectangle2D rect : this.rects)
{
... 阅读全帖 |
|
|
k***e
发帖数: 7933 | 35 网上好像只能用multi-dest来定多段里程票，但是里程就不一样了。
比如单程从芝加哥到上海，日本转机，32500
如果用multi-dest，在日本呆3天，要42500多。
你的意思是说我可以打电话定我在multi-dest查到的里程票，只要32500里程，而且可
以免25块？ 还是我没理解对？ 谢谢 |
|
|
S*A
发帖数: 7142 | 37 The port of 22 is both source port and the dest port.
It depend on which way the packet is going. If the packet
send from your openssh, it is source 22.
In the router you UI want to config the dest port.
it is the first packet that come from external IP with dest port
as 22 to connect to your ssh server. After that, the
router will remember the connection and deal with the
packet on the other direction as well.
The source port in that UI is the port you want to use
on the router. You can set it ... 阅读全帖 |
|
t******c
发帖数: 348 | 38 simple implementation of strcpy is like this:
char *strcpy(char *dest, const char *src)
{
const char *p;
char *q;
for(p = src, q = dest; *p != '\0'; p++, q++)
*q = *p;
*q = '\0';
return dest;
} |
|
w******a
发帖数: 27 | 39 来自主题: Programming版 - 一道笔试题 比如strcpy的原型如下:
char *strcpy(char *dest,char *src);
这里可以用*dest返回值,也可以用函数的返回值把真正的返回值给return出来.
问题是：为什么用dest返回了,这里还用return返回. |
|
b***i
发帖数: 3043 | 40 在VS2015花40多秒，如果wordList有2455个4个字母构成的字符串。但是在leetcode上
只要半秒。为啥呢？
size_t distance(string a, string b) {
size_t c = 0;
for (size_t i = 0;i
if (a[i] != b[i])
if (c++==1)
return c;
return c;
}
size_t size = wordList.size();
size_t from = SIZE_MAX;
size_t dest = SIZE_MAX;
vector> closest; // stores the list of list of shortest
nodes
closest.reserve(size);
for (size_t i = 0;i
vector tmpv;
tmpv.r... 阅读全帖 |
|
c**e
发帖数: 2558 | 41 Assume folder structure below:
C:.
├─source
└─dest
where 'source' contains the .jpg files and destination folder is 'dest'.
C:\> cd source
C:\source> xcopy /S *.jpg ..\dest
will do the job. |
|
w****f
发帖数: 1420 | 42 车取回来了。一切顺利。
MSRP: 46,305 (include dest)
invoice: 43443 (include dest)
Price paid:
41346(EP) + tax (7% of EP instead of invoice) + 125 (doc/reg) - 350 EP bonus
= 44015 (OTD)
Dealer fulfilled gas (~75USD)
No trade.
Did not buy any accessories/warranty/service at the dealer site.(these are
overpriced, at least today)
天亮了上图 |
|
a*****3
发帖数: 10373 | 43 MSRP (dest.除外) - 6% + dest.是个好deal吗?
另外有必要上3.0+S Line吗?我比较喜欢S-Line的外形而已,但比2.0 Premium Plus贵不少.
Thanks. |
|
c*******e
发帖数: 70 | 44 dest fee是厂家收的，名目就是运费，这块的成本dealer没得控制，值不值那个价也得
交，然后一分不少的转嫁给消费者，所以砍价时候你不能说你dealer运一辆车最多也就
花500, dest fee你给我算500好了。
至于厂家怎么算出八九百来运一辆车我就不知道了，估计里面包含了保险等隐含的费用
，没准还有利润。 |
|
d******y
发帖数: 11545 | 45 brian125 Posts: 1,440
December 14 edited December 14
buyhonda
Your numbers once again are wrong. OTD pricing vary's from state to state
due to state taxes , dmv fee's, and doc fee charges.. How you gauge a good
deal is buy knowing the Selling sale price with dest/charge included and
what that dealers doc fee charges are ? Again taxes and tags vary from state
to state and would only be of interest to someone in your buying state if
you provide them with the correct tax and doc fee numbers etc.
LX... 阅读全帖 |
|
N****B
发帖数: 152 | 46 太羡慕你们了， 我拿到的怎么这么高？
2014 Honda CRV Lx Awd
$21295
830 Dest
1548.75 Tax
299 Doc
296 Dmv
7.50 Tire Tax
8.50 online reg
$24284.75 OTD
2014 Honda CRV Ex-l Awd
$25945
830 Dest
1874.25 Tax
299 Doc
296 Dmv
7.50 Tire Tax
8.50 online reg
$29260.25 Total |
|
N****B
发帖数: 152 | 47 2014 Honda CRV Lx Awd
$21295
830 Dest
1548.75 Tax
299 Doc
296 Dmv
7.50 Tire Tax
8.50 online reg
$24284.75 OTD
2014 Honda CRV Ex-l Awd
$25945
830 Dest
1874.25 Tax
299 Doc
296 Dmv
7.50 Tire Tax
8.50 online reg
$29260.25 Total
我拿到的怎么这么高？很羡慕版上大神们的报价，怎么砍？谢谢 |
|
|
|
E**********1
发帖数: 73 | 50 在upstate NY附近，手头比较紧所以准备买base model，找来找去base model挺少的，
都是premium或者limited。难道base model这么小众？
2015 Forester 2.5i， 没有任何option，没有alloy wheel package或者all weather
package，我拿到的报价是$22,450，before tax and title。MSRP＋Dest＝$24,045，
invoice before Dest=$21,904。应该再往下砍多少为好?
多谢！ |
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