m*********k
发帖数: 10521 | 1 [Stock] stk Apr 22 ● 请版主帮着发放108个包子
10*108*1.1=1188
1) 用户:stk 时间: 2010-04-24 07:51:22 金额:188.00
2) 用户:stk 时间: 2010-04-22 22:20:52 金额:1000.00
成功奖励 10 伪币的用户: Dec2008, adots, AGilardino, aizaza, alanguo88, alie, Armageddonar, badcompany, biot, bluejay11, Bridgewatre, Buffett, caishen, cappucino, carbonplay, ChenHil, chhguo, choosewhat, cjiayun, connect4, dall1997, dapply, dcw, dongxie, DQatBR, drik, dzry67, ElliottJr, elmtang, feelalone, feizaopaodao, flemns, freyjaa, |
|
E*******r
发帖数: 2723 | 2 【 以下文字转载自 WBCenter 讨论区 】
发信人: ElliottJr (Froglet), 信区: WBCenter
标 题: 代stk申请发108个包子
发信站: BBS 未名空间站 (Thu Apr 22 23:07:36 2010, 美东)
注：1）版面活动：手续费无；2）代发包子：手续费10%:
2）
代发版面/ID:
stock (包子已经转到版面）
代发事由(主题标题或链接):
stk撒包子给股版兄弟姐妹，下面跟帖前108个各发一个包子不重复。
http://www.mitbbs.com/article_t/Stock/32735949.html
伪币金额:
108x10x110%=1188伪币, 从Stock版面扣。
奖励/代发包子数量: |
|
f****t
发帖数: 1063 | 3 为什么好多MFs里的boa的成分股，都变成了boa pfd stk? 而不是一般的common stocks?
谢谢。 |
|
f****t
发帖数: 1063 | 4 thanks. the issue is that they were common stocks before, not the pfd stk. I
only heard of coverting from pdf to com, not the other way around. So now
it is ok to do com to pfd?
so confused... |
|
f****t
发帖数: 1063 | 5 so common stocks have the freedom to change to pfd stk?
thanks.
preferred
,
think |
|
|
N***i
发帖数: 2063 | 7 【 以下文字转载自 bluechips 俱乐部 】
发信人: liliwater (lyrist), 信区: bluechips
标 题: Netflix Announces New $300 Million Stk Buyback
发信站: BBS 未名空间站 (Sat Jun 12 00:17:55 2010, 美东)
Netflix (NFLX) today said its board has authorized a new plan to repurchase
up to $300 million of its common stock through the end of 2012. This is in
addition to the company’s last authorization in August 2009.
NFLX is up $1.64, or 1.4%, to $120.30. |
|
u*********r
发帖数: 1181 | 8 上面部分是费用，下面部分是各自相应的average annual total return。
非常感谢啊
Gross Expense Ratio：
=======================
TIER I - LIFECYCLE FUNDS
- show details. - hide details.
LIFEPATH 2015 07/05/2006 Blended Fund Investments* N/A 0.1221
% No additional fees apply.
LIFEPATH 2020 06/30/2006 Blended Fund Investments* N/A 0.1092
% No additional fees apply.
LIFEPATH 2025 07/05/2006 Blended Fund Investments* N/A 0.1237
% No additional fees apply.
LIFEPATH 2030 06/3... 阅读全帖 |
|
z*********n
发帖数: 1451 | 9
趁快下班偷偷懒码出来了，是我前面说的简单情况思路的延伸。
直觉告诉我可能有O(n)时间O(1)空间的，但水平有限，只能做到O(n)时间(amortized)
加O(n)空间（worst case)
pair findIJ2(vector const & arr)
{
if (arr.empty())
return make_pair(-1, -1);
int cur_i = 0, cur_j = 0, ans_i = 0, ans_j = 0, last_i = 0;
stack> stk;
for (int k = 1; k < arr.size(); ++ k)
if (arr[k] <= arr[cur_i]) /// use <= instead < here.
{
last_i = cur_i; /// loop to merge the ranges.
while (stk.size()... 阅读全帖 |
|
w****r
发帖数: 69 | 10 来个简单易懂的python版, n1是剩余左括号， n1_是剩余右括号
res = []
def gen(sol, stk, n1, n2, n3, n1_, n2_, n3_):
if n1 == 0 and n2 == 0 and n3 == 0 and n1_ == 0 and n2_ == 0 and n3_ ==
0:
res.append("".join(sol))
return
if n1 > 0:
gen(sol[:] + ["("], stk[:] + [1], n1-1, n2, n3, n1_, n2_, n3_)
if n2 > 0:
gen(sol[:] + ["["], stk[:] + [2], n1, n2-1, n3, n1_, n2_, n3_)
if n3 > 0:
gen(sol[:] + ["{"], stk[:] + [3], n1, n2, n3-1, n1_, n2_, n3_)
if len(stk) ... 阅读全帖 |
|
w****x
发帖数: 2483 | 11 int CalculateContain(int a[], int n)
{
assert(a && n > 0);
int nRet = 0;
stack stk;
for (int i = 0; i < n; i++)
{
if (!stk.empty() && a[stk.top()] < a[i])
{
int nPrev = stk.top();
stk.pop();
while (!stk.empty() && a[stk.top()] <= a[i])
{
int nCur = stk.top();
stk.pop();
int nLimit = a[nCur] <= a[i] ? a[nCur] : a[i];
nRet += (nLimit - a[nPrev]) *... 阅读全帖 |
|
g*********e
发帖数: 14401 | 12 void gen(string in) {
vector res;
stack stk;
if(in[0]=='1')
stk.push("1");
else if(in[0]=='0')
stk.push("0");
else {
stk.push("1"); stk.push("0");
}
while(!stk.empty()) {
string one=stk.top(); stk.pop();
int idx=one.length();
if(idx == in.length()) {
res.push_back(one);
cout<
} else if(in[idx]=='1') {
one+='1';
stk.push(one);
} els... 阅读全帖 |
|
h*****y
发帖数: 218 | 13 在网上发现这个最大直方图求面积的题目的解答，有一个地方没看明白
http://www.sureinterview.com/shwqst/1117001
下面这个代码的stack为什么有一个get的方法？
if (prev.height > stk.get(stk.size() - 2).height) {
======================================================================
public long getMaxArea(int[] histogram) {
long maxArea = 0;
if (histogram == null || histogram.length == 0)
return maxArea;
Stack- stk = new Stack
- ();
stk.push(new Item(Integer.MIN_VALUE, -1));
for (int i = 0; i <... 阅读全帖 |
|
g*******y
发帖数: 1930 | 14 我改了一下:
void findNext(Node *node, stack &stk)){
while(node->left || node->right){
stk.push(node);
node = (node->left)? node->left : node->right;
}
stk.push(node);
}
void post2(Node *root){
if(!root) return;
stack stk;
Node *node, *pre;
findNext(root,stk);
pre = stk.top(); visit(pre); stk.pop();
while(!stk.empty()){
node = stk.top();
if(node->right==0 || node->right == pre){
visit(node); stk.pop(); pre = |
|
Z*****Z
发帖数: 723 | 15 响应大侠号召。写了个直白板的，求拍。
static >Node next(Node tree, Node node) {
if(tree == null){
return null;
}
if(node.right != null){
Node tmp = node.right;
while(tmp.left != null){
tmp = tmp.left;
}
return tmp;
}
Stack> stk = new Stack>();
if(!searchNode(tree, node, stk)){
return null;
... 阅读全帖 |
|
g*******y
发帖数: 1930 | 16 #define MP(i,j) make_pair(i,j)
string s;
int n = s.size();
stack> stk;
void PopStack(){
stk.pop();
if(!stk.empty()){
swap(s[stk.top().first] , s[stk.top().second]);
stk.top().second++;
}
}
void permute(){
stk.push(MP(0,0));
pair x;
while(!stk.empty()){
if(stk.size()==n){
cout<
PopStack();
}else{
int |
|
d***8
发帖数: 1552 | 17 来自主题: JobHunting版 - 谷歌 电面 BinaryTree* findnextnode(BinaryTree *root, int node_val)
{
stack stk;
BinaryTree *cur = root;
//Find match node
while(cur != NULL)
{
stk.push(cur);
if(cur->value == node_val)
break;
else if (cur->value < node_val)
cur = cur->right;
else if(cur->value > node_val)
cur = cur->left;
}
if (cur == NULL)... 阅读全帖 |
|
p****o
发帖数: 46 | 18 第二题如果是, . ' '结束的情况, 用stack
否则的话，结尾还得把stack输出.
确实，一步小心就容易忽略
写个c++的代码
#include
#include
using namespace std;
string strProc(string input){
stack stk;
string result;
for(string::const_iterator it = input.begin(); it!=input.end(); ++it){
if (*it < '9' && *it > '0' && stk.empty()){
result+=*it;
}
else if (*it == ' ' || *it ==',' || *it =='.'){
while (!stk.empty()){
char c= stk.top();
... 阅读全帖 |
|
n*****x
发帖数: 686 | 19 维护一个stack，里面存index。还要记录当前见过的最大值。开始扫数组
比当前最大大就push，比top指的数字小就pop，不然就continue。扫一边数组。
然后开始pop stack，index间隔不是1的就是你要找到的，最后一个元素和-1比。top需
要是n-1。应该算On
vector solve(vector& nums){
vector res(2,-1);
int n = nums.size();
if (n==0) return res;
stack stk;
stk.push(-1);
int max_sofar = nums[0];
for (int i=0; i
while (stk.size()!=1 && nums[stk.top()]>nums[i]) stk.pop();
if (nums[i]>=max_sofar){
max_sofar = nums[i];
... 阅读全帖 |
|
u****d
发帖数: 23938 | 20 ☆─────────────────────────────────────☆
boshihou2 (捉蛇者) 于 (Fri Jul 15 17:06:14 2011, 美东) 提到:
只有coldface 和射天狼，注意到教授要回国。其他人都是凑热闹，求不懂，和凉粉
☆─────────────────────────────────────☆
tooearly (郎教授不炒股) 于 (Fri Jul 15 17:28:35 2011, 美东) 提到:
谢谢博后如此细心。感动。
☆─────────────────────────────────────☆
stlstl (射天狼) 于 (Fri Jul 15 17:41:30 2011, 美东) 提到:
你这是bso你不但细看了原帖，而且细看了回帖
☆─────────────────────────────────────☆
boshihou (过马路) 于 (Fri Jul 15 17:45:07 2011, 美东) 提到:
教授的写法有问题，
第一句就应该说：我要回国了。文章才能更有吸... 阅读全帖 |
|
w****x
发帖数: 2483 | 21 int longestValidParentheses(const char* str) {
assert(str);
int nMax = 0;
const char* p = str;
stack stk;
while (*p != 0)
{
if (*p == '(')
stk.push(p);
else if (*p == ')')
{
if (!stk.empty() && *stk.top() == '(')
{
stk.pop();
nMax = max(p - (stk.empty() ? str-1 : stk.top()), nMax);
}
else stk.push(p);
}
p++;
}
return nMax... 阅读全帖 |
|
w****x
发帖数: 2483 | 22 int longestValidParentheses(const char* str) {
assert(str);
int nMax = 0;
const char* p = str;
stack stk;
while (*p != 0)
{
if (*p == '(')
stk.push(p);
else if (*p == ')')
{
if (!stk.empty() && *stk.top() == '(')
{
stk.pop();
nMax = max(p - (stk.empty() ? str-1 : stk.top()), nMax);
}
else stk.push(p);
}
p++;
}
return nMax... 阅读全帖 |
|
w****x
发帖数: 2483 | 23 这题作的不下5遍了, 说实话, 第一次做还挺不容易的:
============带parent===========================
void InOrderPrint(NODE* pRoot)
{
assert(pRoot);
NODE* pCur = pRoot;
while (pCur != NULL)
{
while (pCur->pLft != NULL)
{
cout<nVal<
pCur = pCur->pLft;
}
cout<nVal<
//The ending condition is tricky but simple
while (pCur != NULL)//use "pCur != NULL" rather than "pCur->pParent
!= NULL"
... 阅读全帖 |
|
w****x
发帖数: 2483 | 24 int longestValidParentheses(string s) {
const char* str = s.c_str();
int nMax = 0;
const char* p = str;
stack stk;
while (*p != 0)
{
if (*p == '(')
stk.push(p);
else if (*p == ')')
{
if (!stk.empty() && *stk.top() == '(')
{
stk.pop();
nMax = max(p - (stk.empty() ? str-1 : stk.top()), nMax);
... 阅读全帖 |
|
w****x
发帖数: 2483 | 25 打印版本，真难做对：
int printMinCost(int a[], int n)
{
if (NULL == a || n <= 1)
return -1;
int recCost[100][100] = {0};
int recSplit[100][100] = {0};
for (int i = 2; i < n; i++)
{
for (int j = 0; j < n-i; j++)
{
int nMin = INT_MAX;
int nSplitIndex = 1;
for (int k = 1; k < i; k++)
{
int nCost = recCost[j][j+k] + recCost[j+k][j+i] + a[j+i] - a
[j];
if (nCost < nMin)
... 阅读全帖 |
|
w****x
发帖数: 2483 | 26 打印版本，真难做对：
int printMinCost(int a[], int n)
{
if (NULL == a || n <= 1)
return -1;
int recCost[100][100] = {0};
int recSplit[100][100] = {0};
for (int i = 2; i < n; i++)
{
for (int j = 0; j < n-i; j++)
{
int nMin = INT_MAX;
int nSplitIndex = 1;
for (int k = 1; k < i; k++)
{
int nCost = recCost[j][j+k] + recCost[j+k][j+i] + a[j+i] - a
[j];
if (nCost < nMin)
... 阅读全帖 |
|
w****x
发帖数: 2483 | 27 一直认为面试出这么难得题真是过分啊！！
================= kth element in young tablet (M+N)log(numeric range)
solution ===========
const int M = 4;
const int N = 4;
int getOrder(int A[M][N], int tg)
{
int c = 0;
int i = 0;
int j = N-1;
while (i < M && j >= 0)
{
if (A[i][j] >= tg)
j--;
else
{
c += j+1;
i++;
}
}
return c+1;
}
int getKthMin(int A[M][N], int k)
{
if (k <= 0 || k > M*N)
return INT_MIN;
int nBe... 阅读全帖 |
|
z****e
发帖数: 9 | 28 Java Code: Not tested yet
find(String s, int si, int partNo, Stack stk, List list)
{
int len = s.length();
if(si>=len)return;
if(partNo==4){
if(si
String ip = s.substring(si);
if(isValidIP(ip)){
stk.push(ip);
list.add(changeToIP(stk);
stk.pop();
}
}
}
else{
for(int i=1;i<4;i++){
if(si+1<=len){
... 阅读全帖 |
|
d******e
发帖数: 164 | 29 ")(()())())(((()))(()()()(()(()(())))(())()((()()(((()())()))(()()())())(())
(()(()()()()))(((()())))(((()))))()()())))(()))))())(((()"
小数据可以过，大数据停在上面这个case。
Run Status: Runtime Error
但是拷出来，自己运行是对的，没有错误。请大家帮忙看看。
class Solution {
public:
int longestValidParentheses(string s) {
int m = 0;
stack stk;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '(' || stk.empty() || s[stk.top()] == ')') {
stk.push(i);
} else {
... 阅读全帖 |
|
j********g
发帖数: 80 | 30 第二个过了Large的 轻拍
class Solution {
public:
vector> findLadders(string start, string end, unordered_
set &dict) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int size=dict.size();
int res=1;
int done=0;
stack stk[2];
unordered_set used;
unordered_map > path;
unordered_set level;
int index=0;
stk[index%2].push(start);
used.insert(start);
vector<... 阅读全帖 |
|
j********g
发帖数: 80 | 31 第二个过了Large的 轻拍
class Solution {
public:
vector> findLadders(string start, string end, unordered_
set &dict) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int size=dict.size();
int res=1;
int done=0;
stack stk[2];
unordered_set used;
unordered_map > path;
unordered_set level;
int index=0;
stk[index%2].push(start);
used.insert(start);
vector<... 阅读全帖 |
|
s****0
发帖数: 117 | 32 贴一个scala的。正在学习，书还没完全看完。初学乍练，多多指教
package myTest
import scala.collection.mutable.Map
import scala.collection.mutable.Stack
class Test3(lst: Array[Int], target: Int, all: Boolean) {
def solve() = {
for (inst <- genInstr(lst.length); op <- ops(lst.length - 1); l <- lst.
permutations) {
val rst = exec(inst, op, l);
if (rst == target) {
println(inst + ", " + op)
}
}
}
def genInstr(n: Int): IndexedSeq[String] = {
if (n == 1)
return Array("g");
for ... 阅读全帖 |
|
g*******y
发帖数: 1930 | 33 void FindLeftmostLeaf(Node *root, stack &stk){
while(root->left || root->right){ //while(当前节点不是叶子)
stk.push(root);
if (root->left){
root = root->left;
}else{
root = root->right;
}
}
stk.push(root);
}
void PostOrder(Node *root){
if(root) return; //check empty tree;
stack stk;
Node *node=root, *parent;
//找子树中最左边的叶子，并把途中的节点放入栈
FindLeftmostLeaf(root,stk);
while(!stk.empty()){
node |
|
J*********a
发帖数: 50 | 34 lz是不是面的时候紧张啊。。。
紧张的时候思路容易混乱，这题可以用stack加个list吧：
public static void printTreeNodeLeafToRoot(TreeNode root) {
if (root == null) return;
Deque stk = new LinkedList<>();
stk.push(root);
List list = new ArrayList<>();
list.add(root);
while (!list.isEmpty()) {
List curr = new ArrayList<>();
for (TreeNode tn : list) {
if (tn.right != null) {
stk.push(tn.right);
... 阅读全帖 |
|
v*****u
发帖数: 1796 | 35 int n=3;
char stk[100];
void printlegal( int ipos, int ileft, int iright)
{
if (ipos==2*n-1)
{
stk[ipos]=')';
for (int i=0; i
printf("%c", stk[i]);
printf("\n");
}else if (ileft==iright)
{
stk[ipos] = '(';
printlegal( ipos+1, ileft-1, iright );
}else if (ileft==0)
{
stk[ipos] = ')';
printlegal( ipos+1, i |
|
g*******y
发帖数: 1930 | 36 November 07
学习了backtrack(回溯法)
之前做了一些回溯的题，比如打印permutation，打印任意n对括号等等，都是瞎蒙的。
还真凑巧，上午做了打印n括号的题，下午就看见有人说到回溯法，想想自己还没系统
学过这个，找了本基础的中文算法书来看了看，虽然书上讲的很浅显，发现自己貌似瞎
蒙还蒙对了思路，呵呵。正好凑巧的是，刚刚看了一点点，网上就有个人问怎么做
Vertex Cover的问题，正好让我来做做练习。
1. 打印任意合法的n对括号：
void printParenthes(int N, int left, int right, stack &stk){
if(left == N && right == N){
printStack(stk);
return;
}
if(left>right){
stk.push(')');
printParenthes(N, left,right+1, stk);
stk.pop();
}
if |
|
x****8
发帖数: 127 | 37 stack?
public static void main(String[] args) {
XmlParser p = new XmlParser("ABDC");
String s;
Stack> stk = new Stack>();
TreeNode root = null;// = new TreeNode();
while((s = p.getNextTag()) != null){
if(p.isStartTag()){
String str = p.getNextTag();//assert is string tag
TreeNode node = new TreeNode(str);
... 阅读全帖 |
|
j******d
发帖数: 2 | 38 class ListNode {
public:
int val;
ListNode *next; // point to the next of self list
ListNode *down; // point to the head of nested list
};
class DeepIterator {
public:
DeepIterator(ListNode *head) : cur(head) {}
bool hasNext(){
return cur != NULL;
}
int next(){
int val = cur->val;
if (cur->down) {
stk.push(cur);
cur = cur->down;
} else {
while (cur->next == NULL && stk.empty() == false) {... 阅读全帖 |
|
j******d
发帖数: 2 | 39 class ListNode {
public:
int val;
ListNode *next; // point to the next of self list
ListNode *down; // point to the head of nested list
};
class DeepIterator {
public:
DeepIterator(ListNode *head) : cur(head) {}
bool hasNext(){
return cur != NULL;
}
int next(){
int val = cur->val;
if (cur->down) {
stk.push(cur);
cur = cur->down;
} else {
while (cur->next == NULL && stk.empty() == false) {... 阅读全帖 |
|
g*********e
发帖数: 14401 | 40 一个粗俗点的办法，就是用一个stack，stack里面每个元素存每个Iteration所有的变
量。
foo(argument(s) arg) {
argument a, b;
// compute a, b from arg
foo(a);
foo(b);
}
stack stk;
while(！stk.empty()) {
argument top=stk.top();
argument b= // compute b from top
argument a= // compute a from top
stk.push(b);
stk.push(a);
} |
|
g********r
发帖数: 89 | 41 Leetcode给的solution，在pop的时候，只要栈顶元素=minStack栈顶元素，就把
minStack的栈顶元素pop掉。如果有duplicates的话，比如stack里面5, 3, 3，
minStack里面是5, 3，在pop第一个3的时候，不应该把minStack里面的3 pop掉吧。否
则Stack里面变成了5, 3, minStack里面是5, 答案是不是有问题？
---- Leetcode solution
public void pop() {
if (stack.pop().equals(minStack.peek())) minStack.pop();
}
-----my solution
void pop() {
if(stk.empty()) return;
int tmp=stk.top();
stk.pop();
if(stk.empty() || (tmp == min_stk.top() && stk.top() > min_stk.top... 阅读全帖 |
|
g********r
发帖数: 89 | 42 void pop() {
if(stk.empty()) return;
int tmp=stk.top();
stk.pop();
if(stk.empty() || (tmp == min_stk.top() && stk.top() > min_stk.top()
))
{
min_stk.pop();
}
} |
|
b******i
发帖数: 914 | 43 大体思路就是记录下所有当前是反括号但是没有正括号pop出来的，和最后剩下的正括号
class Solution {
public:
string balance(string s) {
int n = s.length();
if(n == 0)
return 0;
vector stk;
vector to_remove(n, false);
for(int i = 0; i < n; i++) {
char c = s[i];
if(c == '(') {
stk.push_back(i);
} else {
if(stk.empty())
to_remove[i] = true;
else {
stk.p... 阅读全帖 |
|
w*****h
发帖数: 423 | 44 有人不愿意进去看的话，我贴这
private static List findMaxIsLand(int[][] m)
{
List poslist = new ArrayList();
if(m.length == 0 || m[0].length == 0)return poslist;
int x = m.length, y = m[0].length;
boolean[][] visited = new boolean[x][y];
for(int i = 0; i < x; i++)
{
for(int j = 0; j < y; j++)
{
if(m[i][j] == 0 || visited[i][j])continue;
Stack stk = new Stack阅读全帖 |
|
s******y
发帖数: 200 | 45 能不能也帮我看看？不是很懂每个选了一两个，去年收益-10%， 今年10%左右，正想重
新rebalance,怎么选择呢？
Name Expense ratio
Blended Fund Investments:
VANGUARD TARGET 2055 0.083%
VANGUARD TARGET 2050 0.083%
VANGUARD TARGET 2045 0.083%
VANGUARD TARGET 2040 0.083%
VANGUARD TARGET 2035 0.083%
VANGUARD TARGET 2030 0.083%
VANGUARD TARGET 2025 0.083%
VANGUARD TARGET 2020 0.083%
VANGUARD TARGET 2015 0.083%
VANGUARD TARGET 2010 0.083%
VANGUARD TARGET INC 0.083%
Index Fund Stock Investments:
VANG EXT MKT IDX INS 0.12%
VANG INST INDEX P... 阅读全帖 |
|
i**********t
发帖数: 81 | 46 在fidelity有一个retirement plan account， 请问下面这么多种funds， 该怎么配置
portfolio？ 头都大了。请懂行的同学指点，多谢了。
Blended Investments -- FID ASSET MGR 20% %
Blended Investments -- FID FREEDOM K 2000 %
Blended Investments -- FID FREEDOM K 2005 %
Blended Investments -- FID FREEDOM K 2010 %
Blended Investments -- FID FREEDOM K 2015 %
Blended Investments -- FID FREEDOM K 2020 %
Blended Investments -- FID FREEDOM K 2025 ... 阅读全帖 |
|
J**S
发帖数: 25790 | 47 8分钟内完成一进一出。
Stock Order: SINA-SINA CORPORATION (CAY) COM STK USD0.133
Status Filled at $61.60
Symbol SINA
Description SINA CORPORATION (CAY) COM STK USD0.133
Action Buy
Quantity 600 Shares
Route FDLM
Order Type Limit at $61.60
Time in Force Day
Conditions None
Trade Type Margin
Market Session Standard
Order Date 11/26/2010, 10:50:00 AM ET
Top
Stock Order: SINA-SINA CORPORATION (CAY) COM STK USD0.133
Status Filled at $62.00
Symbol SINA
Description SINA CORPORATION (CAY) COM STK USD0.133
Action ... 阅读全帖 |
|
u****d
发帖数: 23938 | 48 ☆─────────────────────────────────────☆
zijing (紫晶) 于 (Wed Jul 20 12:55:25 2011, 美东) 提到:
QQQ离十年高点很近了，可是看看这两年曾经的那些明星，AKAM，VECO，RVBD，BRCM等
等，从高点跌得多惨，象AAPL这种长青树毕竟是少之又少，所以买个股风险远大于买指
数
☆─────────────────────────────────────☆
whyabk (邓稼先郭永怀) 于 (Wed Jul 20 13:02:26 2011, 美东) 提到:
zijing妹啊，搞系统很累的，累的我都有偏头疼的毛病了。
我劝你早点成个家，慢慢来。
☆─────────────────────────────────────☆
zijing (紫晶) 于 (Wed Jul 20 13:07:30 2011, 美东) 提到:
累吗？我很ENJOY啊，赚钱不是唯一目的，重在参与。炒股票和成家有什么矛盾的地方
吗？
☆─────────────────────────────────... 阅读全帖 |
|
