h*******g
发帖数: 508 | 1 新手 29yrd 刚开403b账户
目标 稳健 or 中等risk
研究了一下选了下面7个
请大家指点下；
fidelity
三个指数fund，其中2，3其实是一样的 2股票更广更多；
1 FID BLUE CHIP GR (FBGRX) ER0.76% 比例是10%
2 SPTN TOT MKT IDX ADV (FSTVX） ER0.07% 20%
3 SPTN 500 INDEX ADV (FUSVX) ER0.07% 20%
下面是sector 4，5都是IT 不过4更喜欢； 6，7是零售和real estate
4 FID SEL TECHNOLOGY (FSPTX) ER0.80% 20%
5 FID SEL SOFTWARE (FSCSX) ER0.79% 10%
6 SPTN REAL ES IDX ADV (FSRVX)ER0.09% 10%
7 FID SEL RETAILING (FSRPX) ER0.83% 10%
没有选bond（有fund推荐么？stock/bond 什么比例合适？对于moderate ... 阅读全帖 |
|
q****4
发帖数: 226 | 2 请大家帮忙看看我的退休计划如何。打算5-6年一次调整。
州里的退休计划在fidelity，可选的基金很少（23只），除开target date Freedom
Fund（有11只），剩下的就只有：
1. FID DIVERSIFD INTL (FDIVX)
2. FID DIVIDEND GR (FDGFX)
3. FID LOW PRICED STK (FLPSX)
4. FID REAL ESTATE INVS (FRESX)
5. FID WORLDWIDE (FWWFX)
6. SPTN EXT MKT IDX ADV (FSEVX)
7. FID FREEDOM INCOME (FFFAX)
8. FID HIGH INCOME (SPHIX)
9. FID INFLAT PROT BOND (FINPX)
10. SPTN US BOND IDX ADV (FSITX)
11. FID RET GOVT MM (FGMXX)
12. MIP CL 1
我计划以下百分比（然后每5-6年看情况调整）：
6 - 50%
4 - 15%
8 - 20%
target Freedo... 阅读全帖 |
|
u******e
发帖数: 65 | 3 Sugget you to hold
Vanguard Total Bond Market Idx InstlPls
Vanguard Total Intl Stock Idx InstlPls
Stock fund often raises with DJIA, bond fund against DJIA.
Rebance them every 3-6 months. |
|
w******1
发帖数: 154 | 4 最近换了一些钱，从2045 retirement 到VANG 500 INDEX ADM， 可是REALIZED G/L是
负数，请问这是什么意思？是投资到2045 retirement 里的钱损失了吗？那从什么时候
开始算起呢？是年初（1月1号）吗？
我还想以后把INTL INDEX和 EXT MKT IDX 里的转到2045 retirement里，可是INTL
INDEX和 EXT MKT IDX 里的钱比刚投入的时候少了，是不是应该等到钱数跟刚投入的时
候一样再转？这样realized就不是负数了？ |
|
s******d
发帖数: 323 | 5 Van anybody help to see if I need to change my allocations. This year's
return so far is only 1.48%. Thanks!
Investment option
Allocation %
Stable Value
Minnesota Life General Account
Money Market
Advantus Money Market
General Bond
BlackRock Strategic Income Opps Instl
Columbia Limited Duration Credit R4
Templeton Global Bond A_________20%
Vanguard Interm-Term Bond Index Adm_________5%
Vanguard Long-Term Investment-Grade Adm
US Stock
Advantus S&P 500/Citigroup Growth
Advantus S... 阅读全帖 |
|
D*****l
发帖数: 357 | 6 麻烦大家指点下401K的选择。。。应该怎么组合呢？本人40岁。谢谢
FID CONTRAFUND K (FCNKX)
05/17/1967
Stock Investments Large Cap 10.40% 17.55% 17.24% 9.95%
HTFD DIV & GRTH R5 (HDGTX)
07/22/1996
Stock Investments Large Cap 4.26% 16.05% 15.61% 8.37%
SPTN 500 INDEX INST (FXSIX)
02/17/1988
Stock Investments Large Cap 7.40% 17.28% 17.31% 7.87%
JANUS ENTERPRISE N (JDMNX)
09/01/1992
Stock Investments Mid-Cap 13.29% 19.29% 18.77% 10.96%
JH DSCPL VAL MDCP R6 (JVMRX)
06/02/1997
Stock Investments... 阅读全帖 |
|
N*****5
发帖数: 35 | 7 现在选的FID FREEDOM K 2050 (FFKHX), 有必要改吗?
50% stock + 50% bond 的组合怎么样? 求推荐, 谢谢!
SPTN EXT MKT IDX ADV (FSEVX) 50%
VANG TOT BD MKT INST (VBTIX) 50%
Fund list如下:
Name/Inception Date Asset Class Category 1 Year 3 Year 5 Year 10 Year/
LOF* Returns As Of Bench- mark
FID GROWTH CO POOL
12/13/2013 Stock Investments Large Cap Growth 21.80% N/A N/A 18.79%
07/31/2015 Show
Benchmark: Russell 3000 Growth 16.37% 18.94% 17.76% 15.46 07/31/2015
TRP BLUE CHIP GR T5
09/30/2009 Stock Investmen... 阅读全帖 |
|
m********u
发帖数: 63 | 8 有如下选择，谢谢！ ER ratio都可够贵的。现在选的是TRP RETIREMENT 2050，ER
ratio = 0.75%。不知道有没有更合适的组合。
Name/Inception Date Asset Class Category Gross Expense Ratio**
Shareholder Fees
AF FUNDMNTL INV R6 (RFNGX)
08/01/1978
Stock Investments Large Cap 0.31% No additional fees apply.
ALZGI NFJ DIV VAL I (NFJEX)
05/08/2000
Stock Investments Large Cap 0.71% No additional fees apply.
MAINSTAY LG CAP GR I (MLAIX)
07/03/1995
Stock Investments Large Cap 0.74% No additional fees apply.
VANG... 阅读全帖 |
|
k***e
发帖数: 556 | 9 I just doubt whether step 4 can be done in nlogn
it is true that the key are sorted. however, the idx corresponds to the key
has no order. Thus you may need to take linear time to find the
corresponding min idx.
I thought I get the nlogn and later found it is wrong ... |
|
l**n
发帖数: 88 | 10 I am a new learner of C++, can you elaborate a little bit? Thanks a lot!
why "pClasses[idx] = new MyClass(int);" works?
what is idx? |
|
c***2
发帖数: 838 | 11 how about doing this for question#5:
design a function to hash each line to an integer between 0 to 1G
create a bit map of 1G
read a line and hash it to an idx, set bit idx to 1
read next line , hash it, if that bit is 1, discard
... |
|
s*******f
发帖数: 1114 | 12 int GetNextPrime(){ // substitute this with more excellent algorithm
static int primes[] = {2,3,5,7,11};
static int idx = 0;
return primes[idx++];
}
void PrintFactors(int n){
if (n < 0){
n = -n;
}
if (n == 0){
return;
}
int prime = 1;
while (prime <= sqrt(double(n))){
prime = GetNextPrime();
while (n % prime == 0){
printf("%d ", prime);
n /= prime;
}
}
if (n > 1){
printf("%d ", n);... 阅读全帖 |
|
f*******t
发帖数: 7549 | 13 平衡括号的题可以用贪心法做吧
#include
#include
#include
#include
#define INVALIDCHAR -1
using namespace std;
char *balance(char *str)
{
int len = strlen(str);
if(len < 1)
return NULL;
char *buff = (char*)calloc(len + 1, 1);
stack left;
for(int i = 0; i < len; i++) {
if(str[i] == '(')
left.push(i);
else if(str[i] == ')') {
if(left.empty()) {
buff[i] = INVALIDCHAR;
co... 阅读全帖 |
|
s*******e
发帖数: 35 | 14 用HASHMAP 找出每个WORD 所在的位置 occurrence map，在做一个反向的由所在位置(
position)到WORD 的reverse occurrence map。 吧这些position 排序，然后递归搜索。
package smartnose;
import java.util.Arrays;
import java.util.*;
public class SubWordSeq {
public static void main(String [] args){
List L = new LinkedList();
L.add("fooo");L.add("barr");L.add("wing");L.add("ding");L.add("wing"
);
String S= "lingmindraboofooowingdingbarrwingmonkeypoundcake";
HashMap<... 阅读全帖 |
|
s*******e
发帖数: 35 | 15 用HASHMAP 找出每个WORD 所在的位置 occurrence map，在做一个反向的由所在位置(
position)到WORD 的reverse occurrence map。 吧这些position 排序，然后递归搜索。
package smartnose;
import java.util.Arrays;
import java.util.*;
public class SubWordSeq {
public static void main(String [] args){
List L = new LinkedList();
L.add("fooo");L.add("barr");L.add("wing");L.add("ding");L.add("wing"
);
String S= "lingmindraboofooowingdingbarrwingmonkeypoundcake";
HashMap<... 阅读全帖 |
|
n*s
发帖数: 752 | 16 4. string transpose
import sys
def transpose(input,i):
mystr = list(input)
mystr[i],mystr[i+1] = mystr[i+1],mystr[i]
return ''.join(mystr)
def str_transpose():
print 'input two strings, separated by blank:'
a, b = sys.stdin.readline().split()
size = len(a)
if size != len(b) or sorted(a) != sorted(b):
return 'no way!'
next = [b]
parent = {b:None}
idx = size -1
notFound = True
while notFound:
newstr = []
for x in next:
... 阅读全帖 |
|
n*s
发帖数: 752 | 17 4. string transpose
import sys
def transpose(input,i):
mystr = list(input)
mystr[i],mystr[i+1] = mystr[i+1],mystr[i]
return ''.join(mystr)
def str_transpose():
print 'input two strings, separated by blank:'
a, b = sys.stdin.readline().split()
size = len(a)
if size != len(b) or sorted(a) != sorted(b):
return 'no way!'
next = [b]
parent = {b:None}
idx = size -1
notFound = True
while notFound:
newstr = []
for x in next:
... 阅读全帖 |
|
D********g
发帖数: 650 | 18 /**
* In an exam you have n problems to work on. i=0, 1, ..., n-1.
The i-th problem takes you x[i] minutes, and give you a score s[i].
You have 60 minutes to work on them. How can you get the best score?
*/
static class DPElement {
int value;
int prevCostIdx; // 0-based cost idx
int takenThisElement; // 0/1, indicating whether taking current
element, idx 0 based.
public DPElement() {
value = 0;
prevCostIdx = 0;
tak... 阅读全帖 |
|
f*******t
发帖数: 7549 | 19 你有except[],从{1,...,n}里去掉它可以得出include[]，复杂度O(n)
然后
idx = rand(include.size())
return include[idx] |
|
f*******t
发帖数: 7549 | 20 你有except[],从{1,...,n}里去掉它可以得出include[]，复杂度O(n)
然后
idx = rand(include.size())
return include[idx] |
|
r*******e
发帖数: 7583 | 21 你这个isValid明显做了很多重复比较
只需要检查row[idx] 跟 row[1..idx-1]是否冲突，一层循环就够了
-i) |
|
f********d
发帖数: 51 | 22 private int getAccessCode(int idx, PlaylistFlagParameter parameter) {
Field field = PlaylistFlagParameter.class.getDeclaredMethod("Access" + idx
);
if (field.getBoolean(parameter)) {
return 1<
}
return 0;
}
private static byte getHopperAccessCode(PlaylistFlagsParametersWrapper
wrapper) {
int bitflag = 0;
for (int i = 0; i < 8; i++) {
bitflag += getAccessCode(i, wrapper);
}
return (byte)bitflag;
} |
|
H*****l
发帖数: 1257 | 23 怎么在O(n^2)时间内既求和又能对结果排序，还可以从求和后的idx恢复原来两个数的
idx呢……？ |
|
a******e
发帖数: 710 | 24 我觉得30分钟做出这两道题很不简单啊。
不过这个循环的判断应该是小于等于号吧？
for (int i = 1; i <= str.size(); ++i)
nPermute *= i;
string correspPermute(string& str, int index)
{
set table;
for (auto& c : str)
table.insert(c);
int nPermute = 1;
for (int i = 1; i < str.size(); ++i)
nPermute *= i;
string res;
--index;
for (int m = index; m >= 0; --m) // 这里不应该是index应该是str.size()
{
nPermute /= m;
int idx = index / nPermute;
auto iter = table.... 阅读全帖 |
|
b******7
发帖数: 92 | 25 这题把gas[i] - cost[i]之后就变成找环形数组的最大子数组和问题。找出最大子数组
和的起始位置idx，若该问题有解，则idx一定是一个解 |
|
p*u
发帖数: 136 | 26 分享一个得到positive feedback的解题思路，代码就不贴了，因为邮件里面说rocket
fuel的engineer会code review
题目网上有，这个链接里面的第3题：http://get-that-job-at-google.blogspot.jp/2013/02/rocketfuel-codesprint-at-iit-bombay.html
首先题目里面给了一个提示：把所有的输入，按照start time均分为K个区间去处理
我的思路用sample input解释一下：
5
2 100 200
3 110 190
4 105 145
1 90 150
5 102 198
1，所有的开始时间和结束时间没有重复的，对输入得到三元组
start|end>，然后根据time排序得到：
idx time r_id type value/ret
--------------------------------------------
0 90 1 start 0 +1
1 100 2 st... 阅读全帖 |
|
p*u
发帖数: 136 | 27 分享一个得到positive feedback的解题思路，代码就不贴了，因为邮件里面说rocket
fuel的engineer会code review
题目网上有，这个链接里面的第3题：http://get-that-job-at-google.blogspot.jp/2013/02/rocketfuel-codesprint-at-iit-bombay.html
首先题目里面给了一个提示：把所有的输入，按照start time均分为K个区间去处理
我的思路用sample input解释一下：
5
2 100 200
3 110 190
4 105 145
1 90 150
5 102 198
1，所有的开始时间和结束时间没有重复的，对输入得到三元组
start|end>，然后根据time排序得到：
idx time r_id type value/ret
--------------------------------------------
0 90 1 start 0 +1
1 100 2 st... 阅读全帖 |
|
r*c
发帖数: 167 | 28 来自主题: JobHunting版 - 上一道小题 The ardendertat.com's code is not always correct. For an example, it will
return 121121121 as the next palin for input 120798387, where there exists a
smaller one: 120808021 .
The following code has no such shortcoming. The key idea is greedy +
heuristic.
#include
#include
#include
#include
#include
using namespace std;
class NextPalinSolution
{
public:
vector NextPalin(const vectorssrc){
vectorsrc(ssrc);
int ... 阅读全帖 |
|
b******i
发帖数: 914 | 29 这题你自己也说了，x, y两个不同的二叉树，node里面存range，
其实就是x, y是两个不同的线段树，要实现线段树的插入和查找的操作。
要注意的地方就是id，比如每个线段树可以用一个递增的id，每次插入增加nodes的时
候就分配新的id。最后返回区域的id就是这两个id, idx, idy综合起来的id。比如如果
各自的id都是int，可以返回一个long id = ( (long)idx << 32) | (long)idy; |
|
k****r
发帖数: 807 | 30 recursive，
用list《List《Integer》》 res纪录结果，
用一个mostleft记录目前最左边的index，
如果目前level小于mostleft，or超出res大小，插最前面，或最后面新的list：
static int mostLeft = 0;
public static void printTreeInVerticalOrder(TreeNode root) {
List> res = new ArrayList<>();
//res.add(new ArrayList());
printHelper(root, 0, res);
printTree(res);
}
public static void printHelper(TreeNode root, int curr, List
Integer>> res) {
if (root == null) return;
els... 阅读全帖 |
|
k****r
发帖数: 807 | 31 recursive，
用list《List《Integer》》 res纪录结果，
用一个mostleft记录目前最左边的index，
如果目前level小于mostleft，or超出res大小，插最前面，或最后面新的list：
static int mostLeft = 0;
public static void printTreeInVerticalOrder(TreeNode root) {
List> res = new ArrayList<>();
//res.add(new ArrayList());
printHelper(root, 0, res);
printTree(res);
}
public static void printHelper(TreeNode root, int curr, List
Integer>> res) {
if (root == null) return;
els... 阅读全帖 |
|
f*******t
发帖数: 7549 | 32 暴力做法蛮简单的吧，所有矩形面积之和减掉矩形相交面积之和
func CalcTotalArea(rectangles []Rectangle) int {
area := 0
for r, idx := range rectangles {
area += (r.x2 - r.x1) * (r.y2 - r.y1)
for i := idx + 1; i < len(rectangles); i++ {
r2 := rectangles[i]
x1 := max(r.x1, r2.x1)
y1 := max(r.y1, r2.y1)
x2 := min(r.x2, r2.x2)
y2 := min(r.y2, r2.y2)
if x1 < x2 && y1 < y2 {
area -= (x2 - x1) * (y2 - y1)
}
}
}
return area
} |
|
d****0
发帖数: 13 | 33 06/15电面，ABC+白哥，一开始互相介绍然后讲一下印象深刻的一个project
Problem 1:
问烂的binary tree level print
Problem 2:
interface Sequence{
int size();
Object get(int idx);
void add(Object o);
boolean remove(int idx);
}
写个class implement这个interface
----------------------------------------------------
今天收到recruiter邮件如下，说虽然是positive feedback，但是：
However, I’ve just been informed that we can no longer move forward with
candidates who require a new H1-B visa. We have reached the number of
headcount approved for f... 阅读全帖 |
|
g******y
发帖数: 143 | 34 Given a decision tree, and lower & upper bounds for each dimension of a d-
dimension space. Find the maximum volume of the d-dimension space and its
lower and upper bounds. 大家有没有优于O(n)的解法？
这题有点绕. 举个两维的例子:
initial range for d0 is (0,10), d1 is (0,10)
Decision tree
idx=1,split=4
/ \
/ \
null idx=0,split=2
d1
10 | |
| |
| |
4 |___|_______________
|
|
|___________________
0 2 10 ... 阅读全帖 |
|
w***u
发帖数: 156 | 35 第一题是不是可以这样：
先sort 一下原来的序列，O(nlogn)
求得 average 数字O(n),
binary search average， 如果直接能在序列中找到这个数，那序列就能直接分成一边
是一个，另外一边就是剩下的数
如果找不到，找到average可以插入的idx，这样把序列分成两半，一边小于average，
另外一边大于average。
大家都减去average， 这样一边就是正数，一边是负数，
问题就变成在两个半边中，找想加等于0的组合。
对于每个半边，纪录sum和idx的对应关系。
这样只要两个sum互为正负对，就能找到partition了 |
|
发帖数: 1 | 36 多个sum为0怎么办？
：第一题是不是可以这样：
：先sort 一下原来的序列，O(nlogn)
：求得 average 数字O(n),
：binary search average， 如果直接能在序列中找到这个数，那序列就能直接分成一
边是一个，另外一边就是剩下的数
：如果找不到，找到average可以插入的idx，这样把序列分成两半，一边小于average，
：另外一边大于average。
：大家都减去average， 这样一边就是正数，一边是负数，
：问题就变成在两个半边中，找想加等于0的组合。
：对于每个半边，纪录sum和idx的对应关系。
：这样只要两个sum互为正负对，就能找到partition了 |
|
发帖数: 1 | 37 多个sum为0怎么办？
：第一题是不是可以这样：
：先sort 一下原来的序列，O(nlogn)
：求得 average 数字O(n),
：binary search average， 如果直接能在序列中找到这个数，那序列就能直接分成一
边是一个，另外一边就是剩下的数
：如果找不到，找到average可以插入的idx，这样把序列分成两半，一边小于average，
：另外一边大于average。
：大家都减去average， 这样一边就是正数，一边是负数，
：问题就变成在两个半边中，找想加等于0的组合。
：对于每个半边，纪录sum和idx的对应关系。
：这样只要两个sum互为正负对，就能找到partition了 |
|
w*****3
发帖数: 18 | 38 Ownership
First Trust Dow Jones Internet ETF 4.82 1.16
4,134,811
277,426
7.19%
Mar 06, 2018
Vanguard Total Stock Mkt Idx Not Rated 2.29 0.01
2,109,721
8,100
0.39%
Jan 31, 2018
Vanguard Extended Market Idx Inv 1.31 0.03
1,206,621
5,048
0.42%
Jan 31, 2018
Polar Capital Global Tech 1.27 0.99
1,171,143
5,400
0.46%
Jan 31, 2018
Sentinel Small Company A 1.16 1.22
1,063,000
0
0.00%
Dec 31, 2017
Total (for Top ... 阅读全帖 |
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|
n********n
发帖数: 2248 | 46 呵呵，看看这篇评论，天然气还是很有希望嘛
Today I’m going to discuss three main reasons why natural gas prices (in
the United States) haven’t participated in the commodity boom of 2010.
The first reason is that natural gas prices are loosely tied to oil prices.
Yes, oil prices rose over the past year, but only by about 10%. In order for
oil prices to significantly buoy natural gas prices, we’ll need to see
another huge ramp up like we did in the summer of 2008. Those higher oil
prices (above the $100 mark – as an es... 阅读全帖 |
|
|
|
c******n
发帖数: 4965 | 49 /* generate a bunch of "random" numbers
taken in serial one by one from a list of numbers in
a given file
*/
class RandomGen(
public RandomGen() throws IOException {
InputStream ios = ..... open "rand.txt" in CLASSPATH;
//
read the file into memory,
}
int idx;
public Integer gen() {
return list_of_numbers[idx++];
}
}
something like the above.
the problem is that the constructor throws IOException, so if I use this
class, and most likely I us... 阅读全帖 |
|
c******n
发帖数: 4965 | 50 if [ $# != 1 ]; then
echo "Must be ape2mp3.sh ape_base_name !"
exit
fi
CPE=$1.ape
CUE=${CPE%.*}.cue
mac "$CPE" temp.wav -d
bchunk -w temp.wav "$CUE" tempwavsegments
perl -e '
open FH , "'"$CUE"'";
chomp ( @names = );
@names=map {s/"//g;s/TITLE *//;s/^ *//g; s/[^\w_\-\. ]//g; $_} grep { $_ =~
m/TITLE/} @names;
print "all titles:\n" , join "\n" ,@names,"\n";
foreach my $index ( 0 .. @names -1 ) {
my $idx = sprintf("%02d", $index);
my $cmd = qq( lame tempwavsegments${idx} |
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