|
|
|
|
|
|
|
|
|
i*****n
发帖数: 514 | 10 pls send your brief intro of your research background.
Abstract:
Complexes of bis-N,N’-pyridylquinolyl-Schiff base iron(II) iodide, [Fe(pmaq
)2]I2, [Fe(paaq)2]I2 and [Fe(ppaq)2]I2, were synthesized and characterized
by EA, IR, UV-Vis spectra, TGA and conductivity measurements. |
|
J*H
发帖数: 1448 | 11 1) i2, you are not welcome here. So do not come in or stay but do get out.
2) i2, if you can't control yourself, go find something else to play with.
Just don't come here. You are not worthy.
3) I don't dislike mainlanders, only the i2 type; I have plenty good friends
who are from the mainland.
Peace! |
|
m**k
发帖数: 4039 | 12 sigh, 我是用比较极端的例子来说明本金还的少不代表这个loan不划算.
不是在拿现在的情况进行对比.
你要拿真正例子来算也是一样的道理, 拿P1(n) 代表现在的monthly payment里的
principal, I1(n)代表interests, P2(n), I2(n)代表refi的. P1(n) > P2(n), I1(n)>
I2(n), P1(n)+I1(n) > P2(n)+I2(n)
我上面的理论依然成立. 想把问题简单化, 你们还非要考虑最复杂的情况. |
|
w********e
发帖数: 7186 | 13 【 以下文字转载自 LosAngeles 讨论区 】
发信人: joyzhao (joyzhao), 信区: LosAngeles
标 题: T420s $955 起；X220 $555 起 （i5 + IPS $631） (转载)
发信站: BBS 未名空间站 (Thu Jun 16 12:06:21 2011, 美东)
发信人: joyzhao (joyzhao), 信区: shopping
标 题: T420s $955 起；X220 $555 起 （i5 + IPS $631）
发信站: BBS 未名空间站 (Mon Jun 13 16:45:02 2011, 美东)
eCoupon USX10XT420S0613 through June 22.
http://shoplenovo.i2.com/SEUILibrary/controller/e/barnesnoblego
（链接如果不work就刷新一下，如果跟spp一样就清空cookies再来）
①Sign in first, or create an account and login
http://shoplenov... 阅读全帖 |
|
d******8
发帖数: 2191 | 14 方法是用setInterval函数
var i1=setInterval( function(){ clearInterval(i1);yourfunction();},10);
var i2=setInterval(function(){ clearInterval(i2);yourfunction();},10);
这样i1,i2对应的任务就是并行的。
由此类推，
var i_array=new Array();
function Parallel(func,datas){
$(datas).each(function(i,v){
i_array[i]=setInterval(function(){
clearInterval(i_array[i]);
window[func](datas[i]);
},10);
});
}
当然这里假设func函数至多接受一个参数。 |
|
|
J**B
发帖数: 204 | 16 Given:
3. interface Animal { void makeNoise(); }
4. class Horse implements Animal {
5. Long weight = 1200L;
6. public void makeNoise() { System.out.println("whinny"); }
7. }
8. public class Icelandic extends Horse
{ 9. public void makeNoise() { System.out.println("vinny"); }
10. public static void main(String[] args) {
11. Icelandic i1 = new Icelandic();
12. Icelandic i2 = new Icelandic(); 12. Icelandic i3 = new Icelandic();
13. i3 = i1; i1 = i2; i2 = null; i3 = i1;
14. } 15. }
When line 14 is r... 阅读全帖 |
|
w*****0
发帖数: 93 | 17 我用miktex2.9自带的texworks. 用feynMF，能显示出来费曼图，用了{fmfgraph*}
，label的内容却无法显示。
我抄了一段别人的例子如下：
\documentclass{article}
\usepackage{feynmf}
\begin{document}
\begin{fmffile}{simplelabels}
\begin{fmfgraph*}(140,125)
\fmfleft{i1,i2}
\fmfright{o1,o2}
\fmflabel{$e^-$}{i1}
\fmflabel{$e^+$}{i2}
\fmflabel{$e^+,\mu^+$}{o1}
\fmflabel{$e^-,\mu^-$}{o2}
\fmflabel{$i\sqrt{\alpha}$}{v1}
\fmflabel{$i\sqrt{\alpha}$}{v2}
\fmf{fermion}{i1,v1,i2}
\fmf{fermion}{o1,v2,o2}
\fmf{photon,label=$\gamma,,Z^0$}{v1,v2}
\end{fmfgraph... 阅读全帖 |
|
Y********x
发帖数: 322 | 18 I used the follows statements to save some variables:
save mydata.txt I1 I2 I3 D1 D2 I4 I5 I6 -ascii -append;
I1,I2,I3,I4,I5,I6 are integer values and D1,D2 are double types.
In one machine, the variables are properly saved in this order; But when I r
un the program in another machine, the order of the data saved are as follow
s:
D1 D2 I1 I2 I3 I4 I5 I6.
FT!!! What the hell is the problem? I could not believe
my eyes! Anybody has clue on this super bizarre problem? |
|
s***n
发帖数: 116 | 19 You derived a paradox, but gave a wrong interpolation to the paradox.
The correct answer to the issue is: there is no practical way to
"combine" two beam into one beam as you described. Writing a wave of
magnitude E as E=E1+E2 does not mean you can practically "combine" two
beams of magnitude E1 and E2 into a beam of magnitude E. In fact, think
of the reverse problem might be easy: if you are given a beam of
magnitude E and try to split it, you would not be able to make it into
two beams of m... 阅读全帖 |
|
b***k
发帖数: 2673 | 20 ☆─────────────────────────────────────☆
yww (petite) 于 (Thu Jan 31 18:33:36 2008) 提到:
I1(t) = B(s) 从0到t的积分
I2(t) = B(s)*B(s) 从0到t的积分
I1已经基本上人人都知道mean和variance了，I2呢？
B为标准布朗运动
☆─────────────────────────────────────☆
woshigsw (砍砍砍) 于 (Thu Jan 31 19:32:48 2008) 提到:
E(l2)=t^2/2
Var(l2)=7*t^4/12
☆─────────────────────────────────────☆
yww (petite) 于 (Thu Jan 31 21:26:14 2008) 提到:
请问I2的var怎么求啊？
☆─────────────────────────────────────☆
laridoff (波拿巴) 于 (Fri Feb 1 00:09:17 2008) 提到: |
|
p*********e
发帖数: 32207 | 21 评估一下转动惯量的话
I=mr^2
如果说要维持compressor进气横截面一定的话那就是面积固定
假设两种情况compressor的几何形状和重量分布成比例
那么m和r成三次关系,也就是说m=c*r^3
如果比较一个大增压器和两个小增压器,直径分别为r1和r2,
那么前者I1=c*r1^5,后者I2=2*c*r2^5
而两者的总横截面积一致,所以r1^2=2*r2^2=>r1=sqrt(2)*r2
那么I1=2^(3/2)*2*c*r2^5=2^(3/2)*I2,也就是2.83*I2
换句话说,在此情况一个大增压器的转动惯量是两个小增压器和的2.83倍
而且这还是在考虑两者几何形状相当的情况,两个增压器的话每个的负载更低
也就是说可以在进气面积与单增压器相当的前提下进一步减重,那么转动惯量会更小
系）
接近
惯量
和轴
应该 |
|
|
|
|
|
|
t*******h
发帖数: 2744 | 27 【 以下文字转载自 USANews 讨论区 】
发信人: thighhigh (Thigh High), 信区: USANews
标 题: poll的结果可能过分偏向Trump了，因为我们30多岁的人基本上不接陌生电话
发信站: BBS 未名空间站 (Mon Sep 26 10:43:10 2016, 美东)
poll的结果可能过分偏向Trump了，因为我们30多岁的人基本上不接陌生电话
Looking at the individual polls, they seem to not have any results for the
18-34 age bracket, which I recall has happened before in CNN polls that show
the race very tight.
http://imgur.com/a/Q0QgV
Edit: Links to polls themselves
Penn: http://i2.cdn.turner.com/cnn/2016/images/09/26/relpa1.pdf Colo: htt... 阅读全帖 |
|
b**r
发帖数: 490 | 28 一般认为R1B 是日尔曼人,凯尔特人中R1B巨多.爱尔兰人中80%是R1B, 几乎在所有民族
中排名第一.
比日尔曼国家德国(R1B 40%)高太多了.
其实日尔曼人很杂,R1B, I1, I2 甚至R1A 在一定区域内都可以称为日尔曼人.
I1,I2 (北欧很多,比如瑞典,金发碧眼为特征)可以比作老欧洲人 , 如果把C,D称为老亚
洲人的话.
老欧洲人和老亚洲人的共同特点是骁勇善战.
现代文明中真正牛逼的是R1B,也是西欧美国的基石 |
|
l*******g
发帖数: 28502 | 29 ☆─────────────────────────────────────☆
amaru (雅歌) 于 (Fri Nov 25 13:47:46 2005) 提到:
it will be a littele while, because I am slow.
That means in heaven, in earth, this world, next world, life or death,
and whatever, I don't want to associate with you anymore.
Is it clear now?
☆─────────────────────────────────────☆
amaru (雅歌) 于 (Fri Nov 25 15:19:02 2005) 提到:
see, the differece lots of arabs and us is when we find we may get hurt
by you with some jews behind, arabs blame jews, and we bl... 阅读全帖 |
|
|
e*i
发帖数: 10288 | 31 Forward this post to your email--if you are using telnet
_=_
_=_ Part 001 of 001 of file 5_off_30_cpn.zip
_=_
begin 666 5_off_30_cpn.zip
M4$L#!!0````(`&!D_$)O_NPBKE```.-W```0````-5]O9F9?,S!?8W!N+G!D
M9NV]=5Q5V]8P#"B*I)2((FR1E-@==)=T2(@@M:6[0T5"!`D+#%04D!`%:02D
MI;M#NKN[O@7F$>X]]]SG?=_OGV?NW]YK[KGF&CW''&.N8E82E^2"
MZYI("%$@",C:P(R$GQ\L9VJ%U;
M!:!=#6A'PI!@:9`V"`I'HD`P.%1'4)`$:V6T!P9TL)`0NMAAK^]A0L))""'?
M"Q1`O%=`5C_:H$@,^D`;"H4XT(9!0/]L@\'0R`-M*!Z>/]O@4`C\... 阅读全帖 |
|
a*****e
发帖数: 51 | 32 lets I1,I2,I3,I4 be the indices of the four boxes.
Take 1,2,3 and 4 pills for I1,I2,I3,I4 respectively. Measure the weight of
total 10 pills, say W. Then (W-4)*10 is the index of the poisonous box. |
|
Z*****Z
发帖数: 723 | 33 Given a Data Structure having first n integers and next n chars. A = i1 i2 i
3 ... iN c1 c2 c3 ... cN.Write an in-place algorithm to rearrange the elemen
ts of the array ass A = i1 c1 i2 c2 ... in cn
O（nlgn)的算法我知道。有没有O（n)的算法？ |
|
l******e
发帖数: 6 | 34 int a1[] = {5,2,1};
int a2[] = {3,2,1};
用4个指针指向两个数组（头和尾分别是（5，2）和（3，2）），同时扫描连个数组并
比较大小，每
次只挪动一个尾指针；直到一个数组结束，再修改头指针。
void output(int* a1, int* a2, int* b, int n)
{
int count = 1;
b[0] = a1[0]+a2[0];
int i1(0), i2(0), j1(1), j2(1);
while (count < n)
{
if (a1[i1]+a2[j2] > a2[i2]+a1[j1])
{
b[count] = a1[i1]+a2[j2];
j2++;
if (j2 == n)
{
j2 = 1;
i1++;
}
}
else
|
|
t****a
发帖数: 1212 | 35 来自主题: JobHunting版 - 两道算法题 1. some ideas here:
a. the sum should be within [1, 27], for each number n we can decompose it
as:
n = i1 + i2 + i3, in which 0<=i<=9, so for n we should have a series of [i1,
i2, i3]
b. the left three digits shouldn't start with 0, so for the left three
digits the i1!=0
c. for each n, any combination of [[i1_1, i2_1, i3_1], [i2_1, i2_2, i3_2]]
fit your requirement. |
|
h*********3
发帖数: 111 | 36 career cup 上的.
* Given an array of integers A, give an algorithm to find the longest
Arithmetic progression in it, i.e find a sequence i1 < i2 < … < ik, such
that
A[i1], A[i2], …, A[ik] forms an arithmetic progression, and k is the
largest possible. The sequence S1, S2, …, Sk is called an arithmetic
progression if
Sj+1 – Sj is a constant
* Given a list of points in the plane, write a program that outputs each
point along with the three other points that are closest to it. These three
points orde... 阅读全帖 |
|
r*******y
发帖数: 1081 | 37 #2, any complexity requirements? Given an array A[1,...,N], starting from A[
1],
find a cycle and determine the smallest index i1 such that A[i1] is in the
cycle. Then starting from A[2] to find the second cycle and determine the
smallest index i2. if i2==i1, this means we find the same cycle and we need
to
ignore this cycle and go ahead starting from A[3]...
找他对应的 |
|
B*******1
发帖数: 2454 | 38 Given an array of integers A, give an algorithm to find the longest
Arithmetic progression in it, i.e find a sequence i1 < i2 < … < ik, such
that
A[i1], A[i2], …, A[ik] forms an arithmetic progression, and k is the
largest possible.
The sequence S1, S2, …, Sk is called an arithmetic progression if
Sj+1 – Sj is a constant
should be DP, but could not figure out the dp function. |
|
g**********y
发帖数: 14569 | 39 i2 = 47, set里是{46}, set.remove(i2) does nothing.
你把Eclipse的debugger打开，看一看就知道了。以后这种问题都可以问debugger. |
|
D*******e
发帖数: 151 | 40 Given an array of integers A, give an algorithm to find the longest
Arithmetic progression in it, i.e find a sequence i1 < i2 < … < ik, such
that A[i1], A[i2], …, A[ik] forms an arithmetic progression, and k is the
largest possible. The sequence S1, S2, …, Sk is called an arithmetic
progression if Sj+1 – Sj is a constant. |
|
c**********e
发帖数: 2007 | 41 Use one loop: (Hehe, nasty again)
for(int i=0;i<3^6;i++) {
i1 = i/243;
i2 = (i/81) % 3;
i3 = (i/27) % 3;
i4 = (i/9) % 3;
i5 = (i/3) % 3;
i6 = i % 3;
cout << s[i1] << s[i2] << s[i3] << s[i4] << s[i5] << s[i6] << endl;
} |
|
z********i
发帖数: 568 | 42 The following algorithm for queue with max is correct to me:
http://www.sureinterview.com/wiki/Shwqst/903001
idea:
1 用数据队列保存所有数据X[0]，X[1]，...
2 用辅助队列保存数据中的X[i1],X[i2],etc. such that
2.1 X[i1]>=X[i2]>=...。
2.2 X[i1]是从X[i1]开始最大的数。
3 enque的时候,删除辅助队列中比要插入的数据小（<）的数据。
4 deque的时候，删除数据队列的第一个数据。同时，如果辅助队列的第一个数据等于数据队列的第一个数据，删除辅助队列的第一个数据。
5 max就是辅助队列的第一个数据
例子一。
数据队列：6 5 4。
辅助队列：6 5 4。
1) max: 辅助队列的第一个数据6.
2) deque：
数据队列：5 4。
辅助队列：5 4
max: 5
3) deque:
数据队列：4。
辅助队列：4
max: 4
例子二。
数据队列：4 5 6。
辅助队列：6。
1) max: 辅助... 阅读全帖 |
|
c****7
发帖数: 4192 | 43 c# 来一个：
public static IEnumerable> P(IEnumerable list) {if (
list.Count() == 1)return new List> { list };return list.
Select((a, i1) => P(list.Where((b, i2) => i2 != i1)).Select(b => (new List
> { a }).Union(b))).SelectMany(c => c);} |
|
b*****u
发帖数: 648 | 44 来自主题: JobHunting版 - 一道设计题 用一个vector存储空闲日期段的开始
一个map 存储日期到演出的mapping
class calendar
{
vector available;
map dateToShow;
}
class performance
{
int date;
int program;
}
// assume that int binarySearch(vectorA, int x) returns the target
index i that A[i] <= x
// cost: log n
// return 0 if available, program id if not
int calendar::check(int date)
{
if (dateToShow.find(date)==dateToShow.end())
return 0;
else
return dateToShow[date];
}
// return true if successfully inserted
bool calendar::ins... 阅读全帖 |
|
b*****u
发帖数: 648 | 45 来自主题: JobHunting版 - 一道设计题 用一个vector存储空闲日期段的开始
一个map 存储日期到演出的mapping
class calendar
{
vector available;
map dateToShow;
}
class performance
{
int date;
int program;
}
// assume that int binarySearch(vectorA, int x) returns the target
index i that A[i] <= x
// cost: log n
// return 0 if available, program id if not
int calendar::check(int date)
{
if (dateToShow.find(date)==dateToShow.end())
return 0;
else
return dateToShow[date];
}
// return true if successfully inserted
bool calendar::ins... 阅读全帖 |
|
z****p
发帖数: 18 | 46 Sorry, I didn't read the problem carefully. Let me give it a second try:
Answer (the answer is not the probability or probability mass function for
the occurrences of the given pattern, but the expected number of occurrence
of the corresponding pattern--we can use that to tell which pattern is more
likely to happen than others):
(1) 993*4*2*2*4*2*2*2*4/(6^8)
(2) 994*2*2*4*2*2*2*4/(6^7)
(3) 995*4*2*2*4*4*4*2/(6^7)
Reasoning:
-Let's solve (1).
-Define an indicator random variable I1, so that I1=1 ... 阅读全帖 |
|
s*******s
发帖数: 1031 | 47 我的代码：
struct itemRecord {
int index;
int val;
itemRecord(int i, int v): index(i), val(v) {}
};
bool itemRecordLess(itemRecord i1, itemRecord i2) {
return i1.val < i2.val;
}
class Solution {
public:
vector twoSum(vector &numbers, int target) {
vector result;
int n = numbers.size();
vector numRecords;
for(int i=0; i
numRe... 阅读全帖 |
|
s****n
发帖数: 70 | 48 Not for sure...
如果用>表示包含关系，那么如果是这种情况
i1 > i2 > i3 > ... > in
用interval tree的话查询i1要遍历所有n个叶节点，查询i2要遍历至少n-1个节点。。
。以此类推，一共要访问O(n^2)叶节点。 |
|
w*******e
发帖数: 285 | 49 我觉得这个没有必要4维，i1+j1 = i2+j2,所以只要3维就够了，然后j2 = i1 + j1 -
i2,最后的结果就是dp[m][n][m] |
|
s******c
发帖数: 1920 | 50 可以和1维那样类似的想法来2维dp
只是每隔子问题的解不能简单用dp_sol[i]来表示,
而是用一个长度n的 vector来index每个子问题的解dp_sol[i1,i2,...i_n]，
i1,i2,...i_n 描述了一个把2d矩阵切成两块的一个界面
i_k表示第k列上这个界面的坐标.
对每一个子问题, 从界面的边界开始尝试添加word
总共有n^n个子问题, 然后每个子问题的递推式可以在n^2时间搞定
(需要preprocess这个矩阵, 计算从每点出发可以cover那些word)
or
of |
|
