Google/News2012. 6. 29. 00:11

어제 새벽에 Moscone Center에서 Google I/O가 열렸습니다..


Google I/O는 애플의 키노트와 똑같다고 보면됩니다..


또한 구글이 나아갈 기술의 방향을 조감해볼 수 있습니다


먼저 어제 나온것들만 요약하면


-안드로이드 4.1 젤리빈 


-오프라인맵


-구글의 레퍼런스 태블릿 넥서스 7 


-구글글래스 


-넥서스 Q




먼저 안드로이드 4.1 젤리빈을 보면..


googleio20123964.jpg



1. 프로젝트 버터

터치감 애니메이션 ㅍ래임 향상으로 더 부드러워진 안드로이드


2. 오프라인 보이스 타이핑

네트워크가 연결 되어 있지 않아도 음성 입력이 가능함


3. 안드로이드 빔 개선

전송 속도가 향상되었스며 기능이 추가됨


4. 상태바 기능강화


더 많은 기능을 설정하고 더 많은 정보를 얻을 수 있음


5. 프리스타일 위젯

위젯 크기를 사용자 마음대로 조절이 가능함


6. Play Store 개선

 1. 어플리케이션 복돌 개선

앱을 암호화 하여 정상적인 구매절차를 걸친 사람만 설치가능

(흐경흐경 안드로이드 루팅 이젠 필수탬 될듯 ㅠ)


 2. 스마트 앱 업데이트

 안드로이드는 어플리케이션 업데이트를 진행할때 해당앱을 처음부터 다시 받아야했기 때문에 시간도 낭비되고 데이터도 낭비 됐는데 

 스마트 업데이트를 통해서 어플리케이션이 수정된 부문만 내려받기 가능


 3. 구글 Play Store에서 즐길 수 있는 콘텐츠가 게임과 앱 이외에도 영화, tv쇼, 메거진 등으로 더 늘어남 

한번 구매한 영화는 클라우드에 저장되어서 보고 싶은 만큼 실컷 볼 수 있음 tv쇼도 그렇게 하기 위해서

여러 미국의 유수의 영화사/방송국과 계약을 맺었음 메거진 같은 경우에는 14동안 맛보기로 구독이 가능

단 한국은 아직 서비스를 이용할 수 없으며 이용 가능 시점도 미지수 ...


7 세이프 모드 추가

전원 버튼 길게 누르고 있으면 됨


8 와이파이 다이렉트 기능추가


역시 백문이 불어일견 밑에는 젤리빈 사진들과 영상들



 세기의 대결 아이스크림 센드위치 vs 젤리빈


jellybeans.jpg

젤리가 와르를...


9.2.png

안드로이드 젤리빈 이스터에그


6-27-2012-jellybean.jpg

업그레이드된 상태바


googlenowlead.jpg

바탕화면과 이번에 새로 추가된 기능


젤리빈 시연 영상


참고

http://forum.xda-developers.com/showthread.php?t=1737340

이건 젤리빈 OTA파일 다운로드 주소 

http://forum.xda-developers.com/showthread.php?p=27991551#post27991551

젤리빈 여러가지 소개(영문)



Google I/O에서 젤리빈 업데이트 기기를 발표했습니다

7월에 갤럭시 넥서스/모토롤라 줌/넥서스s 이 3가지 기종을 발표함(레퍼부심 폭발!)

삼성, lg, htc는 회사에서 뭐 가능하면 해주겠죠?...


- 오프라인 관련


1. 오프라인맵


말 그대로 데이터 통신이 불가능한 곳에서 미리 지도를 받아 놓고 쓸 수 있습니다

해외 여행 가서 쓰면 유용할듯 밑에는 소개 동영상입니다





downloadingmaps.jpg

오프라인 지도 생성중


2. 오프라인 음성인식

Screenshot_2012-06-27-21-31-15.png


--

-구글의 레퍼런스 태블릿 넥서스 7 

tablet-n7-features-ushome-family.png


일단 간단하게 스펙 

CPU : 테그라3 쿼드코어 1.3 ghz

Graphic : 지포스 12코어 프로세서

RAM : 1G

해상도 1280*800(7인치 태블릿)

용량 : 8gb(199$)/16gb(249$)

무게 : 340g


넥서스7 소개 동영상


이번에 Google I/O에서 새롭게 소개된 넥서스7은 1280*800의 해상도를 가진 7인치 태블릿pc로 아이스크림 센드위치 다음버전인 젤리빈을 기본으로 탑제하고있음

가격적인 메리트가 상당히 커서 아이패드랑 이번에 새로 마이크로 소프트에서 공개한 서피스와는 해상도는 다르지만... 치열한 경쟁이 예상됨..

그런데 전면 카메라는 있는데 후면 카메라는 없는게 함정 ㅋㅋ


tablet-gallery-back.png

(후면 카메라 종범)


--

-구글 글래스


-

구글 글래스는 내가 굳이 설명 안해다 다들 알고 있을듯...


구글글래스 데모 시연 영상인데 3:20초 부터 ㄱㄱ 지림...

스카이 다이빙 - 옥상 착륙 - 컨퍼런스장 입갤



---


-넥서스Q

player-q-a.jpg


이번에 새로 등장한 넥서스Q는 구글에서 내놓은 소셜 스트리밍 기기인데 일종의 스마트 TV 셉톱박스(?)

구글 플레이에서 구매한 영화 음악 tv쇼 등을 tv로 스트리밍 할 수 있게 해주는 장치임 apple tv랑 유사함니다

내장 os는 아이스크림 샌드위치이며 cpu는 OMAP 4460프로세서 내장 용량은 16기가 25W출력 앰프가 내장 되어있습니다

가격은 299달러



Nexus Q 소개 동영상



Posted by ikSkyLand
Google/News2011. 7. 27. 20:45


Google의 Chromebook이 드디어 판매가 시작되었습니다.

위의 Buy Now를 눌르면 미국의 아마존 사이트로 링크가 연결됩니다
(http://www.amazon.com/b/ref=os_chrm_54?&node=2858603011) <- 링크주소

 가격은 $349.00 ~ 499.99(\약 366,624 ~525,239)로 6개의 제품이 있습니다

제조는 기존에 알려진것과 동일하게 삼성과 에이서에서 제조했습니다

아래는 간략한 삼성의 Samsung Series 5 3G Chromebook (Titan Silver) 모델에 대한 제품외간의 사진과 제품사양 입니다














 
  • Chrome-based operating system (운영체제 : ChromeOS)
  • Dual-core Intel processor (CPU : 인텔 듀얼-코어 프로세서)
  • SuperBright 12.1" high-resolution, non-glare display ideal for web browsing (액정 : 12.1인치 논글래디스플레이)
  • Boots in less than 10 seconds and resumes instantly (부팅이 10초면 끝)
  • Up to 8.5 hours of continuous use (베터리 : 8.5시간 사용가능)
     
    아래는 크롬북에 대한 간략한 설명들.












    Explore Google Features
     
 

Posted by ikSkyLand
Google/information2011. 5. 22. 10:03

문제 :
{자연대수 e를 풀어서 쓸 때 처음 발견되는 10자리수 솟수(first 10-digit prime found in consecutive digits of e)}.com

풀이 : 
 Google Labs Aptitude Test Partially Answered

1. Solve this cryptic equation, realizing of course that values for M and E could be interchanged. No leading zeros are allowed.

WWWDOT - GOOGLE = DOTCOM

This can be solved through systematic application of logic.  For example, O cannot be equal to 0, since O + O +[1 or  0] = W.  That would make W1, but D + GW, which is not possible.

Here is a slow brute-force method of solution that takes a few minutes on a relatively fast machine:

Off[General :: "spell1"] chars = Characters/@ToLowerCase/@{"WWWDOT", "GOOGLE", "DOTCOM"} ; uchars = Union[Flatten[chars]] ;

eqn = First[#] - Plus @@ Rest[#] &[FromDigits[#, 10] &/@chars] == 0 ; Timing[soln = Select[Permutations[Range[0, 9]], eqn/.Thread[ucharsMost[#]] &]]

{359.3 Second, {{4, 5, 3, 1, 0, 6, 8, 9, 7, 2}, {4, 5, 6, 1, 0, 3, 8, 9, 7, 2}}}

Thread[ucharsMost[#]] &/@soln

{{c4, d5, e3, g1, l0, m6, o8, t ... , d5, e6, g1, l0, m3, o8, t9, w7}}

This gives the two solutions

777589 - 188106 == 589483
777589 - 188103 == 589486

Here is another solution using Mathematica's Reduce command:

eqn = "wwwdot" - "google""dotcom"/.s_StringFromD ... eqs&&#, vars, Integers] &/@eqs, (Unequal @@ vars/.ToRules[#]) =!= False&]//Timing

{96.98 Second, c4&&d5&&e3&&g1&& ... p;&l0&&m3&&o8&&t9&&w7}

A faster (but slightly more obscure) piece of code is the following:

cf = Compile[Evaluate[{#, _Integer} &/@{c, d, e, g, l, m, o, t, w, x}], Module[ {& ...  {d, o, t, c, o} . S ; A - B - mC + e&&A - B - eC + m] ] ;

Transpose[{{c, d, e, g, l, m, o, t, w}, Most[#]}] &/@Module[{perms = Developer`ToPackedArray/@Take[Permutations[Range[0, 9]], All]}, Select[perms, cf @@ #1&]]//Timing

{49.89 Second, {{{c, 4}, {d, 5}, {e, 3}, {g, 1}, {l, 0}, {m, 6}, {o, 8}, {t, 9}, {w, 7}}, {{c, 4}, {d, 5}, {e, 6}, {g, 1}, {l, 0}, {m, 3}, {o, 8}, {t, 9}, {w, 7}}}}

Faster still using the same approach (and requiring ~300 MB of memory):

Transpose[{{c, d, e, g, l, m, o, t, w}, Most[#]}] &/@Compile[{}, Module[{perms = Take[Perm ... d, o, t, c, o} . S ; A - B - mC + e&&A - B - eC + m]]]]][]//Timing

{14.65 Second, {{{c, 4}, {d, 5}, {e, 3}, {g, 1}, {l, 0}, {m, 6}, {o, 8}, {t, 9}, {w, 7}}, {{c, 4}, {d, 5}, {e, 6}, {g, 1}, {l, 0}, {m, 3}, {o, 8}, {t, 9}, {w, 7}}}}

Even faster using the same approach (that does not exclude leading zeros in the solution, but that can easily be weeded out at the end):

eq = Simplify["wwwdot" - "google" - "dotcom"/.s_StringFr ... ers[s]], 10]] ; Join[CoefficientList[eq, #][[2]] &/@Union[Cases[eq, _Symbol, Infinity]], {0}]

{-100, -99900, -1, -100100, -10, -1, -21000, -999, 111000, 0}

Transpose[{{c, d, e, g, l, m, o, t, w}, Most[#]}] &/@Compile[{}, Select[Permutations[Range ... , 9]], {-100, -99900, -1, -100100, -10, -1, -21000, -999, 111000, 0} . #0&]][]//Timing

{6.44 Second, {{{c, 4}, {d, 5}, {e, 3}, {g, 1}, {l, 0}, {m, 6}, {o, 8}, {t, 9}, {w, 7}}, {{c, 4}, {d, 5}, {e, 6}, {g, 1}, {l, 0}, {m, 3}, {o, 8}, {t, 9}, {w, 7}}}}

Here is an independent solution method that uses branch-and-prune techniques:

wwwdot = {w, w, w, d, o, t} ; google = {g, o, o, g, l, e} ; dotcom = {d, o, t, c, o, m} ; vars ... en[{eqn, vareqns, zeroOneConstraints, noLeadingZeros, mustSumToOneConstraints, distinctDigits}] ;

Developer`SetSystemOptions["LinearProgrammingOptions" {"InteriorPointSize"1, "Preprocessing"True}] ;

Off[General :: "spell1"]

wwwdotgoogledotcom[constraints_, vars_, zeroOneVars_] := Module[{allvars = Join[vars, zeroOneV ...  zeroOneVars[[badpos]] 1], stack} ;] ;] ; Sort/@Map[Reverse, solns, {2}] ]

wwwdotgoogledotcom[allInfo, vars, Flatten[newvars]]//Timing

{72.89 Second, {{{c, 4}, {d, 5}, {e, 3}, {g, 1}, {l, 0}, {m, 6}, {o, 8}, {t, 9}, {w, 7}}, {{c, 4}, {d, 5}, {e, 6}, {g, 1}, {l, 0}, {m, 3}, {o, 8}, {t, 9}, {w, 7}}}}

And the winner for overall fastest:

enforceUniqueDigits[l_, k_] := If[Length[Union[Take[l, -k]]] =!= k, Sequence @@ {}, l] <br/>  ...  &/@Select[dgclotem, dotcom[#, 6] + google[#, 6] wwwdot[#, 6] &])//Timing

{2.58 Second, {{c4, d5, e6, g1, l0, m3, oɳ ...  d5, e3, g1, l0, m6, o8, t9, w7}}}

2. Write a haiku describing possible methods for predicting search traffic seasonality.

MathWorld's search engine
seemed slowed this May. Undergrads
prepping for finals.

3.       1
         1 1
         2 1
      1 2 1 1
   1 1 1 2 2 1

What's the next line?  

312211.  This is the "look and say" sequence in which each term after the first describes the previous term: one 1 (11); two 1s (21); one 2 and one 1 (1211); one 1, one 2, and two 1's (111221); and so on.  See the look and say sequence entry on MathWorld for a complete write-up and the algebraic form of a fascinating related quantity known asConway's constant.

RunLengthEncode[x_List] := (Through[{First, Length}[#1]] &)/@Split[x] LookAndSay[n_, d_:1] := NestList[Flatten[Reverse/@RunLengthEncode[#]] &, {d}, n - 1]

FromDigits/@LookAndSay[6]

{1, 11, 21, 1211, 111221, 312211}

4. You are in a maze of twisty little passages, all alike.  There is a dusty laptop here with a weak wireless connection.  There are dull, lifeless gnomes strolling around.  What dost thou do?

    A) Wander aimlessly, bumping into obstacles until you are eaten by a grue.
    B) Use the laptop as a digging device to tunnel to the next level.
    C) Play MPoRPG until the battery dies along with your hopes.
    D) Use the computer to map the nodes of the maze and discover an exit path.
    E) Email your resume to Google, tell the lead gnome you quit and find yourself in whole different world [sic].

In general, make a state diagram.  However, this method would not work in certain pathological cases such as, say, a fractal maze.  For an example of this and commentary, see Ed Pegg's column about state diagrams and mazes.

5. What's broken with Unix?

Their reproductive capabilities.

How would you fix it?

[This exercise is left to the reader.]

6. On your first day at Google, you discover that your cubicle mate wrote the textbook you used as a primary resource in your first year of graduate school. Do you:

    A) Fawn obsequiously and ask if you can have an autograph.
    B) Sit perfectly still and use only soft keystrokes to avoid disturbing her concentration
    C) Leave her daily offerings of granola and English toffee from the food bins.
    D) Quote your favorite formula from the textbook and explain how it's now your mantra.
    E) Show her how example 17b could have been solved with 34 fewer lines of code.

[This exercise is left to the reader.]

7. Which of the following expresses Google's over-arching philosophy?

    A) "I'm feeling lucky"
    B) "Don't be evil"
    C) "Oh, I already fixed that"
    D) "You should never be more than 50 feet from food"
    E) All of the above

[This exercise is left to the reader.]

8. How many different ways can you color an icosahedron with one of three colors on each face?

For an asymmetric 20-sided solid, there are 3^20 possible 3-colorings.  For a symmetric 20-sided object, the Pólya enumeration theorem can be used to obtain the number of distinct colorings.  Here is a concise Mathematica implementation:

Off[General :: "shdw", General :: "spell1"] <<DiscreteMath`Combinato ...  GroupFaces = KSubsetGroup[GroupI, Sort/@f] ; Polya[GroupFaces, colors] ]

ColorMySolid[Icosahedron, colors]

(2 colors^4)/5 + colors^8/3 + colors^10/4 + colors^20/60

%/.colors 3

58130055

What colors would you choose?

[This exercise is left to the reader.]

9. This space left intentionally blank.  Please fill it with something that improves upon emptiness.

For nearly 10,000 images of mathematical functions, see The Wolfram Functions Site visualization gallery.

10. On an infinite, two-dimensional, rectangular lattice of 1-ohm resistors, what is the resistance between two nodes that are a knight's move away?

R[m_, n_] := 1/(2π) Integrate[1/t (1 - ((t - I)/(t + I))^(m + n) ((t - 1)/(t + 1))^Abs[m - n]), {t, 0, ∞}]

R[1, 2]

(8 - π)/(2 π)

This problem is discussed in J. Cserti's 1999 arXiv preprint.  It is also discussed in The Mathematica GuideBook for Symbolics, the forthcoming fourth volume in Michael Trott's GuideBook series, the first two of which were published just last week by Springer-Verlag.  The contents for all four GuideBooks, including the two not yet published, are available on the DVD distributed with the first two GuideBooks.

11. It's 2PM on a sunny Sunday afternoon in the Bay Area.  You're minutes from the Pacific Ocean, redwood forest hiking trails and world class cultural attractions.  What do you do?

[This exercise is left to the reader.]

12. In your opinion, what is the most beautiful math equation ever derived?

There are obviously many candidates.  The following list gives ten of the authors' favorites:

1. Archimedes' recurrence formulaa_ (2 n) = (2 a_n b_n)/(a_n + b_n)b_ (2 n) = (a_ (2 n) b_n)^(1/2)a_n>π>b_na_∞ = b_∞
2. Euler formula^( π) + 10
3. Euler-Mascheroni constantUnderscript[lim, k∞]   (Underoverscript[∑, n = 1, arg3] 1/n - log(k)) 
4. Riemann hypothesisζ (α + β ) 0 and β≠0 implies α1/2
5. Gaussian integral:   ∫_ (-∞)^∞^(-x^2) xπ^(1/2)
6. Ramanujan's prime product formulaUnderoverscript[∏, k = 1, arg3] (p_k^2 + 1)/(p_k^2 - 1) 5/2
7. Zeta-regularized productUnderoverscript[∏, k = 1, arg3] k (2 π)^(1/2)
8. Mandelbrot set recursionz_ (n + 1) z_n^2 + C
9. BBP formulaπUnderoverscript[∑, n = 0, arg3] (-2/(8 n + 4) - 1/(8 n + 5) - 1/(8 n + 6) + 4/(8 n + 1)) (1/16)^n
10. Cauchy integral formulaf(z_0) 1/(2 π ) ∮f(z)/(z - z_0) z

An excellent paper discussing the most beautiful equations in physics is Daniel Z. Freedman's "Some beautiful equations of mathematical physics."  Note that the physics view on beauty in equations is less uniform than the mathematical one.  To quote the not-necessarily-standard view of theoretical physicist P.A.M. Dirac, "It is more important to have beauty in one's equations than to have them fit experiment."

13. Which of the following is NOT an actual interest group formed by Google employees?

    A. Women's basketball
    B. Buffy fans
    C. Cricketeers
    D. Nobel winners
    E. Wine club

[This exercise is left to the reader.]

14. What will be the next great improvement in search technology?

Semantic searching of mathematical formulas.  See http://functions.wolfram.com/About/ourvision.html for work currently underway at Wolfram Research that will be made available in the near future.

15. What is the optimal size of a project team, above which additional members do not contribute productivity equivalent to the percentage increase in the staff size?

    A) 1
    B) 3
    C) 5
    D) 11
    E) 24

[This exercise is left to the reader.]

16. Given a triangle ABC, how would you use only a compass and straight edge to find a point P such that triangles ABP, ACP and BCP have equal perimeters?  (Assume that ABC is constructed so that a solution does exist.)

This is the isoperimetric point, which is at the center of the larger Soddy circle. It is related to Apollonius' problem. The three tangent circles are easy to construct: The circle around C has diameter a + b - c, which gives the other two circles.  A summary of compass and straightedge constructions for the outer Soddy circle can be found in "Apollonius' Problem: A Study of Solutions and Their Connections" by David Gisch and Jason M. Ribando.

17. Consider a function which, for a given whole number n, returns the number of ones required when writing out all numbers between 0 and n.  For example, f(13)=6.  Notice that f(1)=1.  What is the next largest n such that f(n)=n?

The following Mathematica code computes the difference between [the cumulative number of 1s in the positive integers up to n] and [the value of n itself] as n ranges from 1 to 500,000:

data = MapIndexed[#1 - #2[[1]] &, Rest[FoldList[Plus, 0, Table[DigitCount[n, 10, 1], {n, 500000}]]]] ;

<<Graphics`Colors` ListPlot[Take[MapIndexed[{#2[[1]], #1} &, data], {1, -1, 1000}], PlotStyleRed] ;

[Graphics:HTMLFiles/AptitudeTest_58.gif]

The solution to the problem is then the first position greater than the first at which data equals 0:

Position[data, 0]//Flatten

{1, 199981, 199982, 199983, 199984, 199985, 199986, 199987, 199988, 199989, 199990, 200000, 200001}

which are the first few terms of sequence A014778 in the On-Line Encyclopedia of Integer Sequences.

Checking by hand confirms that the numbers from 1 to 199981 contain a total of 199981 1s:

IntegerDigits/@Range[199981]//Flatten//Count[#, 1] &

199981

18.  What is the coolest hack you've ever written?

While there is no "correct" answer, a nice hack for solving the first problem in the SIAM hundred-dollar, hundred-digit challenge can be achieved by converting the limit into the strongly divergent series:

Sum[(-1)^k (2k)^(2k - 1), {k, ∞}]

and then using Mathematica's numerical function SequenceLimit to trivially get the correct answer (to six digits),

Off[SequenceLimit :: "seqlim"] SequenceLimit[FoldList[Plus, 0, N[#, 1000] & @ Table[-(-1)^k (2k)^(2k - 1), {k, 300}]], WynnDegree20]

0.323368

You must tweak parameters a bit or write your own sequence limit to get all 10 digits.

[Other hacks are left to the reader.]

19. 'Tis known in refined company, that choosing K things out of N can be done in ways as many as choosing N minus K from N: I pick K, you the remaining.

This simply states the binomial coefficient identity (N)  (N    )   K            N - K.  

Find though a cooler bijection, where you show a knack uncanny, of making your choices contain all K of mine.  Oh, for pedantry: let K be no more than half N.

'Tis more problematic to disentangle semantic meaning precise from the this paragraph of verbiage peculiar.

20. What number comes next in the sequence: 10, 9, 60, 90, 70, 66, ?

    A) 96
    B) 1000000000000000000000000000000000\
         0000000000000000000000000000000000\
         000000000000000000000000000000000
    C) Either of the above
    D) None of the above

This can be looked up and found to be sequence A052196 in the On-Line Encyclopedia of Integer Sequences, which gives the largest positive integer whose English name has n letters.  For example, the first few terms are ten, nine, sixty, ninety, seventy, sixty-six, ninety-six, ….  A more correct sequence might be ten, nine, sixty, googol, seventy, sixty-six, ninety-six, googolplex.  And also note, incidentally, that the correct spelling of the mathematical term "googol" differs from the name of the company that made up this aptitude test.

The first few can be computed using the NumberName function in Eric Weisstein's MathWorld packages:

<<MathWorld`IntegerSequences`

pairs = Last/@Split[Sort[{StringLength[StringReplace[NumberName[#], {" "->"", "-"->""}]], #} &/@Range[100]], #1[[1]] == #2[[1]] &]

{{3, 10}, {4, 9}, {5, 60}, {6, 90}, {7, 70}, {8, 66}, {9, 96}, {10, 100}, {11, 98}, {12, 78}}

Last/@Take[pairs, 7]

{10, 9, 60, 90, 70, 66, 96}

A mathematical solution could also be found by fitting a Lagrange interpolating polynomial to the six known terms and extrapolating:

pts = {10, 9, 60, 90, 70, 66} ;

newpts = Function[x, Evaluate[InterpolatingPolynomial[pts, x]]]/@Range[7]

{10, 9, 60, 90, 70, 66, 290}

Plot[Evaluate[InterpolatingPolynomial[pts, x]], {x, 0, 7}, PlotStyleRed, Epilog {Red, PointSize[.02], Point/@Transpose[{Range[7], newpts}]}] ;

[Graphics:HTMLFiles/AptitudeTest_76.gif]

21. In 29 words or fewer, describe what you would strive to accomplish if you worked at Google Labs.

[This exercise is left to the reader.]



이 문제를 푼 다음 http://7427466391.com/로 접속 후 2단계 다음 문제 까지 풀면 비로소 구글 연구개발부서인 구글랩 페이지로 연결되며, “구글을 키워나가면서 우리가 배운 점 한가지는 우리가 찾고자 하는 상대방 역시 우리를 찾고 있을 때, 그 사람을 더 쉽게 찾게 된다는 것이다. 우리는 세계 최고의 엔지니어를 찾고 있고, 당신은 여기에 와있다”라는 환영 메시지를 볼 수 있다.

메시지는 “짐작하겠지만 우리는 엄청난 량의 이력서를 매일같이 받고 있기 때문에 이러한 절차를 이용해 ‘신호대 잡음비’를 개선하려 하는 것”이라는 말로 이어진다.

하... 구글에 입사하기 정말 어럽네... 

Posted by ikSkyLand
Google/News2011. 5. 15. 12:59




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Posted by ikSkyLand