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谷歌燒腦面試題目
Going for a job interview is nerve-racking at the best of times, but tech giant Google has become notorious for asking fiendishly tough questions.
即便是招聘高峰期，求職面試也會非常傷神費(fèi)腦，然而科技巨頭谷歌的面試題則因其奇葩刁鉆而異常出名。
The company used to ask head-scratching questions such as 'How many piano tuners are there in the world?' and 'How many golf balls does it take to fill a school bus?'.
谷歌公司曾問過一些令人頭痛不已的面試問題，例如，“世界上有多少位鋼琴調(diào)音師?”以及“一輛校車?yán)镒疃嗄苋嗌兕w高爾夫球?”。
But some of the riddles were so tough they have now been banned from the interview process.
但因有些謎題太過于荒謬古怪，谷歌公司現(xiàn)在已不再將其納入面試流程中。
Recruiter Impact Interview have compiled a list of 140 questions candidates have reported being asked at interviews for various roles at Google, many of which have now been dropped from the process.
招聘面試小冊子匯編140道題，依應(yīng)聘人員所言，在谷歌面試中他們曾以各種角色回答過此類問題。目前其中的許多問題已從面試流程中剔除。
One of the tough teasers that candidates used to be asked was: 'How many piano tuners are there in the entire world?'
應(yīng)聘者們曾經(jīng)回答過這樣一個令人啼笑皆非的難題：“全世界有多少位鋼琴調(diào)音師?”
The riddle is known as a Fermi problem, named after the physicist Enrico Fermi who was known for his ability to make good calculations with very little information or even none.
這類謎題被稱為“費(fèi)米問題”，命名來自物理學(xué)家恩里科·費(fèi)米，他之所以聲名遠(yuǎn)揚(yáng)是因?yàn)樗軌蛟谏倭康慕o定信息甚至沒有信息的情況下進(jìn)行運(yùn)算。
Its purpose is to test a candidates approximation skills as well as their ability at dimensional analysis.
費(fèi)米問題意在考察應(yīng)聘者的估算能力以及量綱分析能力。
The puzzle is solved by multiplying a series of estimates to get to the right answer. So a candidate would have to estimate factors such as how many households have a piano, how often they are tuned to figure out how many piano tunings take place a year.
解決費(fèi)米問題的方法在于通過一系列估算而無限接近正確答案。因此，應(yīng)聘者需要考量一些因素，諸如：擁有鋼琴的家庭戶數(shù)，此類家庭進(jìn)行鋼琴調(diào)音的頻次等，從而得出每年有多少次的鋼琴調(diào)音。
They then need to calculate the average working hours of a piano tuners and the number of jobs they carry out.
隨后，應(yīng)聘者們需要估算出鋼琴調(diào)音師的平均工作時長以及工作量。
The number of piano tunings that take place per year divided by the number per year per piano tuner should yield the answer.
因此，用每年所有家庭需要進(jìn)行鋼琴調(diào)音的次數(shù)除以每年每位鋼琴調(diào)音師的工作量，答案就此誕生。
Google candidates going for a role as product manager also used to be asked: 'How many golf balls can fit in a school bus?
申請谷歌公司產(chǎn)品經(jīng)理職位的應(yīng)聘者也曾被問到：“一輛校車?yán)锬軌蛉菁{多少顆高爾夫球?”
To solve it one must calculate the volume of the average golf ball and the the volume of the average school bus, taking into account factors like the seats and other fixtures taking up space.
為了解決這個問題，應(yīng)聘者必須估算出一顆正常高爾夫球的體積以及一輛正常校車的體積，同時也需將校車內(nèi)占據(jù)空間的座椅和其他固定裝置的體積等因素考慮在內(nèi)。
The answer is the volume of the bus, divided by the volume of the balls.
這個答案就是校車容量除以高爾夫球體積所得的數(shù)字。
As well as questions involving calculation and physics, Google used to ask would-be employees to solve logic puzzles.
谷歌公司的面試問題不僅提及數(shù)學(xué)運(yùn)算與物理學(xué)方面，也曾讓其潛在雇員們解決邏輯難題。
One of the riddles was: 'Every man in a village of 100 married couples has cheated on his wife.
其中的一個謎題是：在一個村莊里，有100對已婚夫婦，每名男子都曾對自己的配偶不忠。
'Every wife in the village instantly knows when a man other than her husband has cheated but does not know when her own husband has.
一旦有男子在外偷吃，村莊里的婦人們便會立刻得知，但是如果這名男子是自己的丈夫，婦人將不會得到消息。
'The village has a law that does not allow for adultery. Any wife who can prove that her husband is unfaithful must kill him that very day.
這個村莊的法律嚴(yán)禁婚外情。任何婦人一旦證實(shí)丈夫有婚外情，就必須當(dāng)日殺了他。
'The women of the village would never disobey this law. One day, the queen of the village visits and announces that at least one husband has been unfaithful. What happens?'
村里的婦人們從來沒有違背過該法規(guī)。有一天，村莊里的女王來訪并宣稱村里至少有一名丈夫背著妻子有婚外情。接下來究竟會發(fā)生什么?
Although opinion varies as to the correct answer, a popular solution is to assume that all the cheating husbands will be executed 100 days after the queen's announcement.
盡管關(guān)于正確答案，眾人各持己見，但仍存有一個廣為流行的解答，即所有背情忘義的丈夫?qū)谂跣枷⒅蟮牡?00天被執(zhí)以死刑。
Every wife in the village instantly knows when a man other than her husband has cheated, so if there's only one unfaithful man she will know it's her husband and will kill him that day.
一旦有男子出軌，村莊里的所有婦人便會即刻得知消息，除非這名男子是自己的丈夫。因此，如果只有一個不忠的男子，那么婦人便得知此人就是自己的丈夫，便會在當(dāng)日使其命喪黃泉。
Bearing in mind that the queen has said that at least one man has cheated, there could be more than a single adulterer.
我們必須記住女王提到，至少有一名男子背叛妻子，這也就意味著該數(shù)目可能大于等于一。
If there are two cheaters, 98 women will know who they are. The next day when the wives who have been cheated on see that nobody has been executed they will realise the only explanation is that their husbands are the culprits, and will kill them.
如果村里有兩名男子背叛妻子，那便有98位婦人知道他們的身份。次日，他們的妻子沒有發(fā)現(xiàn)有人被處以死刑，那么婦人就會明白，唯一的可能就是自己丈夫，隨后殺之。
By that logic, 100 cheating husbands will be executed on the 100th day, as the riddle states that every man in the village has cheated on his wife.
依邏輯推斷，那100名有婚外情的男子會在第100天時通通被處死，正如謎題所述，村莊里的每個男人都曾背叛過自己的妻子。
A riddle about pirates sounds complicated but the answer is in fact quite simple.
有一個關(guān)于海盜的謎題看似玄乎復(fù)雜，但事實(shí)上答案十分簡單。
You’re the captain of a pirate ship, and your crew gets to vote on how the gold is divided up.
假設(shè)你是一艘海盜船的船長，你的船員們就如何分配黃金的問題進(jìn)行投票。
'If fewer than half of the pirates agree with you, you die. How do you recommend apportioning the gold in such a way that you get a good share of the booty, but still survive?
如果你的分配方案得不到半數(shù)支持，你就會被殺死。在自己可以活命的前提下，你如何分配才能拿到最多的財寶?
The solution is to share 51 percent of the treasure.
答案是分掉51%的黃金。
Another logic puzzled candidates were given to solve was: 'How many times a day do the hands of a clock overlap?'
“一天中，時鐘的時針和分針會重合幾次?”這是應(yīng)聘者們被要求回答的又一個邏輯難題。
The hands overlap once an hour, but 11 times in 12 hours and 22 times in a day. This is because the overlap at 12 has already been accounted for.
時針和分針每小時重疊一次，但在12小時內(nèi)會重疊11次，一天之內(nèi)重疊22次。這是因?yàn)樵?2時位置的指針重合已經(jīng)計算在內(nèi)。
The overlaps occur at 12, 1.05, 2.11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49 and 10:55 in the morning and after midday at 12, 1.05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49 and 10:55.
重合時間點(diǎn)分別是上午12:00，1:05，2:11，3:16，4:22，5:27，6:33，7:38，8:44，9:49，10:55以及下午12:00，1:05，2:11，3:16，4:22，5:27，6:33，7:38，8:44，9:49，10:55。
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