2008年1月24日星期四
一條英文題目
一個英文老師出了一道這樣的難題,題目是這樣子的:
____ is better than the god.
____ is worse than the evil.
if you eat ____,you will die.
(三個空格必須是同一個字) 沒有人答的出來
結果......
有一個數學老師用數學的方法解出來了
設：
上帝之善是＋∞
惡魔之惡是－∞
令所求為x
則x＞＋∞
x＜－∞
∴x屬於空集合
∴x=nothing
answer :
nothing is better than the god.（沒有什麼比上帝更好。）
nothing is worse than the evil. （也沒有什麼比惡魔更壞。）
if you eat nothing, you will die（如果你什麼也沒有吃，那麼你就會死！）
____ is better than the god.
____ is worse than the evil.
if you eat ____,you will die.
(三個空格必須是同一個字) 沒有人答的出來
結果......
有一個數學老師用數學的方法解出來了
設：
上帝之善是＋∞
惡魔之惡是－∞
令所求為x
則x＞＋∞
x＜－∞
∴x屬於空集合
∴x=nothing
answer :
nothing is better than the god.（沒有什麼比上帝更好。）
nothing is worse than the evil. （也沒有什麼比惡魔更壞。）
if you eat nothing, you will die（如果你什麼也沒有吃，那麼你就會死！）
2008年1月22日星期二
Microsoft Interview Question 4
Question:
How do you cut a rectangular cake into two equal pieces with one straight cut when someone has already removed a rectangular piece from it? (The removed piece can be of any size or any orientation.)
Answer:
This question definitely has a right answer. It might be argued that it involves a bit of a "trick", but I still like it. The trick is knowing or realizing that any line passing through the center point of a rectangle bisects it. Before you remove the rectangular piece from the cake, there are infinitely many lines which bisect the cake. After you remove the rectangular piece, there is only one - the line which passes through both the center of the cake, and the center of the removed rectangular piece. This line necessarily divides the removed piece in half, and hence the same amount of cake was removed from each half of the remaining portion.
The value in this question is not only seeing if a candidate can compute the answer, but watching them eliminate non-solutions. You would probably realize after a little trial-and-error that such a constraint is not helpful, and that might guide them toward the solution.
P.S. There is another solution - cut the cake in half vertically! (With a single horizontal slice.) I'd say this gets points for creativity, but I'd still want to see the candidate solve the problem the other way.
Question:
The $21 Question
Alan and Vicky have $21 between them. Alan has $20 more than Vicky. How much does each have (you can't use fractions in the answer)?
Answer:
I called this the worst question in the book, based on the fact that it has no answer. I went on to say:
Apparently sometimes people ask questions which have no answer to see how candidates react. This might be helpful in some situations (if you're hiring for a company with a confrontational culture!), but I would never use it; I don't like what it says about me and my company, and I can't imagine what it would say about the candidate, either.
However, I found out that this question does have an answer!
What does this illustrate? That many people apparently doesn't know that dollars are divided up into cents:
A = V + 2000¢
A + V = 2100¢
Hence,
A = 2050¢
V = 50¢
Excellent! When I read this question in the book it was described as having no answer, and it never occurred to me that the book was wrong, and that this question really does have an answer. This is a classic "thinking out of the box" test. Any candidate is going to do the algebra and conclude that there is no integer solution in dollars. Will they then consider shifting units to cents? Very interesting.
How do you cut a rectangular cake into two equal pieces with one straight cut when someone has already removed a rectangular piece from it? (The removed piece can be of any size or any orientation.)
Answer:
This question definitely has a right answer. It might be argued that it involves a bit of a "trick", but I still like it. The trick is knowing or realizing that any line passing through the center point of a rectangle bisects it. Before you remove the rectangular piece from the cake, there are infinitely many lines which bisect the cake. After you remove the rectangular piece, there is only one - the line which passes through both the center of the cake, and the center of the removed rectangular piece. This line necessarily divides the removed piece in half, and hence the same amount of cake was removed from each half of the remaining portion.
The value in this question is not only seeing if a candidate can compute the answer, but watching them eliminate non-solutions. You would probably realize after a little trial-and-error that such a constraint is not helpful, and that might guide them toward the solution.
P.S. There is another solution - cut the cake in half vertically! (With a single horizontal slice.) I'd say this gets points for creativity, but I'd still want to see the candidate solve the problem the other way.
Question:
The $21 Question
Alan and Vicky have $21 between them. Alan has $20 more than Vicky. How much does each have (you can't use fractions in the answer)?
Answer:
I called this the worst question in the book, based on the fact that it has no answer. I went on to say:
Apparently sometimes people ask questions which have no answer to see how candidates react. This might be helpful in some situations (if you're hiring for a company with a confrontational culture!), but I would never use it; I don't like what it says about me and my company, and I can't imagine what it would say about the candidate, either.
However, I found out that this question does have an answer!
What does this illustrate? That many people apparently doesn't know that dollars are divided up into cents:
A = V + 2000¢
A + V = 2100¢
Hence,
A = 2050¢
V = 50¢
Excellent! When I read this question in the book it was described as having no answer, and it never occurred to me that the book was wrong, and that this question really does have an answer. This is a classic "thinking out of the box" test. Any candidate is going to do the algebra and conclude that there is no integer solution in dollars. Will they then consider shifting units to cents? Very interesting.
2008年1月21日星期一
燒紙錢的由來
古時候，有個叫尤文一的秀才，寒窗苦讀十幾年，卻沒能考中舉人。他便棄筆從商，投在大發明家蔡倫的門下學習紙。尤秀才聰明過人，很受蔡倫的器重。蔡倫就把自己的技術全部傳給了尤秀才。
過了幾年，蔡倫死了，尤秀才就繼承了蔡倫的事業，造起紙來。他造的紙又多又好，可當時用紙的人很少，造出的紙賣不出去，庫房裡堆積如山。為此，尤秀才十分犯愁，漸漸地茶飯不進，臥床不起，三天沒過，竟然閉上眼睛死去了。
家裡人頓時哭得死去活來。左鄰右舍知道了這個消息，也都過來幫助料理喪事。尤秀才的妻子哭著對大伙說：「咱們家境不好，沒有甚麼可以陪葬，就把這些紙燒了給他做陪葬吧！」
於是，專門派一個人在尤秀才的靈前燒紙。到了第三天，尤秀才突然坐起來，嘴裡還不停地叫著：「快燒紙，快燒紙。」人們以為尤秀才起屍，都害怕起來。尤秀才卻說：「不要害怕，我是真的活了，是閻王老爺把我放回來。」人們都感到十分奇怪，紛紛尋問根由
尤秀才說：「是你們燒的這些紙救了我。」
這些紙燒化之後，到了陰曹地府就變成了錢。我用這些錢買通了閻王爺，閻王老爺就把我放回來了。」家裡人聽了，無不歡天喜地，就又燒了不少紙。這件事傳出之後，也有人不相信。一個有錢有勢的老員外把尤秀才找去，對他說：「我家用金錢陪葬，不是比紙值錢的多嗎？」尤秀才說：「員外不知，這金銀是人間所用的，決帶不到地獄去，不信，員外可掘開祖墳，那些陪葬的金銀保證分毫沒動。」
員外聽了點頭稱是。於是，買紙的人一下子多起來，尤秀才造出的紙還供不上賣哩。其實，尤秀才並不是真的死而復生，袛不過是為了多賣紙，和妻子商量好設下的一個計策。然而，給死人燒紙的風俗卻一直流傳下來了
過了幾年，蔡倫死了，尤秀才就繼承了蔡倫的事業，造起紙來。他造的紙又多又好，可當時用紙的人很少，造出的紙賣不出去，庫房裡堆積如山。為此，尤秀才十分犯愁，漸漸地茶飯不進，臥床不起，三天沒過，竟然閉上眼睛死去了。
家裡人頓時哭得死去活來。左鄰右舍知道了這個消息，也都過來幫助料理喪事。尤秀才的妻子哭著對大伙說：「咱們家境不好，沒有甚麼可以陪葬，就把這些紙燒了給他做陪葬吧！」
於是，專門派一個人在尤秀才的靈前燒紙。到了第三天，尤秀才突然坐起來，嘴裡還不停地叫著：「快燒紙，快燒紙。」人們以為尤秀才起屍，都害怕起來。尤秀才卻說：「不要害怕，我是真的活了，是閻王老爺把我放回來。」人們都感到十分奇怪，紛紛尋問根由
尤秀才說：「是你們燒的這些紙救了我。」
這些紙燒化之後，到了陰曹地府就變成了錢。我用這些錢買通了閻王爺，閻王老爺就把我放回來了。」家裡人聽了，無不歡天喜地，就又燒了不少紙。這件事傳出之後，也有人不相信。一個有錢有勢的老員外把尤秀才找去，對他說：「我家用金錢陪葬，不是比紙值錢的多嗎？」尤秀才說：「員外不知，這金銀是人間所用的，決帶不到地獄去，不信，員外可掘開祖墳，那些陪葬的金銀保證分毫沒動。」
員外聽了點頭稱是。於是，買紙的人一下子多起來，尤秀才造出的紙還供不上賣哩。其實，尤秀才並不是真的死而復生，袛不過是為了多賣紙，和妻子商量好設下的一個計策。然而，給死人燒紙的風俗卻一直流傳下來了
Microsoft Interview Questions 3
Question:
Pay a person from a gold bar who is working with you for 7 days. (Only 2 Breaks is allowed.) How would you break the gold bar?)
Answer:
Two breaks give you a 4, 2 and a 1.
Day one, give him the 1. Tell him to bring it back the next day. End of day two, take back the 1 and give him the 2. Day 3, he gets the 1 back, but tell him to bring all 3 back the next day. Day 4 he gets the 4 and keeps it. Day 5 he gets the one, but has to bring it back on day 6. Day 6 he gets the 2, Day 7 he finally gets and keeps the 1 for a total of 7.
Question:
An analog clock reads 3:15. What is the angle between the minute hand and hour hand?
Simple Answer:
3 is 1/4 of 12 hours
Hence 1/4 * 360 = 90 degrees
15 is 1/4 of 60 minutes
Hence at 1/4 * 360 = 90 degrees
The hour hand will be 1/4 between the 3 and 4 which means 1/4 * (360 deg/12 hrs) = 7.5
So the angle between = 90 - 90 + 7.5 = 7.5 degrees
Interesting Question:
If a bear walks one mile south, turns left and walks one mile to the east and then turns left again, and walk one mile to the north, and it comes back to the starting point, what is the color of the bear?
Interesting Answer:
The color is white! It was because it is a polar bear and it starts at north pole!
To come back to the starting point, the polar bear would have to have started at the North Pole
Pay a person from a gold bar who is working with you for 7 days. (Only 2 Breaks is allowed.) How would you break the gold bar?)
Answer:
Two breaks give you a 4, 2 and a 1.
Day one, give him the 1. Tell him to bring it back the next day. End of day two, take back the 1 and give him the 2. Day 3, he gets the 1 back, but tell him to bring all 3 back the next day. Day 4 he gets the 4 and keeps it. Day 5 he gets the one, but has to bring it back on day 6. Day 6 he gets the 2, Day 7 he finally gets and keeps the 1 for a total of 7.
Question:
An analog clock reads 3:15. What is the angle between the minute hand and hour hand?
Simple Answer:
3 is 1/4 of 12 hours
Hence 1/4 * 360 = 90 degrees
15 is 1/4 of 60 minutes
Hence at 1/4 * 360 = 90 degrees
The hour hand will be 1/4 between the 3 and 4 which means 1/4 * (360 deg/12 hrs) = 7.5
So the angle between = 90 - 90 + 7.5 = 7.5 degrees
Interesting Question:
If a bear walks one mile south, turns left and walks one mile to the east and then turns left again, and walk one mile to the north, and it comes back to the starting point, what is the color of the bear?
Interesting Answer:
The color is white! It was because it is a polar bear and it starts at north pole!
To come back to the starting point, the polar bear would have to have started at the North Pole
Microsoft Interview Questions 3
Pay a person from a gold bar who is working with you for 7 days. (Only 2 Breaks is allowed.) How would you
Answer:
Answer:
Microsoft Interview Questions 2
Question:
FIND A JAR WITH A CONTAMINATED PILL FROM A SET OF JARS
You have 4 jars of pills. Each pill is a certain weight, except for contaminated pills contained in one jar, where each pill is weight + 1. How could you tell which jar had the contaminated pills in just one measurement?
Answer
Take 1 pill from jar 1, 2 pills from jar 2, 3 pills from jar 3, and 4 pills from jar 4, then weigh all 10 pills together. The total weight will be 10*weight + k, where k = # of contaminated pills that were measured. Thus, if k=1, the contaminated jar is jar 1, or if k=2, the contaminated jar is jar 2, etc.
Another interesting side point. You only need to take pills out of 3 jars (1 from jar1, 2 from jar2, and 3 from jar3). If the weight = 6*weight, then the contaminated jar is jar 1, otherwise proceed as usual. This is how you can determine the contaminated jar in one weighing with the minimal number of pills involved (assuming you can't break up a pill).
Question:
There are 4 women who want to cross a bridge.
They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each woman walks at a different speed. A pair must walk together at the rate of the slower woman's pace.
Woman 1: 1 minute to cross
Woman 2: 2 minutes to cross
Woman 3: 5 minutes to cross
Woman 4: 10 minutes to cross
For example if Woman 1 and Woman 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Woman 4 then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission. What is the order required to get all women across in 17 minutes? Now, what's the other way?
Answer:
1 and 2 go and 1 comes back. Total Time= 3 mins
Question:
IF YOU HAD AN INFINITE SUPPLY OF WATER AND A 5 QUART AND 3 QUART PAIL, HOW WOULD YOU MEASURE EXACTLY
If you had an infinite supply of water and a 5 quart and 3 quart pail, how would you measure exactly 4 quarts?
Answer:
Step 1: Fill the 5 Qaurt first and then empty it in the 3 Quart. So in the 5 Quart is filled with 2 Quart of water and the 3 Quart is fuly filled.
Step 2: Empty the 3 Quart
Step 3: Empty the water (2 Quarts) that is filled in the 5 Quart into the 3 Quart. So the 5 Quart is empty and the 3 Quart is filled with 2 Quarts of water.
Step 4: Now fill the 5 Quart fully with water and pour it in the 3 Quart till the 3 Quart becomes fully filled. Now the 5 Quart is filled with exactly 4 Quarts of water.
then 3 and 4 go and 2 comes back. Total Time in this = 12 mins
lastly 1 and 2 go. Total Time: 2 mins
Net total time to cross= 3+12+2= 17 mins
FIND A JAR WITH A CONTAMINATED PILL FROM A SET OF JARS
You have 4 jars of pills. Each pill is a certain weight, except for contaminated pills contained in one jar, where each pill is weight + 1. How could you tell which jar had the contaminated pills in just one measurement?
Answer
Take 1 pill from jar 1, 2 pills from jar 2, 3 pills from jar 3, and 4 pills from jar 4, then weigh all 10 pills together. The total weight will be 10*weight + k, where k = # of contaminated pills that were measured. Thus, if k=1, the contaminated jar is jar 1, or if k=2, the contaminated jar is jar 2, etc.
Another interesting side point. You only need to take pills out of 3 jars (1 from jar1, 2 from jar2, and 3 from jar3). If the weight = 6*weight, then the contaminated jar is jar 1, otherwise proceed as usual. This is how you can determine the contaminated jar in one weighing with the minimal number of pills involved (assuming you can't break up a pill).
Question:
There are 4 women who want to cross a bridge.
They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each woman walks at a different speed. A pair must walk together at the rate of the slower woman's pace.
Woman 1: 1 minute to cross
Woman 2: 2 minutes to cross
Woman 3: 5 minutes to cross
Woman 4: 10 minutes to cross
For example if Woman 1 and Woman 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Woman 4 then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission. What is the order required to get all women across in 17 minutes? Now, what's the other way?
Answer:
1 and 2 go and 1 comes back. Total Time= 3 mins
Question:
IF YOU HAD AN INFINITE SUPPLY OF WATER AND A 5 QUART AND 3 QUART PAIL, HOW WOULD YOU MEASURE EXACTLY
If you had an infinite supply of water and a 5 quart and 3 quart pail, how would you measure exactly 4 quarts?
Answer:
Step 1: Fill the 5 Qaurt first and then empty it in the 3 Quart. So in the 5 Quart is filled with 2 Quart of water and the 3 Quart is fuly filled.
Step 2: Empty the 3 Quart
Step 3: Empty the water (2 Quarts) that is filled in the 5 Quart into the 3 Quart. So the 5 Quart is empty and the 3 Quart is filled with 2 Quarts of water.
Step 4: Now fill the 5 Quart fully with water and pour it in the 3 Quart till the 3 Quart becomes fully filled. Now the 5 Quart is filled with exactly 4 Quarts of water.
then 3 and 4 go and 2 comes back. Total Time in this = 12 mins
lastly 1 and 2 go. Total Time: 2 mins
Net total time to cross= 3+12+2= 17 mins
Microsoft Interview Questions 1
Question:
Find the defective ball out of 8 ballsYou have 8 balls. One of them is defective and weighs less than others. You have a balance to measure balls against each other. In 2 weightings how do you find the defective one?
Answer
1.Picks 3 balls on the left and 3 balls on the right, if they are balanced, pick the other 2 balls to measure.
2. If 3 left balls is lesser weights in the first step, pick two out of three to be measured again, if they are balanced, you know the last ball is the defective one. Easy! Right?
Question:
CRAZY GUY ON THE AIRPLANE
A line of 100 airline passengers is waiting to board a plane. they each hold a ticket to one of the 100 seats on that flight. (for convenience, let's say that the nth passenger in line has a ticket for the seat number n.)
Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. all of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. if it is occupied, they will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
First Way of Doing It:
Let P(i) = probability that the ith person after the crazy guy loses their seat. What we want to know is P(99).
P(1) = 1/100 (the crazy guy picks seat 1). Notice that it is not 1/99 - the crazy guy could pick his own seat since he selects purely by random.
P(2) = 1/100 (crazy guy picks his seat) + P(1)*1/99 (crazy guy picks the first guy's seat, and the first guy picks the second guy's seat).
Following this sequence, we get the following recursive relationship:
P(n) = P(n-1)(1/(101-n))
If we start to work this relationship out, we get a surprising answer:
P(1) = 1/100
P(2) = 1/99
P(3) = 1/98
and, more generally,
P(n) = 1/(101-n) ( for n<100)
Thus, P(99) = 1/(101-99) = 1/2
Second Way of Doing It:
Let's start simply with 2 people. There a only 2 possible permutations (2x1) which are;
1 2
2 1
person 2 sits in his seat 1 out of the 2 possible combinations or 1/2.
Now 3 people. There are a total of 6 permutations (3x2x1) however 2 are not possible in this scenario as each person will sit in their seat if it is available. Therefore, apart from position 1, you cannot have a higher number in a particular position number. The impossible combinations I have marked with **
1 2 3
1 3 2 **
2 1 3
3 1 2
2 3 1 **
3 2 1
person 3 sits in his seat 2 out of the 4 possible combinations 2/4 = 1/2.
Now 4 people. There are a total of 24 permutations (4x3x2x1) and in this case there are 16 combinations that are not possible (marked **)
1 2 3 4
1 2 4 3 **
1 3 2 4 **
1 3 4 2 **
1 4 2 3 **
1 4 3 2 **
2 1 3 4
2 1 4 3 **
3 1 2 4
3 1 4 2 **
4 1 2 3
4 1 3 2
2 3 1 4 **
2 4 1 3 **
3 2 1 4
3 4 1 2 **
4 2 1 3
4 3 1 2 **
2 3 4 1 **
2 4 3 1 **
3 2 4 1 **
3 4 2 1 **
4 2 3 1
4 3 2 1 **
person 4 sits in his seat 4 out of the 8 possible combinations 4/8 = 1/2.
Question:
100 DOORS IN A ROW
You have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
Question: what state are the doors in after the last pass? which are open which are closed?
• The state of the door depends on the number of times it will be visited as follows:
if the number of times it is visited is even - the door will end up closed
otherwise - the door will end up open
• The number of times each door is visited is directly related to the number of dividers the number (door number) has (including 1 and itself).
• For example, every prime number door will be visited twice, and hence ends up closed
• Now, for every divisor of a number there must exist another divisor - that other devisor is the result of dividing with the first divisor! So, divisors for a number come in pairs... except for... when the number is a square of a divisor.
• If the number is a square of a divisor, then it means it has one divisor - the square root - that does not have a pair. Dividing the number by it's square root yields the square root itself, and not a new divisor.
• hence numbers that have an integer square root have an odd number of divisors, while numbers that do not have an integer square root do not.
• Therefore only 1,4,9,16,25,36,49,64,81,100 will be opened! and you don't need to write a computer program to solve this (which was certainly not the purpose of the question in an interview setting)
Find the defective ball out of 8 ballsYou have 8 balls. One of them is defective and weighs less than others. You have a balance to measure balls against each other. In 2 weightings how do you find the defective one?
Answer
1.Picks 3 balls on the left and 3 balls on the right, if they are balanced, pick the other 2 balls to measure.
2. If 3 left balls is lesser weights in the first step, pick two out of three to be measured again, if they are balanced, you know the last ball is the defective one. Easy! Right?
Question:
CRAZY GUY ON THE AIRPLANE
A line of 100 airline passengers is waiting to board a plane. they each hold a ticket to one of the 100 seats on that flight. (for convenience, let's say that the nth passenger in line has a ticket for the seat number n.)
Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. all of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. if it is occupied, they will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
First Way of Doing It:
Let P(i) = probability that the ith person after the crazy guy loses their seat. What we want to know is P(99).
P(1) = 1/100 (the crazy guy picks seat 1). Notice that it is not 1/99 - the crazy guy could pick his own seat since he selects purely by random.
P(2) = 1/100 (crazy guy picks his seat) + P(1)*1/99 (crazy guy picks the first guy's seat, and the first guy picks the second guy's seat).
Following this sequence, we get the following recursive relationship:
P(n) = P(n-1)(1/(101-n))
If we start to work this relationship out, we get a surprising answer:
P(1) = 1/100
P(2) = 1/99
P(3) = 1/98
and, more generally,
P(n) = 1/(101-n) ( for n<100)
Thus, P(99) = 1/(101-99) = 1/2
Second Way of Doing It:
Let's start simply with 2 people. There a only 2 possible permutations (2x1) which are;
1 2
2 1
person 2 sits in his seat 1 out of the 2 possible combinations or 1/2.
Now 3 people. There are a total of 6 permutations (3x2x1) however 2 are not possible in this scenario as each person will sit in their seat if it is available. Therefore, apart from position 1, you cannot have a higher number in a particular position number. The impossible combinations I have marked with **
1 2 3
1 3 2 **
2 1 3
3 1 2
2 3 1 **
3 2 1
person 3 sits in his seat 2 out of the 4 possible combinations 2/4 = 1/2.
Now 4 people. There are a total of 24 permutations (4x3x2x1) and in this case there are 16 combinations that are not possible (marked **)
1 2 3 4
1 2 4 3 **
1 3 2 4 **
1 3 4 2 **
1 4 2 3 **
1 4 3 2 **
2 1 3 4
2 1 4 3 **
3 1 2 4
3 1 4 2 **
4 1 2 3
4 1 3 2
2 3 1 4 **
2 4 1 3 **
3 2 1 4
3 4 1 2 **
4 2 1 3
4 3 1 2 **
2 3 4 1 **
2 4 3 1 **
3 2 4 1 **
3 4 2 1 **
4 2 3 1
4 3 2 1 **
person 4 sits in his seat 4 out of the 8 possible combinations 4/8 = 1/2.
Question:
100 DOORS IN A ROW
You have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
Question: what state are the doors in after the last pass? which are open which are closed?
• The state of the door depends on the number of times it will be visited as follows:
if the number of times it is visited is even - the door will end up closed
otherwise - the door will end up open
• The number of times each door is visited is directly related to the number of dividers the number (door number) has (including 1 and itself).
• For example, every prime number door will be visited twice, and hence ends up closed
• Now, for every divisor of a number there must exist another divisor - that other devisor is the result of dividing with the first divisor! So, divisors for a number come in pairs... except for... when the number is a square of a divisor.
• If the number is a square of a divisor, then it means it has one divisor - the square root - that does not have a pair. Dividing the number by it's square root yields the square root itself, and not a new divisor.
• hence numbers that have an integer square root have an odd number of divisors, while numbers that do not have an integer square root do not.
• Therefore only 1,4,9,16,25,36,49,64,81,100 will be opened! and you don't need to write a computer program to solve this (which was certainly not the purpose of the question in an interview setting)
2008年1月15日星期二
電腦最忌的 １４ 個小動作
１、大力敲擊Enter鍵
這個恐怕是人所共有的通病了，因為Enter鍵通常是我們完成一件事情時，最後要敲擊的一個鍵，大概是出於一種勝利的興奮感，每個人在輸入這個Enter鍵時總是那麼大力而爽快地敲擊。本人的多個鍵盤就是這樣報廢的，最先不看見字的是AWSD（呵呵，心知肚明），最先不能使用的按鍵卻是Enter。
2、在鍵盤上面吃零食，喝飲料
這個習慣恐怕是很普遍了，我看到很多人都是這樣的，特別是入迷者更是把電腦台當成飯桌來使用。我想你要是拆一回你的鍵盤，也許同樣的行為就會減少的，你可以看到你的鍵盤就像水積岩一樣，為你平時的習慣，保留了很多的“化石”，飯粒、餅乾渣、頭髮等等比比皆是，難怪有人說：公用機房裏的鍵盤比公廁還髒。同時這樣的碎片還可能進入你的鍵盤裏面，堵塞你鍵盤上的電路，從而造成輸入困難。飲料的危害就更加厲害了，一次就足以毀滅你的鍵盤。就是你的鍵盤僥倖沒有被毀滅，恐怕打起字來，也是粘粘糊糊很不好過。
3、光碟總是放在光碟機裏(還有看VCD時,暫停後出玩或吃飯!!!)
很多人總是喜歡把光碟放在光碟機裏，特別是CD碟，其實這種習慣是很不好的。光碟放在光碟機裏，光碟機會每過一段時間，就會進行檢測，特別是燒錄機，總是在不斷的檢測光碟機，而高倍速光碟機在工作時，電機及控制部件都會產生很高的熱量，為此光碟機廠商們一直在極力想辦法解決。
雖然現在已有幾種方法能將光碟機溫度控制在合理的範圍內，但如果光碟機長時間處於工作狀態，那麼，即使再先進的技術也仍無法有效控制高溫的產生。熱量不僅會影響部件的穩定性，同時也會加速機械部件的磨損和鐳射頭的老化。所以令光碟機長時間工作，實在是不智之舉，除非你想把你的光碟和光碟機煮熟。
4、關了機又馬上重新啟動
經常有人一關機就想起來光碟沒有拿出來，或者還有某個事情沒有完成等等，筆者就是其中一個，可以說有同樣毛病的人還是很多的。很多人反應迅速，在關閉電源的剛剛完成就能想起來，然後就伸出手來開機；更有DIY好手，總是動作靈敏，關機，十秒鐘處理完故障，重新開機；殊不知這樣對電腦危害有多大。
首先，短時間頻繁脈衝的電壓衝擊，可能會損害電腦上的積體電路；其次，受到傷害最大的是硬碟，現在的硬碟都是高速硬碟，從切斷電源到碟片完全停止轉動，需要比較長的時間。如果碟片沒有停轉，就重新開機，就相當於讓處在減速狀態的硬碟重新加速。長此下去，這樣的衝擊一定會使得你的硬碟一命歸西的。
5、開機箱蓋運行
開機箱蓋運行一看就知道是DIY們常幹的事情。的確開了機箱蓋，是能夠使得CPU涼快一些，但是這樣的代價是以犧牲其他配件的利益來實現的。因為開了機箱蓋，機箱裏將失去前後對流，空氣流將不再經過記憶體等配件，最受苦的是機箱前面的光碟機和硬碟們，失去了對流，將會使得他們位於下部的電路板產生的熱量變成向上升，不單單散不掉，還用來加熱自己，特別是燒錄機，溫度會比平時高很多。
不信你比較一下開不開機箱蓋的光碟機溫度。開機箱蓋還會帶來電磁輻射，噪音等危害，而且會使得機箱中的配件更加容易髒，帶來靜電的危害，並阻礙風扇的轉動。同時，讓其他隱患有機可乘，比如你在電腦前邊喝茶邊觀看一部片子，一個爆笑的鏡頭使你將口中的清茶悉數噴進了敞開的機箱內……
6、用手摸螢幕
其實無論是CRT或者是LCD都是不能用手摸的。電腦在使用過程中會在元器件表面積聚大量的靜電電荷。最典型的就是顯示器在使用後用手去觸摸顯示幕幕，會發生劇烈的靜電放電現象，靜電放電可能會損害顯示器，特別是脆弱的LCD。
另外，CRT的表面有防強光、防靜電的AGAS(Anti-GlareAnti -Static)塗層，防反射、防靜電的ARAS(Anti- ReflectionAnti- Static)塗層，用手觸摸，還會在上面留下手印，不信你從側面看顯示器，就能看到一個個手印在你的螢幕上，難道你想幫公安局叔叔們的忙，提前提取出傷害顯示器“兇手”的指紋嗎？同時，用手摸顯示器，還會因為手上的油脂破壞顯示器表面的塗層。
LCD顯示器比CRT顯示器脆弱很多，用手對著LCD顯示幕指指點點或用力地戳顯示幕都是不可取的，雖然對於CRT顯示器這不算什麼大問題，但LCD顯示器則不同，這可能對保護層造成劃傷、損害顯示器的液晶分子，使得顯示效果大打折扣，因此這個壞習慣必須改正，畢竟你的LCD顯示器並不是觸摸屏。
7、一直使用同一張牆紙或具有靜止畫面的屏保
無論是CRT或者是LCD的顯示器，長時間顯示同樣的畫面，都會使得相應區域的老化速度加快，長此下去，肯定會出現顯示失真的現象。要是你有機會看看機房裏的電腦，你就會發現，很多上面已經有了一個明顯的畫面輪廓。何況人生是多姿多彩的，何必老是用同一副嘴臉呢？
8、把光碟或者其他東西放在顯示器上
顯示器在正常運轉的時候會變熱。為了防止過熱，顯示器會吸入冷空氣，使它通過內部電路，然後將它從頂端排出。不信你現在摸摸你放在上面的光碟，是不是熱熱的象?O餅？若你總是把光碟或紙張放在顯示器上頭；更加誇張的是讓你家貓咪冬天時在上頭蜷著睡覺，當顯示器是溫床，這會讓熱氣在顯示器內部累積的。那麼色彩失真、影像問題、甚至壞掉都會找上你的顯示器。
9、拿電腦主機來墊腳
如果想要殺死你的臺式電腦，那麼開車帶它去越野兜風，或是背著它去爬山、蹦迪，那樣會更快一些；你的這種方法震動太小了，要比較長的時間才能出成績。如果你願意堅持下去，估計取得的第一個成績就是產生一出個圓滿歸西的是硬碟吧，死因是硬碟壞道。
10、電腦與空調、電視機等家用電器使用相同的電源插座
這是因為帶有電機的家電運行時會產生尖峰、浪湧等常見的電力污染現象，會有可能弄壞電腦的電力系統，使你的系統無法運作甚至損壞。同時他們在啟動時，也會和電腦爭奪電源，電量的小幅減少的後果是可能會突然令你的系統重啟或關機。
11、給你的電腦抽二手煙
就像香煙、雪茄或微小煙粒會傷害你的肺一樣，煙也可能會跑進你的軟盤機並危及資料。煙霧也可能會覆蓋CD-ROM、DVD驅動器的讀取頭，造成讀取錯誤。煙頭煙灰更有可能使得你的印表機和掃描器品質大大的下降。
12、不停的更換驅動程式
很多的DIY很喜歡不斷的更新驅動程式，雖然更新驅動程式有可能提升性能和相容性，但是?ˇA當的新版本可能會引起硬體功能的異常，在舊版本運轉正常的時候建議不要隨意升級驅動。先仔細閱讀驅動的README文件，對你有好處。就是像顯卡這樣更新換代迅速的硬體最好不要總是追新，不要隨便使用最新版的驅動程式，應該使用適合自己硬體情況的驅動程式，因為每一代的驅動程式都是針對當時市面上最流行的顯卡晶片設計，老晶片就不要隨便使用新的驅動，更不要隨便使用測試版的驅動，測試版的驅動就先留給網站的編輯們去測試他們的系統。
13、裝很多測試版的或者共用版的軟體
追新一族總是喜歡在自己的機子用上最新的軟體，和驅動程式一樣，更新程式有可能提升性能、增加功能和相容性，但是不適當的新版本可能會引起系統的異常。特別是測試版的程式，更是害處更多，既然沒有推出正式版，就說明該軟體還存在著很多不確定的BUG，這些小蟲就像定時炸彈一樣，隨時可能在你的系統中爆炸，損壞你的系統。
共用版的軟體有一些過一段時間（或次數）就會失效，要是你的系統通過共用版軟體更改了某方面的功能，而共用版軟體又因為失效而無法運行，那麼你的系統就不能回到你想要的狀態了；還有就是使用了共用版的軟體來建立的資料或者文檔，因為共用版軟體失效，而無法打開。所以安裝共用版時應當注意共用版提供使用的次數或者時間，以免無法還原系統和丟失資料
14、在系統運行中進行非正常重啟
在系統運行時，進行非正常重啟（包括按機箱上的重啟鍵、電源鍵和Ctrl Alt Del），可能使得系統丟失系統檔、存檔錯誤以及丟失設置等。本來windows是提供了磁片掃描工具，可以糾正部分出錯的檔，但是因為掃描需要一段比較長的時間，很多人都會中斷他的工作，經常出現這樣的情況，還有可能使得硬碟上的資料的出錯幾率和次數大大增加，從而使得整個系統崩潰。
這個恐怕是人所共有的通病了，因為Enter鍵通常是我們完成一件事情時，最後要敲擊的一個鍵，大概是出於一種勝利的興奮感，每個人在輸入這個Enter鍵時總是那麼大力而爽快地敲擊。本人的多個鍵盤就是這樣報廢的，最先不看見字的是AWSD（呵呵，心知肚明），最先不能使用的按鍵卻是Enter。
2、在鍵盤上面吃零食，喝飲料
這個習慣恐怕是很普遍了，我看到很多人都是這樣的，特別是入迷者更是把電腦台當成飯桌來使用。我想你要是拆一回你的鍵盤，也許同樣的行為就會減少的，你可以看到你的鍵盤就像水積岩一樣，為你平時的習慣，保留了很多的“化石”，飯粒、餅乾渣、頭髮等等比比皆是，難怪有人說：公用機房裏的鍵盤比公廁還髒。同時這樣的碎片還可能進入你的鍵盤裏面，堵塞你鍵盤上的電路，從而造成輸入困難。飲料的危害就更加厲害了，一次就足以毀滅你的鍵盤。就是你的鍵盤僥倖沒有被毀滅，恐怕打起字來，也是粘粘糊糊很不好過。
3、光碟總是放在光碟機裏(還有看VCD時,暫停後出玩或吃飯!!!)
很多人總是喜歡把光碟放在光碟機裏，特別是CD碟，其實這種習慣是很不好的。光碟放在光碟機裏，光碟機會每過一段時間，就會進行檢測，特別是燒錄機，總是在不斷的檢測光碟機，而高倍速光碟機在工作時，電機及控制部件都會產生很高的熱量，為此光碟機廠商們一直在極力想辦法解決。
雖然現在已有幾種方法能將光碟機溫度控制在合理的範圍內，但如果光碟機長時間處於工作狀態，那麼，即使再先進的技術也仍無法有效控制高溫的產生。熱量不僅會影響部件的穩定性，同時也會加速機械部件的磨損和鐳射頭的老化。所以令光碟機長時間工作，實在是不智之舉，除非你想把你的光碟和光碟機煮熟。
4、關了機又馬上重新啟動
經常有人一關機就想起來光碟沒有拿出來，或者還有某個事情沒有完成等等，筆者就是其中一個，可以說有同樣毛病的人還是很多的。很多人反應迅速，在關閉電源的剛剛完成就能想起來，然後就伸出手來開機；更有DIY好手，總是動作靈敏，關機，十秒鐘處理完故障，重新開機；殊不知這樣對電腦危害有多大。
首先，短時間頻繁脈衝的電壓衝擊，可能會損害電腦上的積體電路；其次，受到傷害最大的是硬碟，現在的硬碟都是高速硬碟，從切斷電源到碟片完全停止轉動，需要比較長的時間。如果碟片沒有停轉，就重新開機，就相當於讓處在減速狀態的硬碟重新加速。長此下去，這樣的衝擊一定會使得你的硬碟一命歸西的。
5、開機箱蓋運行
開機箱蓋運行一看就知道是DIY們常幹的事情。的確開了機箱蓋，是能夠使得CPU涼快一些，但是這樣的代價是以犧牲其他配件的利益來實現的。因為開了機箱蓋，機箱裏將失去前後對流，空氣流將不再經過記憶體等配件，最受苦的是機箱前面的光碟機和硬碟們，失去了對流，將會使得他們位於下部的電路板產生的熱量變成向上升，不單單散不掉，還用來加熱自己，特別是燒錄機，溫度會比平時高很多。
不信你比較一下開不開機箱蓋的光碟機溫度。開機箱蓋還會帶來電磁輻射，噪音等危害，而且會使得機箱中的配件更加容易髒，帶來靜電的危害，並阻礙風扇的轉動。同時，讓其他隱患有機可乘，比如你在電腦前邊喝茶邊觀看一部片子，一個爆笑的鏡頭使你將口中的清茶悉數噴進了敞開的機箱內……
6、用手摸螢幕
其實無論是CRT或者是LCD都是不能用手摸的。電腦在使用過程中會在元器件表面積聚大量的靜電電荷。最典型的就是顯示器在使用後用手去觸摸顯示幕幕，會發生劇烈的靜電放電現象，靜電放電可能會損害顯示器，特別是脆弱的LCD。
另外，CRT的表面有防強光、防靜電的AGAS(Anti-GlareAnti -Static)塗層，防反射、防靜電的ARAS(Anti- ReflectionAnti- Static)塗層，用手觸摸，還會在上面留下手印，不信你從側面看顯示器，就能看到一個個手印在你的螢幕上，難道你想幫公安局叔叔們的忙，提前提取出傷害顯示器“兇手”的指紋嗎？同時，用手摸顯示器，還會因為手上的油脂破壞顯示器表面的塗層。
LCD顯示器比CRT顯示器脆弱很多，用手對著LCD顯示幕指指點點或用力地戳顯示幕都是不可取的，雖然對於CRT顯示器這不算什麼大問題，但LCD顯示器則不同，這可能對保護層造成劃傷、損害顯示器的液晶分子，使得顯示效果大打折扣，因此這個壞習慣必須改正，畢竟你的LCD顯示器並不是觸摸屏。
7、一直使用同一張牆紙或具有靜止畫面的屏保
無論是CRT或者是LCD的顯示器，長時間顯示同樣的畫面，都會使得相應區域的老化速度加快，長此下去，肯定會出現顯示失真的現象。要是你有機會看看機房裏的電腦，你就會發現，很多上面已經有了一個明顯的畫面輪廓。何況人生是多姿多彩的，何必老是用同一副嘴臉呢？
8、把光碟或者其他東西放在顯示器上
顯示器在正常運轉的時候會變熱。為了防止過熱，顯示器會吸入冷空氣，使它通過內部電路，然後將它從頂端排出。不信你現在摸摸你放在上面的光碟，是不是熱熱的象?O餅？若你總是把光碟或紙張放在顯示器上頭；更加誇張的是讓你家貓咪冬天時在上頭蜷著睡覺，當顯示器是溫床，這會讓熱氣在顯示器內部累積的。那麼色彩失真、影像問題、甚至壞掉都會找上你的顯示器。
9、拿電腦主機來墊腳
如果想要殺死你的臺式電腦，那麼開車帶它去越野兜風，或是背著它去爬山、蹦迪，那樣會更快一些；你的這種方法震動太小了，要比較長的時間才能出成績。如果你願意堅持下去，估計取得的第一個成績就是產生一出個圓滿歸西的是硬碟吧，死因是硬碟壞道。
10、電腦與空調、電視機等家用電器使用相同的電源插座
這是因為帶有電機的家電運行時會產生尖峰、浪湧等常見的電力污染現象，會有可能弄壞電腦的電力系統，使你的系統無法運作甚至損壞。同時他們在啟動時，也會和電腦爭奪電源，電量的小幅減少的後果是可能會突然令你的系統重啟或關機。
11、給你的電腦抽二手煙
就像香煙、雪茄或微小煙粒會傷害你的肺一樣，煙也可能會跑進你的軟盤機並危及資料。煙霧也可能會覆蓋CD-ROM、DVD驅動器的讀取頭，造成讀取錯誤。煙頭煙灰更有可能使得你的印表機和掃描器品質大大的下降。
12、不停的更換驅動程式
很多的DIY很喜歡不斷的更新驅動程式，雖然更新驅動程式有可能提升性能和相容性，但是?ˇA當的新版本可能會引起硬體功能的異常，在舊版本運轉正常的時候建議不要隨意升級驅動。先仔細閱讀驅動的README文件，對你有好處。就是像顯卡這樣更新換代迅速的硬體最好不要總是追新，不要隨便使用最新版的驅動程式，應該使用適合自己硬體情況的驅動程式，因為每一代的驅動程式都是針對當時市面上最流行的顯卡晶片設計，老晶片就不要隨便使用新的驅動，更不要隨便使用測試版的驅動，測試版的驅動就先留給網站的編輯們去測試他們的系統。
13、裝很多測試版的或者共用版的軟體
追新一族總是喜歡在自己的機子用上最新的軟體，和驅動程式一樣，更新程式有可能提升性能、增加功能和相容性，但是不適當的新版本可能會引起系統的異常。特別是測試版的程式，更是害處更多，既然沒有推出正式版，就說明該軟體還存在著很多不確定的BUG，這些小蟲就像定時炸彈一樣，隨時可能在你的系統中爆炸，損壞你的系統。
共用版的軟體有一些過一段時間（或次數）就會失效，要是你的系統通過共用版軟體更改了某方面的功能，而共用版軟體又因為失效而無法運行，那麼你的系統就不能回到你想要的狀態了；還有就是使用了共用版的軟體來建立的資料或者文檔，因為共用版軟體失效，而無法打開。所以安裝共用版時應當注意共用版提供使用的次數或者時間，以免無法還原系統和丟失資料
14、在系統運行中進行非正常重啟
在系統運行時，進行非正常重啟（包括按機箱上的重啟鍵、電源鍵和Ctrl Alt Del），可能使得系統丟失系統檔、存檔錯誤以及丟失設置等。本來windows是提供了磁片掃描工具，可以糾正部分出錯的檔，但是因為掃描需要一段比較長的時間，很多人都會中斷他的工作，經常出現這樣的情況，還有可能使得硬碟上的資料的出錯幾率和次數大大增加，從而使得整個系統崩潰。
2008年1月14日星期一
2008年1月5日星期六
人生到底在追求甚麼
一個美國商人坐在墨西哥海邊一個小漁村的碼頭上，
看著一個墨西哥漁夫划著一艘小船靠岸。
小船上有好幾尾大黃鰭鮪魚，
這個美國商人對墨西哥漁夫能抓這麼高檔的魚恭維了一番，
還問要多少時間才能抓這麼多？
墨西哥漁夫說，才一會兒功夫就抓到了。
美國人再問，你為甚麼不待久一點，好多抓一些魚？
墨西哥漁夫覺得不以為然：
這些魚已經足夠我一家人生活所需啦！
美國人不以為然，幫他出主意，他說：
我是美國哈佛大學企管碩士，我倒是可以幫你忙！
你應該每天多花一些時間去抓魚，到時候你就有錢去買條大一點的船。
自然你就可以抓更多魚，再買更多漁船。
然後你就可以擁有一個漁船隊。
到時候你就不必把魚賣給魚販子，而是直接賣給加工廠。
然後你可以自己開一家罐頭工廠。
如此你就可以控制整個生產、加工處理和行銷。
然後你可以離開這個小漁村，搬到墨西哥城，再搬到洛杉磯，
最後到紐約。在那裡經營你不斷擴充的企業。
墨西哥漁夫問：這又花多少時間呢？
美國人回答：十五到二十年。
墨西哥漁夫問：然後呢？
美國人大笑著說：
然後你就可以在家當皇帝啦！
時機一到，你就可以宣佈股票上市，把你的公司股份賣給投資大眾。
到時候你就發啦！你可以幾億幾億地賺！
墨西哥漁夫問：然後呢？
美國人說：
到那個時候你就可以退休啦！
你可以搬到海邊的小漁村去住。
每天睡到自然醒，出海隨便抓幾條魚，
跟孩子們玩一玩，再跟老婆睡個午覺，
黃昏時，晃到村子裡喝點小酒，跟哥兒們玩玩吉他囉！
墨西哥漁夫疑惑的說：我現在不就是這樣了嗎？
人的一生，到底在追求甚麼？
那你的一生，到底在追求甚麼？
看著一個墨西哥漁夫划著一艘小船靠岸。
小船上有好幾尾大黃鰭鮪魚，
這個美國商人對墨西哥漁夫能抓這麼高檔的魚恭維了一番，
還問要多少時間才能抓這麼多？
墨西哥漁夫說，才一會兒功夫就抓到了。
美國人再問，你為甚麼不待久一點，好多抓一些魚？
墨西哥漁夫覺得不以為然：
這些魚已經足夠我一家人生活所需啦！
美國人不以為然，幫他出主意，他說：
我是美國哈佛大學企管碩士，我倒是可以幫你忙！
你應該每天多花一些時間去抓魚，到時候你就有錢去買條大一點的船。
自然你就可以抓更多魚，再買更多漁船。
然後你就可以擁有一個漁船隊。
到時候你就不必把魚賣給魚販子，而是直接賣給加工廠。
然後你可以自己開一家罐頭工廠。
如此你就可以控制整個生產、加工處理和行銷。
然後你可以離開這個小漁村，搬到墨西哥城，再搬到洛杉磯，
最後到紐約。在那裡經營你不斷擴充的企業。
墨西哥漁夫問：這又花多少時間呢？
美國人回答：十五到二十年。
墨西哥漁夫問：然後呢？
美國人大笑著說：
然後你就可以在家當皇帝啦！
時機一到，你就可以宣佈股票上市，把你的公司股份賣給投資大眾。
到時候你就發啦！你可以幾億幾億地賺！
墨西哥漁夫問：然後呢？
美國人說：
到那個時候你就可以退休啦！
你可以搬到海邊的小漁村去住。
每天睡到自然醒，出海隨便抓幾條魚，
跟孩子們玩一玩，再跟老婆睡個午覺，
黃昏時，晃到村子裡喝點小酒，跟哥兒們玩玩吉他囉！
墨西哥漁夫疑惑的說：我現在不就是這樣了嗎？
人的一生，到底在追求甚麼？
那你的一生，到底在追求甚麼？
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