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The answer to all questions is 42
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Waarom wil je de naald vinden? Weet je zeker dat je een naald zoekt? Wie heeft de naald daar neergelegd? Waarom wil je meerdere manieren weten? Wat is je doel? Menos
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Ik zou naar de supermarkt gaan en een nieuwe naald kopen. Dat is een stuk sneller en makkelijker. Menos
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it was really usefully .thanks
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its asuch a important information ......... thank you .............
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really helpfull information....thank u:)
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A better answer: Split the 8 pennies into 3 groups of 3,3,2 pennies. Weight the first two groups of 3 pennies each. Case 1) - They weight the same. Therefore, take third group of 2 pennies and find the lighter coin. Case 2) Group 1 weights more than Group 2. Take group 2 (3 pennies) and pick any 2 out of 3. If they weight the same, then the third penny is lighter. Answer is trivial if they don't weight the same. It works for any scenario. Always split the groups in 3. Menos
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Neil has it right. Dave - the question says less than three steps.
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I Can do it in one step: Look at the year on the pennies. If 7 pennies weigh the same it means that all 7 were manufactured either before or after 1982, when the mint changed the composition of the coin from solid copper to copper with a zinc core. the one that weighs differently was minted on the opposite side of 1982 as the 7 other pennies. Menos
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yes, firstly i will do marriage
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I would rather invite my client ti my wedding. afrer giving him a feast will set up the meeting there itself. Menos
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First I have one question my girlfriend getting married me or another person??? Suppose she getting married another person my answer is I feel 💯 because ture feeling and then I attended client meeting successfully ... Why because she think am not important person in her life .. so that reason of she getting married another person... I think client meeting more important .. because my client give me a full trust with me . Menos
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This sounds like one geared not so much towards getting the right answer, but getting to it the right way. If you think a bit and say "one", the interviewer will know you did it the brute-force way, doing the math. You'd get at the answer faster, and probably impress them more, if you think instead how many times a ten will be produced in doing that math, rather than what the actual result of the math will be. Menos
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http://www.purplemath.com/modules/factzero.htm
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To generate a zero, we need a (5,2) factor pair. For any given number N, we have at least N/2 number of multiples 2, so the number of zeroes can be determined by the count of numbers that have 5 as a factor (i.e we have more 2s than 5s) Roughly, we can count N/5 numbers that are multiples of 5, add to that numbers that are multiples of 5^2 (these will have two 5 factors) i.e N/25, add to that numbers that are multiples of 5^3 (these will have three 5 factors) and so on. For eg: 10! -> 2 multiples of 5 -> 2 zeroes 100! -> 20 multiples of 5 + 4 multiples of 25 -> 24 zeros 500! -> 100 multiples of 5 + 20 multiples of 25 + 4 multiples of 125 -> 124 zeros 1000! -> 200 multiples of 5 + 40 multiples of 25 + 8 multiples of 125 (5^3) + 1 multiple of 625 (5^4) -> 249 zeros Menos
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I think that at the faster solution you mean int getAlignedValue_Fast(int pageSize, int valueToAlign) { return valueToAlign & ~(pageSize-1); } Note: There is a difference between !(pageSize-1) and ~(pageSize-1) ~(0x11) is 0xee !(0x11) is 0 Menos
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I just wanted to point out that the "faster solution" only works if the pageSize is assumed to be a power of 2. For example, suppose pageSize = 10 (or 01010 in binary), and valueToAlign = 24 (or 11000 in binary), then the fast method would give 16, but it should be 20. Anyways, thanks for posting the question and solution. Menos
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@observer I see how the mask works out for the alignment, why it is works mathematically? Thanks Menos
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It can modeled as binary symmetric channel. As to my understanding, channel capacity can be acheived with perfect feedback and simple retransmission scheme, so i guess the answer is 1-H(0.1). Menos
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10/(1.23456...)=8.1 packets per sec
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Interviewer was correct. With probability 0.1 you have one retransmission, with probability (0.1)^2 you have two, etc., since you can also lose the retransmitted packets, reducing the throughput to approximately 0.89. Menos
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I have learnt from my college project how to take decision in group and work as team.we make several important points to deal with it and work on as a team.my college project is wideband spectrum sensing using sub-nyquist techniques for cognitive radio network.so,we learn how to do a multitasking work and also gets motivated about the project. Yes I am willing to travel. Menos
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In college project I learnt many things while making a project with my team members first thing I learnt to solve a complicated things without getting a irritated when i am not able to solve the complexity with my team members even teachers help us when we stuck while making a college project we learnt many skills it's very helpful for us. Yes I can travel if I got the opportunity to get something or to learn something new. Menos
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I have learn Android app development in my college project...yes I am willing to travel with this company.. Menos
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i think depends on demand and oppertunity.
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I think practically work experience is very helpful after diploma/degree otherwise dificulty. Menos
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selse job reqverment