Enterprise Social Record的动态

?? Check out the latest Cocktails and Crime Analysis holiday episode! Merry Christmas, Happy Holidays and Happy New Year! Watch: https://lnkd.in/efZbdgwr #CTandCA #crimeanalysis #crimeanalyst #intelligenceanalysis #intelligenceanalyst #leapodcasts

https://lnkd.in/guRpnUcU Tired of watching the industry taking it's lead from others when there are great examples out there as to how to take control ourselves? One such way is to take charge of one's own evaluation rather than relying on what current market mechanisms tell you.

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#FoodAsMore ?? #SingaporeStyle #XieXie ?? Chang Ling L. for sharing this on #LinkedIn. I so consistently #authenticAlly admire #Singapore ???? in multiple dimensions, including #professionally. #DiversityIsYourStrength #UnityIsYourPower #LKY and #LKYlegacy #Respect #gLocally #ThankYou ?? #LiveLongAndProsper ??

Hello everyone, #31 - ???????????????? ???????????? 411 ?????? ???????????????? 137 ?????????????????? - Mohd Fahad Feel free to ask if there are any queries and do share your ideas too. 1. ???????????? ?????????????? 411 (??) ?????????? ???????????????????? ???????? ?????????????? ??-???????????????????? ?? ? Brute force for every starting point find all the end points that satisfy our answer. Time complexity will just be O(n*n). (??) ?????????????? ???????????? ?????????? ???????? ?????? ???????????? ? This can be solved using dp. For any state i , dp1[i]=max(dp1[i-1]+a[i],dp2[i-2]+a[i]) this gives us the maximum between taking the same drink again and changing drink options. (dp1[i] represents the maximum possible answer we can have by taking ith first drink from 0 to i, similary dp2[i] represent the same for second drink, a[i] represent first drink, b[i] represent second). (??) ???????? ?????? ?????????????? ???????????????????? ?????????????????? ???? ?? ? This was greedy question. Now for n=1,3,9 we can just print all n values’9’ the max possible. Meanwhile for 5 we have to end with ‘0’ or ‘5’ since cant end with ‘0’ then end and start with ‘5’. Now comes 2,4,8 for these you have to keep in mind that last 1,2,3 digits will be divided by them respectively. For 7 we have to see that for n=1,n=2 we have to print 7 and 77 else for odd number the pattern will be 99999x99999 and for even 99999xx999999, now comes case for n=6, for divisibility of 2 we do 8999999998 now to make is divisible by 3, if n is odd we can just reduct the middle element making is 8999998999998, or if nis even we can make 89999977999998. 2. ???? ???????????? ?????????????? 137 (??,??) ???????? ?????? ?????????? ???? ??-???????? ?????????????????? ?? ?????? ???? ? Keep a check on where the consecutiveness breaks using prefix sum. Now for all sub array i can just check if there is any break between the end point of subarray if not then the max element is push in ans or else -1. (??,??) ?????????????? ?????????? ?????? ???? ?????????????? ?????????? ?????????? ?? ?????? ???? ? I will put all the values in a vector and sort it descending as precomputation with their positions. Run a n*n loop finding the first and second row we will take now get the maximum from row i and maximum possible form row j. For the third position I will start running a loop from highest value up until i find a value that is not in same row and column as our other two already chosen value, check the answer and break it. Context - I am taking the initiative to write all my intuitions and solutions for the cp contests from platforms like Codeforces, Codechef, Leetcode and Atcoder. As I am preparing for the upcoming placement session this initiative is intended to learn to share my thinking and help others and learn myself too. Inputs that will help to make this better are highly appreciated from everyone. https://lnkd.in/g_puUr4t Thank you.

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When I was just turning 21, the most exotic stuff you could find in most grocery stores was Guinness, Fat Tire, or if you went to Central Market in Austin you could find Boddington's. Homebrewing grew in popularity because there was a taste segmentation that was NOT being served by the likes of Anheuser-Busch or the mega brewers who dominated. This created an ideal market for Goose Island, which promoted creativity in brewing - and the guys who spun off from there made a big splash! Fostering that creativity is why we have so many great beer choices today. Booms aren't supposed to last forever - their social role is to bring better technologies and choices to the masses. Investors see booms as opportunities, but ultimately that's over with - that's just the incentive to create the better result for the consumer. Once [whichever] industry reaches steady state, it's hum drum in the P&L and expansion department often -- but the result is a better world for consumers. And so it went with craft beer. The boom is over - but so what? Who needs an ongoing boom forever? The consumer's situation is improved. Its work is done.

Here’s a classic quant riddle (read this to understand how quants think): Let’s say you’re having a party and shopping for drinks. If 1 bottle of beer costs $1, how much does 100 bottles cost? Most people would say $100, but that’s wrong. Here’s why: — Even if 1 bottle costs $1, the store may give you a discount for buying in bulk so all 100 bottles only costs $80. In this case you get wholesale pricing, which works in your favor! — It could also go the other way: What if the store is the only one open and knows you really need the beer? They might say “sorry, we usually can’t sell you all of our bottles, but for $200 we’ll let you have it.” Now they’re using your order size against you cause it shows you really need the beer! — The price of the 100 bottles is now anywhere from $80-$200. If you left the house with just $100 like most people would, there’s a risk you get to the store and walk out short. That’s why it’s so important to understand non-linearity. Paul Wilmott talks about this more in the documentary “Quants - The Alchemists of Wall Street”. You can watch it for free (highly recommend): — Want to become a quant but don’t know where to start? We teach traders how to level up their coding and math skills. Check out our free materials and join the community at https://buff.ly/46fg3z8