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Is the certificate useful? Useful! In most cases, the certificate used to deal with the collective subconscious is still very useful, what is the collective subconscious? It is useful for everyone to think that it is useful, and everyone thinks it is useless. We grew up, out of the birth certificate (ID card? ), "at least" has endured three kinds of collective unconscious destruction: education, work and marriage.Among them, education is to refer to t

In fact, background development is nearing the end of the current background team as a secondary participation in the application of the interface debuggingOf course, the interface debugging will occur a lot of inexplicable situation, need to assist to solveBriefly describe the problems encountered:1, mobile phone number verification rules, the old mobile phone verification regular expression is not support the new number, now operators open a lot of new numbers, leading to the rule invalidation

, after all, now people like to call the outside unit for the experts, their units are identified as brick home!Do not give customers too many choices, more choices, he felt that you do not have the ability to let him to the dilemma, learning Apple Black Technology, violence to solve the general technical problems---a lot of technical problems, in fact, is the problem of human nature, human behavior problems.This article is from "behind the scenes player" blog, reprint please contact the author!

Can linux of the USB flash drive really save the mobile office work? -- Linux general technology-Linux technology and application information. For details, refer to the following section. The bandit was first launched. Recently, I always felt that there was a problem with the mobile hard disk, and there was always a problem where files were damaged and could not be copied. I thought there was a bad track. I

remaining B (n,k)? Take B (4,3) as an example to see if we can use B (3,?). Solve. B (4,3) = 11, can be divided into two categories: one is x4 in pairs appear, a class is x4 into a single appearance. Because k=3, so any 3 points can not shatter, namely: Α+β≤b (3,3). And because for 2α, X4 is in pairs appear, so, x1,x2,x3 any two points must not shatter, otherwise, plus X4, there will be three points are shatter. namely: Α≤b (3,2). Can be proved by mathema

= [Random.randrange (0, 100000, 1) forIinchRange (0, 100000)] 9L4 = [Random.randrange (0, 100000, 1) forIinchRange (0, 100000)] TenL5 = [Random.randrange (0, 100000, 1) forIinchRange (0, 100000)] OneL6 = [Random.randrange (0, 100000, 1) forIinchRange (0, 100000)] AL7 = [Random.randrange (0, 100000, 1) forIinchRange (0, 100000)] -s =time.time () - forLinch[L1, L2, L3, L4, L5, L6, L7]: the f (L) - Print "%s seconds"% (Time.time ()-s) -s =time.time () - map (f, [L1, L2, L3, L4, L5, L6,