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Judging levels based on user points
lv1:1~50lv2:51~110lv3:111~180lv4:181~260......依次类推直到lv100
How to quickly and efficiently low based on the user's points to determine the level? If the words to judge a few data can also, if to 100 if, the efficiency is reduced, do you have any good method?
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Judging levels based on user
Recently in writing a small algorithm, need to deal with decimals, the first thought is to determine whether there is a decimal point in the number of decimal points to the end of several .... It's too much trouble to write a half, because a lot of value to deal with too much trouble ...
I finally wrote this to use, and I feel better.
public static Boolean isNumber1 (String str) {//Judge integral retur
1265. four-point common reference time limit: 1 second space limit: 65536 kb score: 0 gives four points in three-dimensional space (points and points are not the same ), determine whether the four vertices are in the same plane (the four vertices are collocated ). If it is a common area, "yes" is output; otherwise, "no
用户表:用户序号、用户等级、用户积分等级表:等级序号、等级名称、积分下限、积分上线1、v1、0、602、v2、61、1803、v3、181、2704、v4、271、1000
Known integrals, if the level and integral have a certain regularity, you can quickly determine the user's level according to the algorithm;So there is no regularity, so how to quickly determine the level?
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用户表:用户序号、用户等级、用户积分等级表:等级序号、等级名称、积分下限、积分上线1、v1、0、602、v2、61、1803、v3、181、2704、v4、271、1000
Known integ
value elegance, extensibility, and coding knowledge levels, it's recommended that you use the final code I gave earlier.Lecture on the roving summary reportSeemingly very simple problem, but there is so deep water ah, if not everyone's discussion, flatly will not get such results, and learn so much knowledge.There is no perfect code, we look forward to the better, continue to discuss it, perhaps after the communication collision will have a better solution!(Microsoft: Crouching trough, see you
http://lavasoft.blog.51cto.com/62575/79864/Deep java keyword instanceof2008-06-02 07:50:43Tags: Java keywords casual workplace instanceof original works, allow reprint, please be sure to use hyperlinks in the form of the original source of the article, author information and this statement. Otherwise, the legal liability will be investigated. http://lavasoft.blog.51cto.com/62575/79864 in-depth java keyword instanceof instanceof keyword is used to determine
to the "topological sequence" problem, because in the re-brush this topic, with a topological sequence can be done。 There are also various other practices on the Web. But the question is abstracted as "whether the number of points that are judged to be 0 is unique" can be AC. In fact, the condition of being able to produce a championship is: "Only one person has not lost the"---"Only one point in the 0. In this problem, you need to pay attention to t
. Net to determine whether an object is a numerical type. Summary (high nutrition content, including final code and running points),. net Value
After the previous article was published, it triggered a positive discussion and played a leading role. Thank you for your participation.
Excuse me: this problem is much harder than it looks.
There is still no correct answer to the discussion, but I have summarized
http://poj.org/problem?id=1064"The main topic"Given the length of a few ropes, ask if you want to cut out the length of the K-bar rope, which is the most common K-rope how long?IdeasTwo points. Set the nether to 0, the upper bound is set to the sum of all the rope length, each time to take the mid, judging if each rope length for mid-cut out whether the number of ≥k, meet the lower border is set to mid, otherwise the upper border is set to mid. Becaus
terminal to do is color, git log > Git.log; Cat Git.log; There's no color. This time we need to use the script command, and then use the above-written small program to demonstrate the good:
$./foo | CatNon-terminal$ SCRIPT-QC./foo | CatTerminal
The script command can fool the Foo command so that it thinks its standard output is a terminal. The-c option of the script command is dedicated to this:
-C COMMANDRun the COMMAND rather than an interactive shell. This
Idea: Diamond + a Right anglestruct node{ double x, y;} D[30];d ouble getdis (node A,node b) { return sqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y));} BOOL IsEqual (double x,double y) { if (x-y>=-1e-8x-y Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Determine whether a square can be formed by a given 4 points
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