mass multiplayer

Gfm200 thermal gas Mass FlowmeterGas measurement experts have been manufacturing flow products for 20 years.● Direct measurement of medium quality flow without temperature and pressure compensation● Large Diameter Pipeline measurement Optimization● It is not sensitive to particulate matter in the medium, and the effect can be ignored.● The proprietary flow algorithm is suitable for high humidity gases and high temperature gases without affecting the a

,: so, assuming the sample library, there is 2^34 data (17.1 billion data), Assuming that the data is evenly distributed, the maximum number of rows returned for each 16-bit (16 01-digit randomly composed combination of 2^16) is 2^34/2^16=2^ (34-16) = 262,144 candidate results, 4 16-bit truncated indexes, and the total result is: 4*262144= 1048576, about more than 1 million, through the reduction of dimensional processing, the original need to compare 17.1 billion times, now only need to compare

", "" "+" account Expired "," 43 "= > "IP Address Restrictions", "+" and "content contains sensitive words" ); $smsapi= "http://api.smsbao.com/"; $username= ' kill lui gung dog '; $user= "Test";//SMS Platform Account $pass=MD5("test888");//SMS Platform Password $str= "Super Play Machine" How are you? You are a respected user. "$v[' name ']. "?? Your verification code is: ".$v[' Code ']; $content=$str;//SMS content to send $phone=$v[' Phome '];//mobile phone number to

FileReader (New File ("F:/number.txt"));String text = null;First read the first n data, build the heapdo {Text = Reader.readline ();if (text! = null) {array[index++] = integer.parseint (text);}} while (text! = NULL index Minheap heap = new Minheap (array);//Initialize Heapfor (int i:heap.heap) {System.out.print (i + "");}Heap. Buildminheap ();//Build a small top heapSystem.out.println ();System.out.println ("After building the small top heap:");for (int i:heap.heap) {System.out.print (i + "");

Problem Solving Ideas:1. Record the molecular weight with a double array2. Store the string in a character array, scan from the back, always remember "letter", the next character to determine whether it is a number, if it is a number: With an integer record, this code with the sum, subscript + +.Judge to see if the number is recorded, that is, whether the sum has entered the while loop and is assigned, if not assigned, the next character is not a number, and the W (total record) value is directl

million integers to see the relative bitmap in the 01,01, and if the change is 00, the 10,10 remains the same. After the stroke is finished, look at the bitmap, and the corresponding bit is 01 integer output.Scenario 2: You can also use a method similar to the 1th question to divide a small file. Then, in the small file, find the integers that are not duplicated and sort them. Then merge and take care to remove the duplicate elements.to 4 billion non-repeating integers of unsigned int, not orde

Test instructions: Give you u and l string 3 consecutive U for dangerous goods, give you x length of string, to find the number of dangerousIdea: RecursionNumber of security solutions: m[i]=m[i-1]+m[i-2]+m[i-3] Fibonacci sequence Upgrade (╯‵-′) ╯︵┻━┻1#include 2#include 3 using namespacestd;4 intN;5 #defineMAXN 516 Long LongM[MAXN];7 BOOLdatecin ()8 {9 if(SCANF ("%d", n)! =EOF)Ten { One if(n==0) A return false; - return true; - }ac the return false; - } -

-(UIImage *) Resizeimage: (UIImage *) imageWithquality: (cginterpolationquality) qualityRate: (cgfloat) rate{UIImage *resized =nil;cgfloat width = image. size. width * rate;cgfloat height = image. size. height * rate;Uigraphicsbeginimagecontext(cgsizemake(width, height));Cgcontextref context = Uigraphicsgetcurrentcontext();Cgcontextsetinterpolationquality(context, quality);[ImageDrawinrect:cgrectmake(0,0, width, height)]; resized = Uigraphicsgetimagefromcurrentimagecontext();Uigraphicsendimagec

Describe A known positive integer n is the product of two different prime numbers, trying to find the larger prime number. Input The input has only one row and contains a positive integer n. For 60% of data, 6≤n≤1000. For 100% of data, 6≤n≤2*10^9. Output The output has only one row and contains a positive integer p, which is the larger prime number. Sample input 21st Sample o

] print (' 6: ', _dataread) if (_dataread==message): #返回发送的文本内容及相应的发送成功的提示后, end Program Print (' 7:ok ') e=e+1 lcd_5110.lcd_write_string (' OK is: ' +str (E), 0,5) pyb.de Lay (24000) r=10 e=03. Effectsend a text message to all the mobile numbers listed in number, with the content tpyboard_gps. Here you can control the delay time, such as adding a random number, I will not write more, their own play it.

The most happy thing is to give leadership scoring, when you just pen, your bank card without hesitation sounded the text message, a sum of money just into the account, boiling hot, embarrassed to give a look of the evaluation it? On the stage said, to rely on the party and the State responsible attitude, with a pragmatic attitude towards leadership, earnestly fill in. The moment can not make a decision to fill in the field, back to the unit after a good thinking and re-fill. Just hand it in bef

it is easy to get a warning:For example, I have this warning when I perform the following partitioning method (parted)/dev/sdb Mkpart primary 0 2000GB"The resulting partition is not a properly aligned for best performance."Finding data from the Web can be known as a warning because of an unreasonable start-sector location in the partition.Here's how to fix it:View the values of several files separately# cat/sys/block/sdb/queue/optimal_io_size0# cat/sys/block/sdb/queue/minimum_io_size4096# Cat/s

]++; T/=Prime[i]; } CNT++; } } if(t!=1) {factor[cnt][0]=T; factor[cnt][1]=1; CNT++; }}ll sum (ll p,ll N) {if(p==0)return 0; if(n==0)return 1; if(n1) return((1+pow_mod (p,n/2+1))%mod*sum (p,n/2)%mod)%MOD; Else return((1+pow_mod (p,n/2+1))%mod*sum (p,n/2-1) +pow_mod (p,n/2)%mod)%MOD;}intMain () {intb; IsPrime (); while(~SCANF ("%d%d",a,b) {getfactor (a); ll ans=1; for(intI=0; i) {ans*= (SUM (factor[i][0],b*factor[i][1])%MOD); Ans%=MOD; } printf ("%i64d\n", ans); } retur

, management complex; In addition, we also made shards, do not do read and write separation necessary; and we are almost a master one from the node configurationProblem 3:redis Fork is mainly for the Rdb bar, remove is for what?Answer: Fork is not friendlyQuestion 4: How can I ensure that the RDB snapshot is accurate without fork and that there are other cow mechanisms?A: There are other ways, this is still in the exploratory phase, but the goal is not to forkIssue 5: There are many problems wit

16 bits of Simhash are the same. Choose 16-bit, 8-bit, 4-bit, you can test your choice according to your own data, although the smaller the number the more accurate, but the space will become larger. The storage space divided into 4 16-bit segments is 4 times times the size of a separate simhash storage space. Previously calculated 5000w data is 382 Mb, expanded 4 times times 1.5G or so, can also accept6. Simhash Storage and indexing1. When the text content is longer, the use of simhash accurac

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};BitSetbitSet=new BitSet(6);//将数组内容组bitmapfor(inti=0;i{bitSet.set(array[i],true);}System.out.println(bitSet.size());System.out.println(bitSet.get(3));}} The corresponding bit bit if there is a corresponding integer then pass bitset.get (x) will return true, reverse false. where x is the bitmap position subscript.Well, bitmap is here. The next blog is about the "Tiger Balm"-hash algorithm, which deals with massive amounts of data, and its application in the MapReduce framework.Transfer f

for (1+ans[i]^1+ans[i]^2+.....+ans[i]^num[i])%mod= (num[i]+1)%mod; Sum= (sum* (num[i]+1))%MoD; Continue; } A=pow (ans[i],num[i]+1); A= (A-1)%MoD; Extend (Ans[i]-1, mod,x,y);//because Ans[i] is a prime number, ans[i]-1 is even, so ans[i]-1 and 9901 coprimexx= (x%mod+mod)%MoD; A= ((a%mod) * (xx%mod))%MoD; Sum= (sum*a)%MoD; } printf ("%i64d\n", sum);}intMain () { while(SCANF ("%i64d%i64d", a,b)! =EOF) { if(a==0) {printf ("0\n"); Continue; } Else if(a==1|| b==0) {printf ("1\n"); Conti

Http://www.cnblogs.com/rausen/p/4138233.html#include "Sieve method for prime number" "mass factor decomposition" bzoj2721 [Violet 5] Cherry blossom

Title: http://acm.hdu.edu.cn/showproblem.php?pid=5288Test instructions: interval [l,r], if lAnalysis: Find out the contribution of each number to the answer. For each a[i], the position of the left A[i] nearest and can be rounded out a[i] l[i] and the right away from a[i] nearest and can be complete a[i] r[i], then a[i] The contribution to the answer is (r[i]-i) * (I-l[i]). How to find L[i], first the position of each number in the input order exists in the array, the enumeration of a[i] factor,