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  • Why do users have to enter a 7-digit twitter PIN to grant my application access?

    - by Tony
    I am implementing some ruby on rails code tweet stuff for my users. I am creating the proper oauth link...something like http://twitter.com/oauth/authorize?oauth_token=y2RkuftYAEkbEuIF7zKMuzWN30O2XxM8U9j0egtzKv But after my test account grants access to twitter, it pulls up a page saying "You've successfully granted access to . Simply return to and enter the following PIN to complete the process. 1234567" I have no idea where the user should enter this PIN and why they have to do that. I don't think this should be a necessary step. Twitter should be redirecting the user to the callback URL I provided in the application settings. Does anyone know why this is happening? UPDATE I found this article that states I need to send my users to this URL (note "authenticate" instead of "authorize"): http://twitter.com/oauth/authenticate?oauth_token=y2RkuftYAEkbEuIF7zKMuzWN30O2XxM8U9j0egtzKv I made the change but Twitter redirects the user to the authorize path after he clicks "Allow" which then gives him the 7 digit PIN again!

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  • What is the best way to find the digit at n position in a decimal number?

    - by Elijah
    Background I'm working on a symmetric rounding class and I find that I'm stuck with regards to how to best find the number at position x that I will be rounding. I'm sure there is an efficient mathematical way to find the single digit and return it without having to resort to string parsing. Problem Suppose, I have the following (C#) psuedo-code: var position = 3; var value = 102.43587m; // I want this no ? (that is 5) protected static int FindNDigit(decimal value, int position) { // This snippet is what I am searching for } Also, it is worth noting that if my value is a whole number, I will need to return a zero for the result of FindNDigit. Does anyone have any hints on how I should approach this problem? Is this something that is blaringly obvious that I'm missing?

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  • How can you get the first digit in an int (C#)?

    - by Dinah
    In C#, what's the best way to get the 1st digit in an int? The method I came up with is to turn the int into a string, find the 1st char of the string, then turn it back to an int. int start = Convert.ToInt32(curr.ToString().Substring(0, 1)); While this does the job, it feels like there is probably a good, simple, math-based solution to such a problem. String manipulation feels clunky. Edit: irrespective of speed differences, mystring[0] instead of Substring() is still just string manipulation

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  • Bash alias with piping

    - by n8felton
    I'm not exactly sure what I'm doing wrong with this one. I'm trying to run the command alias localip='ip -4 -o addr show eth0 | egrep -o '([[:digit:]]{1,3}\.){3}[[:digit:]]{1,3}' | head -n 1' If I run the command ip -4 -o addr show eth0 | egrep -o '([[:digit:]]{1,3}\.){3}[[:digit:]]{1,3}' | head -n 1 I get the result I expect, however, when trying to create an alias with the command, I get -bash: syntax error near unexpected token `(' Any help would be appreciated. TIA.

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  • Binary on the Coat of Arms of the Governor General of Canada

    - by user132636
    Can you help me further this investigation? Here is about 10% of the work I have done on it. I present it only to see if there are any truly curious people among you. I made a video a few weeks ago showing some strange things about the Governor General's Coat of Arms and the binary on it. Today, I noticed something kinda cool and thought I would share. Here is the binary as it appears on the COA: 110010111001001010100100111010011 As DEC: 6830770643 (this is easily found on the web) Take a close look at that number. What do you notice about it? It has a few interesting features, but here is the one no one has pointed out... Split it down the middle and you have 68307 70643. The first digit is double the value of the last digit. The second digit is double the second last digit. The third digit is half of the third to last digit. And the middle ones are even or neutral. At first, I thought of it as energy. ++-nnnn+-- But actually you can create something else with it using the values. 221000211. See how that works. You may be asking why that is significant. Bare with me. I know 99% are rolling their eyes. 221000211 as base3 gives you this as binary: 100011101000111 100011101000111 as HEX is 4747, which converts to "GG". Initials of Governor General. GG.ca is his website. When you convert to base 33 (there are 33 digits in the original code) you get "GOV" Interesting? :D There is a lot more to it. I'll continue to show some strange coincidences if anyone is interested. Sorry if I am not explaining this correctly. By now you have probably figured out that I have no background in this. Which is why I am here. Thank you.

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  • How to generate a random unique string with more than 2^30 combination. I also wanted to reverse the process. Is this possible?

    - by Yusuf S
    I have a string which contains 3 elements: a 3 digit code (example: SIN, ABD, SMS, etc) a 1 digit code type (example: 1, 2, 3, etc) a 3 digit number (example: 500, 123, 345) Example string: SIN1500, ABD2123, SMS3345, etc.. I wanted to generate a UNIQUE 10 digit alphanumeric and case sensitive string (only 0-9/a-z/A-Z is allowed), with more than 2^30 (about 1 billion) unique combination per string supplied. The generated code must have a particular algorithm so that I can reverse the process. For example: public static void main(String[] args) { String test = "ABD2123"; String result = generateData(test); System.out.println(generateOutput(test)); //for example, the output of this is: 1jS8g4GDn0 System.out.println(generateOutput(result)); //the output of this will be ABD2123 (the original string supplied) } What I wanted to ask is is there any ideas/examples/libraries in java that can do this? Or at least any hint on what keyword should I put on Google? I tried googling using the keyword java checksum, rng, security, random number, etc and also tried looking at some random number solution (java SecureRandom, xorshift RNG, java.util.zip's checksum, etc) but I can't seem to find one? Thanks! EDIT: My use case for this program is to generate some kind of unique voucher number to be used by specific customers. The string supplied will contains 3 digit code for company ID, 1 digit code for voucher type, and a 3 digit number for the voucher nominal. I also tried adding 3 random alphanumeric (so the final digit is 7 + 3 digit = 10 digit). This is what I've done so far, but the result is not very good (only about 100 thousand combination): public static String in ="somerandomstrings"; public static String out="someotherrandomstrings"; public static String encrypt(String kata) throws Exception { String result=""; String ina=in; String outa=out; Random ran = new Random(); Integer modulus=in.length(); Integer offset= ((Integer.parseInt(Utils.convertDateToString(new Date(), "SS")))+ran.nextInt(60))/2%modulus; result=ina.substring(offset, offset+1); ina=ina+ina; ina=ina.substring(offset, offset+modulus); result=result+translate(kata, ina, outa); return result; } EDIT: I'm sorry I forgot to put the "translate" function : public static String translate(String kata,String seq1, String seq2){ String result=""; if(kata!=null&seq1!=null&seq2!=null){ String[] a=kata.split(""); for (int j = 1; j < a.length; j++) { String b=a[j]; String[]seq1split=seq1.split(""); String[]seq2split=seq2.split(""); int hint=seq1.indexOf(b)+1; String sq=""; if(seq1split.length>hint) sq=seq1split[hint]; String sq1=""; if(seq2split.length>hint) sq1=seq2split[hint]; b=b.replace(sq, sq1); result=result+b; } } return result; }

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  • Recognizing terminals in a CFG production previously not defined as tokens.

    - by kmels
    I'm making a generator of LL(1) parsers, my input is a CoCo/R language specification. I've already got a Scanner generator for that input. Suppose I've got the following specification: COMPILER 1. CHARACTERS digit="0123456789". TOKENS number = digit{digit}. decnumber = digit{digit}"."digit{digit}. PRODUCTIONS Expression = Term{"+"Term|"-"Term}. Term = Factor{"*"Factor|"/"Factor}. Factor = ["-"](Number|"("Expression")"). Number = (number|decnumber). END 1. So, if the parser generated by this grammar receives a word "1+1", it'd be accepted i.e. a parse tree would be found. My question is, the character "+" was never defined in a token, but it appears in the non-terminal "Expression". How should my generated Scanner recognize it? It would not recognize it as a token. Is this a valid input then? Should I add this terminal in TOKENS and then consider an error routine for a Scanner for it to skip it? How does usual language specifications handle this?

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  • jQuery - Compatibility Problem with Internet Explorer 7 and Opera

    - by Marius
    Hello there, I have this counter which counts + 1 every time somebody shares content from the site. When it happens, the social icon that was clicked will bounce. It works in Firefox,Chrome, IE8, and Opera, however the bouncing animation is wrong in opera. $.fn.countExternal = function(animSpeed, num) { // for each counter this.each(function(){ // select all the digit containers var span = $(this).children(); // count the num of digit containers var len = $(span).length; // get the current count u = $(span).text(); // copy variable and add increment(s) v = num + ''; // foreach digit container... for (i=v.length - 1; i >= 0; i--) { // ...check which digits are not affected by the increment(s) if (v.charAt(i) == u.charAt(i)) { break; } } // slice from the total number of digit containers the digits containers which needs updating. slce = len - (v.length - (i + 1)) var updates = $(span).slice(slce); // loop through each digit container and fade out ... $(updates).fadeTo(animSpeed, 0,function(){ $(updates).each(function(index){ f = i + 1 + index; // ...then pick the right digit and update the digit... $(this).text(v.charAt(f)); // ...before fading back in. Cycle complete. $(this).fadeTo(animSpeed, 1); }); }); }); }; }) (jQuery); Demo (NSFW) is here (look underneath the social sharing icons). Any idea how I can solve the IE, and possibly the Opera compatibility problem? Thank you for your time.

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  • What is the AssemblyFileVersion used for in C#?

    - by robUK
    Hello, In the assemblyInfo.cs I have AssemblyVersion and AssemblyFileVersion. Normally I just increment the AssemblyVersion like this. 1st digit: Major change 2nd digit: Minor change 3rd digit: bug fixes 4rd digit: Subversion revision number However, I am wondering what is the AssemblyFileVersion for, and when do I need to increment. Should it be the same as the assemblyVersion? Should I just comment it out, if I am not using it? Many thanks for any advice,

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  • How i find the greatest number from six or more digit number?

    - by Rajendra Bhole
    Hi,thanks in advance, I have the code in which i want to find out the greatest number from six numbers, the code as follows, if( (pixels->r == 244 || pixels->g == 242 || pixels->b == 245) || (pixels->r == 236 || pixels->g == 235 || pixels->b == 233) || (pixels->r == 250 || pixels->g == 249 || pixels->b == 247) || (pixels->r == 253 || pixels->g == 251 || pixels->b == 230) || (pixels->r == 253 || pixels->g == 246 || pixels->b == 230) || (pixels->r == 254 || pixels->g == 247 || pixels->b == 229)) { numberOfPixels1++; NSLog( @"Pixel data1 %d", numberOfPixels1); } if( (pixels->r == 250 || pixels->g == 240 || pixels->b == 239) ||(pixels->r == 243 || pixels->g == 234 || pixels->b == 229) || (pixels->r == 244 || pixels->g == 241 || pixels->b == 234) || (pixels->r == 251 || pixels->g == 252 || pixels->b == 244) || (pixels->r == 252 || pixels->g == 248 || pixels->b == 237) || (pixels->r == 254 || pixels->g == 246 || pixels->b == 225)) { numberOfPixels2++; NSLog( @"Pixel data2 %d", numberOfPixels2); } if( (pixels->r == 255 || pixels->g == 249 || pixels->b == 225) ||(pixels->r == 255 || pixels->g == 249 || pixels->b == 225) || (pixels->r == 241 || pixels->g == 231 || pixels->b == 195) || (pixels->r == 239 || pixels->g == 226 || pixels->b == 173) || (pixels->r == 224 || pixels->g == 210 || pixels->b == 147) || (pixels->r == 242 || pixels->g == 226 || pixels->b == 151)) { numberOfPixels3++; NSLog( @"Pixel data3 %d", numberOfPixels3); } if( (pixels->r == 235 || pixels->g == 214 || pixels->b == 159) ||(pixels->r == 235 || pixels->g == 217 || pixels->b == 133) || (pixels->r == 227 || pixels->g == 196 || pixels->b == 103) || (pixels->r == 225 || pixels->g == 193 || pixels->b == 106) || (pixels->r == 223 || pixels->g == 193 || pixels->b == 123) || (pixels->r == 222 || pixels->g == 184 || pixels->b == 119)) { numberOfPixels4++; NSLog( @"Pixel data4 %d", numberOfPixels4); } if( (pixels->r == 199 || pixels->g == 164 || pixels->b == 100) ||(pixels->r == 188 || pixels->g == 151 || pixels->b == 98) || (pixels->r == 156 || pixels->g == 107 || pixels->b == 67) || (pixels->r == 142 || pixels->g == 88 || pixels->b == 62) || (pixels->r == 121 || pixels->g == 77 || pixels->b == 48) || (pixels->r == 100 || pixels->g == 49 || pixels->b == 22)) { numberOfPixels5++; NSLog( @"Pixel data5 %d", numberOfPixels5); } if( (pixels->r == 101 || pixels->g == 48 || pixels->b == 32) ||(pixels->r == 96 || pixels->g == 49 || pixels->b == 33) || (pixels->r == 87 || pixels->g == 50 || pixels->b == 41) || (pixels->r == 64 || pixels->g == 32 || pixels->b == 21) || (pixels->r == 49 || pixels->g == 37 || pixels->b == 41) || (pixels->r == 27 || pixels->g == 28 || pixels->b == 46)) { numberOfPixels6++; NSLog( @"Pixel data6 %d", numberOfPixels6); } I have to find out greatest from numberOfPixels1....numberOfPixels6 from above code. There are any optimum way to find out the greatest number?

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  • regex: trim all strings directly preceeded by digit except if string belongs to predefined set of st

    - by Geert-Jan
    I've got addresses I need to clean up for matching purposes. Part of the process is trimming unwanted suffices from housenumbers, e.g: mainstreet 4a --> mainstreet 4. However I don't want: 618 5th Ave SW --> 618 5 Ave SW in other words there are some strings (for now: st, nd, rd, th) which I don't want to strip. What would be the best method of doing this (regex or otherwise) ? a wokring regex without the exceptions would be: a = a.replaceAll("(^| )([0-9]+)[a-z]+($| )","$1$2$3"); //replace 1a --> 1 I thought about first searching and substiting the special cases with special characters while keeping the references in a map, then do the above regex, and then doing the reverse substitute using the reference map, but I'm looking for a simpler solution. Thanks

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  • How to make validation for a textbox that accept only comma(,) & digit in c# web application?

    - by prateeksaluja20
    Hello Experts, I am working on a website.I am using C# 2008.I want to make a text box that accept only numbers & comma(,). for example-919981424199,78848817711,47171111747 or there may be a single number like 919981424199. I was able to do one thing My text box only containing number by using this Regular Expression validation.in its property-Validation Expression i wrote "[0-9]+". This is working but now my requirement is to send bulk SMS & each number is separated by (,). I tried a lot but not getting the ans.so please help me to sort out this problem. Thanks in advance.

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  • How to make validation for a textbox that accept only comma(,) & digit in asp.net application?

    - by prateeksaluja20
    I am working on a website. I am using C# 2008. I want to make a text box that accept only numbers & comma(,). for example-919981424199,78848817711,47171111747 or there may be a single number like 919981424199. I was able to do one thing My text box only containing number by using this Regular Expression validation.in its property-Validation Expression i wrote "[0-9]+". This is working but now my requirement is to send bulk SMS & each number is separated by (,). I tried a lot but not getting the ans. so please help me to sort out this problem.

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  • How do I check the validity of the Canadian Social Insurance Number in C#?

    - by user518307
    I've been given the assignment to write an algorithm in C# that checks the validity of a Canadian Social Insurance Number (SIN). Here are the steps to validate a SIN. Given an example Number: 123 456 782 Remove the check digit (the last digit): 123456782 Extract the even digits (2,4,6,8th digith): 12345678 Double them: 2 4 6 8 | | | | v v v v 4 8 12 16 Add the digits together: 4+8+1+2+1+6 = 22 Add the Odd placed digits: 1+3+5+7 = 16 Total : 38 Validity Algorithm If the total is a multiple of 10, the check digit should be zero. Otherwise, Subtract the Total from the next highest multiple of 10 (40 in this case) The check digit for this SIN must be equal to the difference of the number and the totals from earlier (in this case, 40-38 = 2; check digit is 2, so the number is valid) I'm lost on how to actually implement this in C#, how do I do this?

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  • Is your Credit Card Number valid?

    - by Rekha
    The credit card numbers may look like some random unique 16 digits number but those digits inform more than what we think it could be. The first digit of the card is the Major Industry Identifier: 1 and 2 -  Airlines 3  – Travel and Entertainment 4 and 5 -  Banking and Financial 6 – Merchandizing and Banking 7 – Petroleum 8 – Telecommunications 9 – National assignment The first 6 digits represent the Issuer Identification Number: Visa – 4xxxxx Master Card – 51xxxx & 55xxxx The 7th and following digits, excluding the last digit, are the person’s account number which leads to trillion possible combinations if the maximum of 12 digits is used. Many cards only use 9 digits. The final digit is the checksum or check digit. It is used to validate the card number using Luhn algorithm. How To Validate Credit Card Number? Take any credit card number, for example 5588 3201 2345 6789. Step 1: Double every other digit from the right: 5*2      8*2      3*2      0*2      2*2      4*2      6*2      8*2 ————————————————————————- 10        16        6          0          4          8      12        16 Step 2: Add these new digits to undoubled digits. All double digit numbers are added as a sum of their digits, so 16 becomes 1+6 = 7: Undoubled digits:       5          8          2          1          3          5          7          9 Doubled Digits:          10       16         6          0          4          8         12         16 Sum:  5+1+0+8+1+6+2+6+1+0+3+4+5+8+7+1+2+9+1+6 = 76 If the final sum is divisible by 10, then the Credit Card number is valid, if not, the number is invalid or fake!!! Hence the example is a fake number? via mint  cc and image credit This article titled,Is your Credit Card Number valid?, was originally published at Tech Dreams. Grab our rss feed or fan us on Facebook to get updates from us.

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  • Rotate a linked list

    - by user408041
    I want to rotate a linked list that contains a number. 123 should be rotated to 231. The function created 23 but the last character stays empty, why? typedef struct node node; struct node{ char digit; node* p; }; void rotate(node** head){ node* walk= (*head); node* prev= (*head); char temp= walk->digit; while(walk->p!=NULL){ walk->digit=walk->p->digit; walk= walk->p; } walk->digit=temp; } How I create the list: node* convert_to_list(int num){ node * curr, * head; int i=0,length=0; char *arr=NULL; head = NULL; length =(int) log10(((double) num))+1; arr =(char*) malloc((length)*sizeof(char)); //allocate memory sprintf (arr, "%d" ,num); //(num, buf, 10); for(i=length;i>=0;i--) { curr = (node *)malloc(sizeof(node)); (curr)->digit = arr[i]; (curr)->p = head; head = curr; } curr = head; return curr; }

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  • Project Euler 51: Ruby

    - by Ben Griswold
    In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 51.  I know I started back up with Python this week, but I have three more Ruby solutions in the hopper and I wanted to share. For the record, Project Euler 51 was the second hardest Euler problem for me thus far. Yeah. As always, any feedback is welcome. # Euler 51 # http://projecteuler.net/index.php?section=problems&id=51 # By replacing the 1st digit of *3, it turns out that six # of the nine possible values: 13, 23, 43, 53, 73, and 83, # are all prime. # # By replacing the 3rd and 4th digits of 56**3 with the # same digit, this 5-digit number is the first example # having seven primes among the ten generated numbers, # yielding the family: 56003, 56113, 56333, 56443, # 56663, 56773, and 56993. Consequently 56003, being the # first member of this family, is the smallest prime with # this property. # # Find the smallest prime which, by replacing part of the # number (not necessarily adjacent digits) with the same # digit, is part of an eight prime value family. timer_start = Time.now require 'mathn' def eight_prime_family(prime) 0.upto(9) do |repeating_number| # Assume mask of 3 or more repeating numbers if prime.count(repeating_number.to_s) >= 3 ctr = 1 (repeating_number + 1).upto(9) do |replacement_number| family_candidate = prime.gsub(repeating_number.to_s, replacement_number.to_s) ctr += 1 if (family_candidate.to_i).prime? end return true if ctr >= 8 end end false end # Wanted to loop through primes using Prime.each # but it took too long to get to the starting value. n = 9999 while n += 2 next if !n.prime? break if eight_prime_family(n.to_s) end puts n puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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  • MatheMagics - Guess My Age - Method 2

    - by PointsToShare
    © 2011 By: Dov Trietsch. All rights reserved MatheMagic – Guess My Age – Method 2 The Mathemagician stands on the stage and asks an adult to do the following: ·         Do the next few steps on your calculator, or the calculator in your phone, or even on a piece of paper. ·         Do it silently! Don’t tell me the results until I ask for them directly ·         Multiply your age by 2. ·         Add 7 to the result ·         Multiply the result by 5. ·         Tell me the result. I will nonetheless immediately tell you what your age is. How do I do this? Let’s do the algebra. Let A denote your age (2A + 7) 5 = 10A + 35 so it is of the 3 digit form XY5 Now make two numbers out of the result - The last digit and the number before it. The Last digit is obviously 5, the other 2 (or 3 for a centenarian) and this number is the age + 3. Example: I am 76 years old and here is what happens when I do the steps 76 x 2 = 152 152 + 7 = 159 159 x 5 = 795 This is made of 79 and 5. And … 79 – 3 = 76 A note to the socially aware mathemagician – it is safer to do it with a man. The chances of a veracious answer are much, much higher! The trick may be accomplished on any 2 or 3 digit number, not just one’s age, but if you want to know your date’s age, it’s a good way to elicit it. That’s All Folks PS for more Ageless “Age” mathemagics go to www.mgsltns.com/games.htm and also here: http://geekswithblogs.net/PointsToShare/archive/2011/11/15/mathemagics---guess-my-age-method-1.aspx

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  • C++ Program performs better when piped

    - by ET1 Nerd
    I haven't done any programming in a decade. I wanted to get back into it, so I made this little pointless program as practice. The easiest way to describe what it does is with output of my --help codeblock: ./prng_bench --help ./prng_bench: usage: ./prng_bench $N $B [$T] This program will generate an N digit base(B) random number until all N digits are the same. Once a repeating N digit base(B) number is found, the following statistics are displayed: -Decimal value of all N digits. -Time & number of tries taken to randomly find. Optionally, this process is repeated T times. When running multiple repititions, averages for all N digit base(B) numbers are displayed at the end, as well as total time and total tries. My "problem" is that when the problem is "easy", say a 3 digit base 10 number, and I have it do a large number of passes the "total time" is less when piped to grep. ie: command ; command |grep took : ./prng_bench 3 10 999999 ; ./prng_bench 3 10 999999|grep took .... Pass# 999999: All 3 base(10) digits = 3 base(10). Time: 0.00005 secs. Tries: 23 It took 191.86701 secs & 99947208 tries to find 999999 repeating 3 digit base(10) numbers. An average of 0.00019 secs & 99 tries was needed to find each one. It took 159.32355 secs & 99947208 tries to find 999999 repeating 3 digit base(10) numbers. If I run the same command many times w/o grep time is always VERY close. I'm using srand(1234) for now, to test. The code between my calls to clock_gettime() for start and stop do not involve any stream manipulation, which would obviously affect time. I realize this is an exercise in futility, but I'd like to know why it behaves this way. Below is heart of the program. Here's a link to the full source on DB if anybody wants to compile and test. https://www.dropbox.com/s/6olqnnjf3unkm2m/prng_bench.cpp clock_gettime() requires -lrt. for (int pass_num=1; pass_num<=passes; pass_num++) { //Executes $passes # of times. clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &temp_time); //get time start_time = timetodouble(temp_time); //convert time to double, store as start_time for(i=1, tries=0; i!=0; tries++) { //loops until 'comparison for' fully completes. counts reps as 'tries'. <------------ for (i=0; i<Ndigits; i++) //Move forward through array. | results[i]=(rand()%base); //assign random num of base to element (digit). | /*for (i=0; i<Ndigits; i++) //---Debug Lines--------------- | std::cout<<" "<<results[i]; //---a LOT of output.---------- | std::cout << "\n"; //---Comment/decoment to disable/enable.*/ // | for (i=Ndigits-1; i>0 && results[i]==results[0]; i--); //Move through array, != element breaks & i!=0, new digits drawn. -| } //If all are equal i will be 0, nested for condition satisfied. -| clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &temp_time); //get time draw_time = (timetodouble(temp_time) - start_time); //convert time to dbl, subtract start_time, set draw_time to diff. total_time += draw_time; //add time for this pass to total. total_tries += tries; //add tries for this pass to total. /*Formated output for each pass: Pass# ---: All -- base(--) digits = -- base(10) Time: ----.---- secs. Tries: ----- (LINE) */ std::cout<<"Pass# "<<std::setw(width_pass)<<pass_num<<": All "<<Ndigits<<" base("<<base<<") digits = " <<std::setw(width_base)<<results[0]<<" base(10). Time: "<<std::setw(width_time)<<draw_time <<" secs. Tries: "<<tries<<"\n"; } if(passes==1) return 0; //No need for totals and averages of 1 pass. /* It took ----.---- secs & ------ tries to find --- repeating -- digit base(--) numbers. (LINE) An average of ---.---- secs & ---- tries was needed to find each one. (LINE)(LINE) */ std::cout<<"It took "<<total_time<<" secs & "<<total_tries<<" tries to find " <<passes<<" repeating "<<Ndigits<<" digit base("<<base<<") numbers.\n" <<"An average of "<<total_time/passes<<" secs & "<<total_tries/passes <<" tries was needed to find each one. \n\n"; return 0;

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  • How to calculate checksum?

    - by Patel Rikin
    I m developing instrument driver and i want to know how to calculate checksum of frame. Explanation: Expressed by characters [0-9] and [A-F]. Characters beginning from the character after [STX] and until [ETB] or [ETX] (including [ETB] or [ETX]) are added in binary. The 2-digit numbers, which represent the least significant 8 bits in hexadecimal code, are converted to ASCII characters [0-9] and [A-F]. The most significant digit is stored in CHK1 and the least significant digit in CHK2. This is sample frame : <STX>2Q|1|2^1||||20011001153000<CR><ETX><CHK1><CHK2><CR><LF> and i want to know what is value of chk1 and chk2 and i am new in this so i m totally blank about how to calculate checksum i am not getting above 3rd and 4th point. can any one provide sample code for c#. Please help me.

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  • Project Euler 4: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 4.  As always, any feedback is welcome. # Euler 4 # http://projecteuler.net/index.php?section=problems&id=4 # Find the largest palindrome made from the product of # two 3-digit numbers. A palindromic number reads the # same both ways. The largest palindrome made from the # product of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of # two 3-digit numbers. import time start = time.time() def isPalindrome(s): return s == s[::-1] max = 0 for i in xrange(100, 999): for j in xrange(i, 999): n = i * j; if (isPalindrome(str(n))): if (n > max): max = n print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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