@Fay Zan Basically you don’t want user to input more than 9 digits for your field, This can be achieved using two ways programmatically or using layout views properties, In you xml simple give this attribute to your EditText android:maxLength=”9″ OR programmatically by checking the length of your field. for example suppose the id of … Read more
Mine’s similar to jprofitt’s but I divided them into peaks and valleys, so I can do some more stuff with it. I think his loop is much more cleaner than mine is, but I just wanted to test it out for myself.Don’t judge meeeeee This script just plots the points out and selects the peaks … Read more
Use a List (e.g. ArrayList) instead. And before passing it to the Graph API convert it back to Numbers[] array with toArray(). There’s also an example how to loop over the list in the toArray link. solved Add and retrieve elements with Numbers array in Java
You are actually getting a new random number every loop. Your problem is here: … if DeathDoor == door: … You are comparing an integer to a string, which always yields False. The correct way is if DeathDoor == int(door) Also resetting RandomNum and DeathDoor to zero is unecessary. Since you are just beginning to … Read more
I assume you want JS, despite the example being in gods know what const isLeapYearHelper = (year) => (year % 400 == 0 ? true : year % 100 == 0 ? false : year % 4 == 0 ? true : false) const isLeapYear = (str) => { const date = new Date(str) const … Read more
You have issues in your second loop, I don’t get the why you used the variable tongchiahet, The main thing that you have to use is a break statement, which helps you to stop iteration of second loop if so % chia == 0 in between iterations: Here is a working Example for you, And … Read more
Try this: string Input = “919982115672”; Console.Write(Search(Input)); public string Search(string Input) { Dictionary<string, string> PreDefinedSearchAndTexts = new Dictionary<string, string>(); PreDefinedSearchAndTexts.Add(“8211”, “Hello”); PreDefinedSearchAndTexts.Add(“82”, “My name is inigo montya”); PreDefinedSearchAndTexts.Add(“9821”, “You killed my father”); PreDefinedSearchAndTexts.Add(“5672”, “Prepare to die”); StringBuilder sb = new StringBuilder(); foreach(string Key in PreDefinedSearchAndTexts.Keys) { if(Input.Contains(Key)) { sb.Append(Key).Append(” : “).AppendLine(PreDefinedSearchAndTexts[Key]); } } return sb.ToString(); … Read more
First for future reference you should probably post that on code review, unless there is a specific problem then you should create a Minimal, Complete, and Verifiable post. There is nothing inherently wrong other than you do not check that N>0 which could lead to an infinite loop, j should be going to i not … Read more
You need to use some combinatory logic to do this. A straightforward way consist in converting your 10 bytes number into BCD representation first (Binary-coded decimal), then convert your BCD number into an ASCII string, which is quite simple. Have a look at this for example: https://en.wikipedia.org/wiki/Double_dabble 4 solved C language, How can I Convert … Read more
A very simple test, which works because Python integers are unlimited in resolution: return float(x) == int(float(x)) The result from raw_input isn’t a number, it’s a string. That’s why it has to be converted to float first. 11 solved How to test if a number is an integer or a float that only has a … Read more
i agree with @shmosel solution. But it’s true just when second number has only 3 digits. boolean compareTail(long one, long two) { int digits =(int) Math.log10(two)+1; long formated = one % ((long) Math.pow(10, digits)); return formated == two; } 0 solved how to test if Long ends with a number pattern on JAVA?
I think, this is program you are searching for. public class Palindrome { public static void main(String[] args) { for(int i=0;i<=25000;i++){ int number=i; int reverse=0; int temp=0; while(number>0){ temp=number%10; number=number/10; reverse=reverse*10+temp; // System.out.println(rev); if(i==reverse){ if(reverse%2!=0){ System.out.println(reverse); } } } } } } 1 solved Odd Palindrome generator (0 to 25000)
With simple logic ‘4111111111111111’.match(/.{1,4}/g) Other way var str=”4111111111111111″; var parts = []; for (var i = 0; i < str.length; i += 4) { parts.push(str.substring(i, i + 4)); } console.log(parts); solved Jquery – Split a string(credit card number) in 4 groups [duplicate]
Use BigInteger or BigDecimal class and make use of their divide() method. 3 solved Java how to divide very large numbers [closed]
There are 4 things I’d like to mention. Here’s some code and its output: def srk_func(words): current = [] lastc = [] for x in words: if len(current) == 0: current.append(int(x)) elif len(current) == 1: if current[0] < int(x): current.append(int(x)) else: if len(current) >= len(lastc): lastc = current current[:] = [] current.append(int(x)) elif len(current) >= … Read more