Server Side Programming Articles - Page 1314 of 2646

In this tutorial, we are going to write a program to evaluate the equation 2^(2^A) % B.We are going to find the value of the equation using a recursive function. Let's see the steps to solve the problem.Write a recursive function that takes 2 arguments A and B.If A is 1, then return 4 % B as 2^(2^1) % B = 4 % B.Else recursively call the function with A-1 and b.Return the result^2%B.Print the solutionExampleLet's see the code. Live Demo#include using namespace std; long long solveTheEquation(long long A, long long B) {    // 2^(2^1) % B = 4 ... Read More

In this tutorial, we are going to solve the below problem.Given an array of integers, we have to find two numbers A and B. All the remaining numbers in the array are the divisors of A and B.If a number is a divisor of both A and B, then it will present twice in the array.Let's see the steps to solve the problem.The max number in the array is one of the numbers from A and B. Let's say it is A.Now, B will be the second-largest number or the number which is not a divisor of A.ExampleLet's see the ... Read More

We can set window size using PhantomJS and Selenium webdriver in Python. To work with the PhantomJS, we should create a driver object of the webdriver.PhantomJS class.Then pass the path of the phantomjs.exe driver file as a parameter to the Class. Next, to set the window size, we shall use the set_window_size method and pass the dimensions as parameters to the method.To obtain the window size of the browser, we can use the get_window_size method.Syntaxdriver.set_window_size(800, 1000) print(driver.get_window_size())Examplefrom selenium import webdriver #set phantomjs.exe path driver = webdriver.PhantomJS(executable_path="C:\phantomjs.exe") driver.maximize_window() #launch URL driver.get("https://www.tutorialspoint.com/index.htm") #set new window size driver.set_window_size(800, 880) #obtain window size print(driver.get_window_size()) ... Read More

In this tutorial, we are going to solve the equation (ab)%m where a is a very large number.The equation (ab)%m=(a%m)*(a%m)...b_times. We can solve the problem by finding the value of a%m and then multiplying it b times.Let's see the steps to solve the problem.Initialize the numbers a, b, and m.Write a function to find the a%m.Initialize the number with 0.Iterate over the number in string format.Add the last digits to the number.Update the number with number modulo them.Get the value of a%m.Write a loop that iterates b times.Multiply the a%m and modulo the result with m.Print the result.ExampleLet's see the ... Read More

In this tutorial, we are going to solve the following problem.Given a string containing only characters a and b, our task is to delete the sub-string ab from the string. And print the remaining string.Here, the idea is very simple to solve the problem. Every string with only a's and b's will shrink to either a's or b's at the end.Let's see the steps to solve the problem.Initialize the string.Initialize two counter variables for a and b.Iterate over the given string.Count the a's and b'sFind the maximum from the a and b frequencies.Print the difference between the two.ExampleLet's see the ... Read More

Given an array arr[] containing integer elements. The goal is to count the number of Arithmetic Progression subsequences inside arr[]. The range of elements inside arr[] are [1, 1000000].Empty sequence or single element will also be counted.Let us understand with examples.For ExampleInput - arr[] = {1, 2, 3}Output - Count of AP (Arithmetic Progression) Subsequences in an array are: 8Explanation - The following subsequences will form AP:-{}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}Input - arr[] = {2, 4, 5, 8}Output - Count of AP (Arithmetic Progression) Subsequences in an array are: 12Explanation - The following subsequences ... Read More

The following problem is an example of daily newspaper crossword, here we are given a 2-Dimensional character array and the problem statement is to find out the given word from the 2-Dimensional character array maze.The search algorithm includes finding the individual characters from Top-to-Bottom, Right-to-Left and vice-versa but not diagonally.Let us understand with examplesInput- String word-LAYS2D String[ ] -  { "LOAPYS", "KAYSOT", "LAYSST",  "MLVAYS", "LAYSAA", "LAOYLS" };Output- Count of number of given strings in 2D character array: 7Explanation - As we are given with the string array of words and from them, we have to find the word “LAYS” which can ... Read More

We are given a tree which can have endless levels, a variable child which will store the number of children a node can have, a variable weight which will store the weight associated with the path and a variable path that will store the path and task is to calculate the count of number of paths which has weights equals to the X and there must be an at least one edge with the given weight.For ExampleInput - int child = 4, weight = 4, path = 4; Output - Count of number of paths whose weight is exactly X and ... Read More

Given three variables size, max_val, last_element as input. The goal is to find the count of different arrays that can be formed in a manner such that they have size elements, have elements between 1 and max_val and the first element is always 1 and the last element is always max_val. Also make sure that no two consecutive elements are the same.Let us understand with examples.For ExampleInput - size = 5, max_val = 3, last_element = 3Output - Count of arrays having consecutive element with different values are: 5Explanation - The arrays will be:-[ 1, 2, 3, 1, 3 ], ... Read More

We are given with an integer number let's say, N which is considered as a divisor and it will be divided with the numbers starting from the 1 - N and the task is to calculate the count of those divisors which have more number of set bits than the quotient when divided with the given number N.For ExampleInput - int N = 6Output - Count of divisors having more set bits than quotient on dividing N are: 5Explanation - Firstly, we will divide the number N with the numbers starting from 1 - N and calculate the set bits of ... Read More