Find LCM of Rational Number in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:59:16

195 Views

Here we will see how to find the LCM of Rational numbers. We have a list of rational numbers. Suppose the list is like {2/7, 3/14, 5/3}, then the LCM will be 30/1.To solve this problem, we have to calculate LCM of all numerators, then gcd of all denominators, then the LCM of rational numbers, will be like −$$LCM =\frac{LCM\:of\:all\:𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟𝑠}{GCD\:of\:all\:𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟𝑠}$$Example Live Demo#include #include #include using namespace std; int LCM(int a, int b) {    return (a * b) / (__gcd(a, b)); } int numeratorLCM(vector vect) {    int result = vect[0].first;    for (int i = 1; i ... Read More

Find if nCr is Divisible by Given Prime in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:52:50

209 Views

Suppose there are three variables N, R and P. The N and R are used to get the NCR and P is a prime. We have to find whether NCR is divisible by P. Suppose we have some numbers N = 7, R = 2 and P = 3, then 7C2 = 21, this is divisible by 3, so the output will be true.We know that NCR = N! / (R! * (N – R)! ). We will use Legendre Formula to largest power of P, which divides any N!, R! and (N – R)! in order to NCR to ... Read More

Find If a Point Lies Inside a Circle in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:46:31

3K+ Views

Suppose, one circle is given (the center coordinate and radius), another point is also given. We have to find whether the point is inside the circle or not. To solve it, we have to find the distance of the given point from circle center. If that distance is less or equal to the radius, then that is inside the circle, otherwise not.Example Live Demo#include #include using namespace std; bool isInsideCircle(int cx, int cy, int r, int x, int y) {    int dist = (x - cx) * (x - cx) + (y - cy) * (y - cy);    if ( dist

Install Lynis Linux Auditing Tool in CentOS

Lakshmi Srinivas
Updated on 21-Oct-2019 07:46:27

643 Views

Lynis is an open source and a powerful auditing tool for Unix-like operating systems. It scans system for protection knowledge, common approach know-how, any pre-installed software’s and to be had knowledge, configuration mistakes, safety disorders, consumer accounts without password, unsuitable file permissions, firewall auditing and so forth.Understanding LynisLinux performs individual test cases to secure your Linux system. To display test report, follow the below cases –Determine operating systemSearch for available tools and utilitiesCheck for Lynis updateRun tests from enabled pluginsRun security tests per categoryReport status of security scanDuring the test cases, all scanned details are stored in a log file ... Read More

Find HCF of Two Numbers Without Recursion or Euclidean Algorithm in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:43:18

773 Views

As we know, the HCF or GCD can be calculated easily using the Euclidean Algorithm. But here we will see how to generate GCD or HCF without using the Euclidean Algorithm, or any recursive algorithm. Suppose two numbers are present as 16 and 24. The GCD of these two is 8.Here the approach is simple. If the greater number of these two is divisible by the smaller one, then that is the HCF, otherwise starting from (smaller / 2) to 1, if the current element divides both the number, then that is the HCF.Example Live Demo#include using namespace std; int ... Read More

Find GCD of Factorial of Elements in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:39:53

435 Views

Suppose we have an array A, with N elements. We have to find the GCD of factorials of all elements of the array. Suppose the elements are {3, 4, 8, 6}, then the GCD of factorials is 6. Here we will see the trick. As the GCD of two numbers, is the greatest number, which divides both of the numbers, then GCD of factorial of two numbers is the value of factorial of the smallest number itself. So gcd of 3! and 5! is 3! = 6.Example Live Demo#include using namespace std; long fact(int n){    if(n arr[i])   ... Read More

Minimum Positive Integer Divisible by C and Not in Range in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:36:23

141 Views

Suppose we have three integers A, B, and C. We have to find one minimum integer X, such that X mod C = 0, and X is not in the range [A, B]. If the values of A, B and C are 5, 10 and 4 respectively, then the value of X will be 4. Let us see the steps to get the solution −Steps −If C is not in the range [A, B], then return C as a resultOtherwise get the first multiple of C, which is greater than B, then return that valueExample Live Demo#include using namespace std; ... Read More

Install Docker on CentOS 7

Lakshmi Srinivas
Updated on 21-Oct-2019 07:34:12

512 Views

Docker is an open-source project that automates the deployment of application inside the software container. The container allows the developer to package up all project resources such as libraries, dependencies, assets etc. Docker is written in Go Programming language and is developed by Dotcloud. It is basically a container engine which uses the Linux Kernel features like namespaces and control groups to create containers on top of an operating system and automates the application deployment on the container.Installing DockerBefore installing Docker on CentOS, it is required to update CentOS packages. To update the packages, use the following commands-# yum -y ... Read More

Minimum Operations to Make GCD of Array a Multiple of K in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:33:39

174 Views

Suppose we have an array arr and another value k. We have to find a minimum number of operations to make the GCD of the array equal to the multiple of k. In this case, the operation is increasing or decreasing the value. Suppose the array is like {4, 5, 6}, and k is 5. We can increase 4 by 1, and decrease 6 by 1, so it becomes 5. Here a number of operations is 2.We have to follow these steps to get the result −Steps −for all elements e in the array, follow steps 2 and 3if e ... Read More

Minimum Manipulations to Make Two Strings Anagram in C++

Arnab Chakraborty
Updated on 21-Oct-2019 07:28:24

499 Views

Suppose we have two strings of equal length, we have to find a minimum number of alterations required to make two strings anagram, without deleting any character. The Anagram is two strings that have the same set of characters. Suppose two strings are “HELLO”, and “WORLD” here number of required changes is 3, as three characters are different in this case.The idea is simple, we have to find the frequency of each character in the first string, then go through the second string, if characters in the second string are present, in the frequency array, then decrease the frequency value. ... Read More

Advertisements