Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
-
Economics & Finance
C++ Articles
Page 508 of 597
C++ program to rearrange all elements of array which are multiples of x in increasing order
We are given an integer type array as ‘int arr[]’ and an integer type variable as ‘x’. The task is to rearrange all the elements of an array in such a manner that they will be divisible by a given integer value ‘x’ and the arrangement should be in an increasing order.Let us see various input output scenarios for this −Input − int arr[] = {4, 24, 3, 5, 7, 22, 12, 10}, int x = 2Output − Rearrangement of all elements of array which are multiples of x 2 in decreasing order is: 4 10 3 5 7 12 22 24Explanation −we ...
Read MoreRecursive sum of digit in n^x, where n and x are very large in C++
We are given positive integer variables as ‘num’ and ‘x’. The task is to recursively calculate the num ^ x then add the digits of a resultant number till the single digit isn’t achieved and the resultant single digit will be the output.Let us see various input output scenarios for this −Input − int num = 2345, int x = 3Output − Recursive sum of digit in n^x, where n and x are very large are: 8Explanation − we are given positive integer values as num and x with the values as 2345 and power as 3. Firstly, calculate 2345 ^ 3 i.e. ...
Read MoreRecursive program to print all numbers less than N which consist of digits 1 or 3 only in C++
We are given an integer variable as N storing the positive integer type value. The task is to recursively print all the numbers less than given value N having digit 1, 3 or the combination of both.Let us see various input output scenarios for this −Input − int num = 40Output − Recursive program to print all numbers less than N which consist of digits 1 or 3 only are: 33 31 13 11 3 1Explanation − we are given a positive integer value as 40 stored in a variable num. Now, we will recursively find out all the numbers containing digits 1, ...
Read MoreRearrange array such that arr[i] >= arr[j] if i is even and arr[i]<=arr[j] if i is odd and j < i in C++
We are given an integer array containing odd and even integer values. The task is to rearrange an array in such a manner that arr[i] should be greater than or equals to arr[j] based on the condition that value at index arr[i] should be even and if value at arr[i] is odd then arr[i] should be less than equals to arr[j].Let us see various input output scenarios for this −Input − int arr[] = {5, 9, 10, 12, 32, 35, 67, 89}Output − Array after rearranging elements are: 12 32 10 35 9 67 5 89Explanation − we are given an array with ...
Read MoreMultiples of 3 and 5 without using % operator in C++
We can find the multiples using % operator without any hurdles. But, the problem states that we can't use % operator.Here, we make use of the + operator. We can get the multiples by adding 3 or 5 to the previous multiple. Let's see an example.Input15Output1 2 3 - Multiple of 3 4 5 - Multiple of 5 6 - Multiple of 3 7 8 9 - Multiple 3 10 - Multiple of 5 11 12 - Multiple of 3 13 14 15 - Multiple of both 3 and 5AlgorithmInitialise the number n.Initialise two number to keep track of next ...
Read MoreNumber of pairs with Bitwise OR as Odd number in C++
Given an array, we have to find the number of pairs whose Bitwise OR is an odd number. Let's see the example.Inputarr = [1, 2]Output1 There is only one pair whose Bitwise OR is an odd number. And the pair is (1, 2).AlgorithmInitialise the array with random numbers.Initialise the count to 0.Write two loops to get the pairs of the array.Compute the bitwise OR between every pair.Increment the count if the result is an odd number.Return the count.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getOddPairsCount(int arr[], int n) { int count ...
Read MoreNumber of pairs whose sum is a power of 2 in C++
Given an array, we have to find the number of pairs whose sum is a power of 2. Let's see the example.Inputarr = [1, 2, 3]Output1 There is only one pair whose sum is a power of 2. And the pair is (1, 3).AlgorithmInitialise array with random numbers.Initialise the count to 0.Write two loops to get all the pairs of the array.Compute the sum of every pair.Check whether the sum is a power of 2 or not using bitwise AND.Increment the count if the count is a power of 2.Return the count.ImplementationFollowing is the implementation of the above algorithm in ...
Read MoreNumber of pairs from the first N natural numbers whose sum is divisible by K in C++
Given numbers N and K, we have to count the number of pairs whose sum is divisible by K. Let's see an example.InputN = 3 K = 2Output1 There is only one pair whose sum is divisible by K. And the pair is (1, 3).AlgorithmInitialise the N and K.Generate the natural numbers till N and store them in an array.Initialise the count to 0.Write two loops to get all pairs from the array.Compute the sum of every pair.If the pair sum is divisible by K, then increment the count.Return the count.ImplementationFollowing is the implementation of the above algorithm in C++#include ...
Read MoreNumber of ordered points pair satisfying line equation in C++
The line equation that should be satisfied is y = mx + c. Given an array, m, and c, we have to find the number of order points satisfying the line equation. Let's see an example.Inputarr = [1, 2, 3] m = 1 c = 1Output2The pairs satisfying the line equation are2 1 3 2AlgorithmInitialise the array, m, and c.Write two loops to get all pairs from the array.Check whether the pair is satisfying the line equation or not.We can check the whether the equation is satisfied or not by substituting the values into the line equation.If the pair satisfies ...
Read MoreNumber of non-negative integral solutions of sum equation in C++
In this tutorial, we are going to write a program that finds the number non-negative integral solution of sum equation.The sum equation is x + y + z = n. You are given the number n, you need to find the number of solutions for the equation. Let's see an example.Input2Output6Solutions are0 0 2 0 1 1 0 2 0 1 0 1 1 1 0 2 0 0AlgorithmInitialise the number m.Initialise the count to 0.Write three nested loops to get all the combinations of three numbers.Check the validation of the equation.If the current numbers satisfies the equation, then increment ...
Read More