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C++ Articles
Page 284 of 597
Check if a number is divisible by 23 or not in C++
Here we will see one program, that can check whether a number is divisible by 23 or not. Suppose a number 1191216 is given. This is divisible by 23.To check the divisibility, we have to follow this rule −Extract the last digit of the number/truncated number every timeadd 7 * (last digit of the number calculated previous) to the truncated numberRepeat these steps as long as necessary.17043, so 1704 + 7*3 = 1725 1725, so 172 + 7 * 5 = 207 207, this is 9 * 23, so 17043 is divisible by 23.Example#include #include using namespace std; ...
Read MoreCheck if a number is divisible by 41 or not in C++
Here we will see one program, that can check whether a number is divisible by 41 or not. Suppose a number 104413920565933 is given. This is divisible by 41.To check the divisibility, we have to follow this rule −Extract the last digit of the number/truncated number every timesubtract 4 * (last digit of the number calculated previous) to the truncated numberRepeat these steps as long as necessary.30873, so 3087 - 4*3 = 3075 3075, so 307 - 4 * 5 = 287 287, so 28 – 4 * 7 = 0 So, 30873 is divisible by 41.Example#include #include ...
Read MoreSort elements of the array that occurs in between multiples of K in C++
Suppose we have an array A, and another integer K. We have to sort the elements that are in between any two multiples of K. Suppose A is like [2, 13, 3, 1, 21, 7, 8, 13, 12], and K = 2. The output will be [2, 1, 3, 7, 13, 21, 8, 13, 12]. Here multiple of 2 are 2, 8 and 12, the elements in between 2 and 8 are 13, 3, 1, 21, 7, they will be sorted as 1, 3, 7, 13, 21, the elements between 8 and 12 is only 13, so that is already ...
Read MoreMedian and Mode using Counting Sort in C++
Consider we have an array of size n, we have to find the Median and Mode using the counting sort technique. This technique is useful, when the array elements are in limited range. Suppose the elements are {1, 1, 1, 2, 7, 1}, then the Mode is 1, and Median is 1.5. Let us see what is Median and what is Mode −Median is the middle number in a sorted list of numbersMode is the element whose occurrences are maximum in the listTo get the Median and Mode, we have to follow these steps −Assume that the size of the ...
Read MoreMinimum Cost To Make Two Strings Identical in C++
Suppose we have two strings A and B, and another two cost values like CostA, and CostB. We have to find the minimum cost to make A and B identical. We can delete characters from string, the cost for deleting from string A is CostA, and cost for deleting from string B is CostB. Cost of removing all characters from a string is same. Suppose the string A = “wxyz”, B = “wyzx”, CostA is 10 and CostB is 20. So the output will be 30. If we delete x from both the strings, then A and B will be ...
Read MoreMinimum Index Sum for Common Elements of Two Lists in C++
Suppose two person wants to choose different cities, they have listed out the cities in different list, we have to help the, to find common choices. So we need to find those cities, those are marked by both of them.This operation is very similar to the set intersection property, we will take two lists as set, then perform the set intersection to get the common elements.Example#include #include #include using namespace std; vector commonInterest(string set1[], int n1, string set2[], int n2) { vector v(min(n1, n2)); vector::iterator it; // Sorting both the list sort(set1, set1 ...
Read MoreFind Corners of Rectangle using mid points in C++
Suppose we have a rectangle ABCD, but we have only the coordinates of the mid points P and Q, and the length of the rectangle L.Our task is to find the coordinates of A, B, C and D using the coordinates of P and Q, and the length of side L. For example, if P is (1, 0), and Q is (1, 2), and L is 2, then A, B, C, D will be respectively (0, 0), (0, 2), (2, 2). (2, 0).There can be three cases that can occur.The rectangle is horizontal, so AD and BC are parallel to ...
Read MoreFind the GCD of N Fibonacci Numbers with given Indices in C++
Here we have to find the GCD of n Fibonacci terms with the given indices. So at first we have to get the maximum index, and generate Fibonacci terms, some Fibonacci terms are like this: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ….. The index is starts from 0. So the element at 0th index is 0. If we have to find gcd of Fibonacci terms at indices {2, 3, 4, 5}, then the terms are {1, 2, 3, 4}, so GCD of these numbers are 1.We will use one interesting approach to do this task. To ...
Read MoreFind the Rotation Count in Rotated Sorted array in C++
Consider we have an array, which is rotated sorted array. We have to find number of rotations are required to sort the array. (We will consider rotation right to left.)Suppose the array is like: {15, 17, 1, 2, 6, 11}, then we have to rotate the array two times to sort. The final order will be {1, 2, 6, 11, 15, 17}. Here output is 2.The logic is simple. If we notice, we can see that the number of rotation is same as the value of index of minimum element. So if we get the minimum element, then its index ...
Read MoreFinding LCM of more than two (or array) numbers without using GCD in C++
We have an array A, we have to find the LCM of all elements without using the GCD operation. If the array is like {4, 6, 12, 24, 30}, then the LCM will be 120.The LCM can be calculated easily for two numbers. We have to follow this algorithm to get the LCM.getLCM(a, b) −begin if a > b, then m := a, otherwise m := b while true do if m is divisible by both a and b, then return m m := m + ...
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