- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Found 7346 Articles for C++
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
304 Views
Here we will see if one number can be represented as sum of two or more consecutive numbers or not. Suppose a number is 12. This can be represented as 3+4+5.There is a direct and easiest method to solve this problem. If a number is power of 2, then it cannot be expressed as sum of some consecutive numbers. There are two facts that we have to keep in mind.Sum of any two consecutive numbers is odd, then one of them will be odd, another one is even.Second fact is 2n = 2(n-1) + 2(n-1).Example Live Demo#include using namespace std; ... Read More
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
97 Views
Here we will see, if we can represent a number as sum of two non-zero powers of 2. So we will check the given number N can be represented as (2x + 2y) where x, y > 0. Suppose a number is 10, this can be represented as 23 + 21.The approach is simple. There are two cases. If the number n is even, it can be represented as 2x. Where x > 0. Another case is that is N is odd, it can never be represented as sum of powers of 2. We cannot use power as 0, so ... Read More
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
554 Views
In this section we will see, how to check a line segment is passes through the origin or not. We have two coordinate points to represent the endpoints of the line segment.The approach is simple. If we can form the equation of the straight line, and by placing (0, 0) into the equation, and the equation satisfies, then the line passes through the origin.Suppose the points are and So the equation of line passes through these two lines is −$$y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)*\lgroup x-x_{1}\rgroup+c$$Putting x = 0 and y = 0, we get$$x_{1}\lgroup y_{2}-y_{1}\rgroup=y_{1}\lgroup x_{2}-x_{1}\rgroup$$Example Live Demo#include using namespace std; bool checkPassOrigin(int x1, int ... Read More
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
212 Views
Here we will see how to check a number is divisible by 9 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 9, if the sum of digits is divisible by 9.Example Live Demo#include using namespace std; bool isDiv3(string num){ int n = num.length(); long sum = accumulate(begin(num), end(num), 0) - '0' * n; if(sum % 9 == 0) return true; return false; } int main() { string num = "630720"; if(isDiv3(num)){ cout
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
359 Views
Here we will see how to check a number is divisible by 8 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 8, if the number formed by last three digits are divisible by 8.Example Live Demo#include using namespace std; bool isDiv8(string num){ int n = num.length(); int last_three_digit_val = (num[n-3] - '0') * 100 + (num[n-2] - '0') * 10 + ((num[n-1] - '0')); if(last_three_digit_val % 8 == 0) return true; return false; } int main() { string num = "1754586672360"; if(isDiv8(num)){ cout
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
86 Views
Here we will see how to check a number is divisible by 75 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 75, when the number is divisible by 3 and also divisible by 25. if the sum of digits is divisible by 3, then the number is divisible by 3, and if last two digits are divisible by 25, then the number is divisible by 25.Example Live Demo#include using namespace std; bool isDiv75(string num){ int n = num.length(); long sum = accumulate(begin(num), end(num), ... Read More
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
503 Views
Here we will see how to check a number is divisible by 5 or not. In this case the number is very large number. So we put the number as string.To check whether a number is divisible by 5, So to check divisibility by 5, we have to see the last number is 0 or 5.Example Live Demo#include using namespace std; bool isDiv5(string num){ int n = num.length(); if(num[n - 1] != '5' && num[n - 1] != '0') return false; return true; } int main() { string num = "154484585745184258458158245285265"; if(isDiv5(num)){ cout
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
370 Views
Here we will see how to check a number is divisible by 3 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 3, if the sum of digits is divisible by 3.Example Live Demo#include using namespace std; bool isDiv3(string num){ int n = num.length(); long sum = accumulate(begin(num), end(num), 0) - '0' * n; if(sum % 3 == 0) return true; return false; } int main() { string num = "3635883959606670431112222"; if(isDiv3(num)){ cout
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
153 Views
Here we will see how to check a number is divisible by 25 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 25, when the last two digits are 00, or they are divisible by 25.Example Live Demo#include using namespace std; bool isDiv25(string num){ int n = num.length(); int last_two_digit_val = (num[n-2] - '0') * 10 + ((num[n-1] - '0')); if(last_two_digit_val % 25 == 0) return true; return false; } int main() { string num = "451851549333150"; if(isDiv25(num)){ cout
![Arnab Chakraborty](https://www.tutorialspoint.com/assets/profiles/123881/profile/60_2940367-1661162113.png)
152 Views
Here we will see how to check a number is divisible by 20 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 20, when that is divisible by 10, and after dividing 10, the remaining number is divisible by 2. So the case is simple. If the last digit is 0, then it is divisible by 10, and when it is divisible by 10, then the second last element is divisible by 2, the number is divisible by 20.Example Live Demo#include using namespace std; bool isDiv20(string ... Read More