![Trending Articles on Technical and Non Technical topics](/images/trending_categories.jpeg)
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
Find if the given number is present in the infinite sequence or not in C++
Suppose we have three integers a, b and c. Suppose in an infinite sequence, a is the first term, and c is a common difference. We have to check whether b is present in the sequence or not. Suppose the values are like a = 1, b = 7 and c = 3, Then the sequence will be 1, 4, 7, 10, …, so 7 is present in the sequence, so the output will be ‘yes’.
To solve this problem, we have to follow these two steps −
When c = 0, and a = b, then print yes, and if a is not same as b, then return no
When c > 0, then for any non-negative integer k, the equation will be b = a + k*c must be satisfied. So (b-a)/c will be a non-negative integer.
Example
#include<iostream> using namespace std; void isBInSequence(int a, int b, int c){ if (a == b) cout << "Yes"; if ((b - a) * c > 0 && (b - a) % c == 0) cout << "Yes"; else cout << "No"; } int main() { int a = 1, b = 7, c = 3; cout << "The answer is: "; isBInSequence(a, b, c); }
Output
The answer is: Yes
Advertisements