Download CSE 1502 Spring 2005 Project 1: Finding Prime Numbers with C++ and more Study Guides, Projects, Research Computer Science in PDF only on Docsity! CSE 1502 Spring 2005 Introduction to Software Development with C++ Instructor: Venkatesh Ramamoorthy Project 1 Turn in by 31-March-2005 Problem Statement For a given positive number p, write a C++ program that finds and prints all of the prime numbers between 2 and p (both inclusive). A prime number is a number that can only be divided by one and itself (e.g. the numbers 2, 3, 5, 7, 11, 13, 17, . . .). Note that 0 and 1 are not in the example list; this is because 0 and 1 are neither prime nor composite. Note also, that 2 is the only even prime number. Inputs The positive number p up to which the first several prime numbers are required to be displayed. Outputs The list of prime numbers 2, 3, 5, 7, . . . p, if p is prime. If p is not prime, p is excluded from this list. Validations Based upon the discussion in the problem statement, what validations can you think of, that need to be applied on the input p? Test values The list of prime numbers up to, and including, 107, are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107 File Name Ensure that the file-name in which you are turning in this program, is called primes.cpp Project Name The project name in which this Project is turned in is project01. Acknowledgments This assignment originally appeared in a specialized form in “Absolute C++” by Walter Savich, (Instructor’s compli- mentary copy), Pearson Education, Inc., Addison-Wesley, Problem 4, Page 93. Hints One way to solve this problem is to use a doubly-nested loop. The outer loop runs with a counter value from 2 to p, while the inner loop checks to see whether the counter value for the outer loop is prime. That leaves us with the following question: How do we “check” to see if a number is prime or not? One way to decide whether a number p is prime is to loop from 2 to (p − 1); if any of these numbers evenly divides p, then p cannot be prime. If none of the values from 2 to (p − 1) evenly divide p, then p must be prime. (Note that there are several easy ways to make this algorithm more efficient).