8/22/2023 0 Comments A list of prime numbers up to 100![]() ![]() both inner and outer loops are checking only within possible limits. the even numbers are not checked even once throughout the process. Why this code performs better than already accepted ones: Checkout the results for different N values in the end. ![]() My code takes significantly lesser iteration to finish the job. \ If n is greater than sqrt (number), the number is not equally divisible by n. For each number from 2 to 100, find Remainder Number n, where n ranges from 2 to sqrt (number). How would I need to change this code to the way my book wants it to be? int main () So I did try changing my 2nd loop to for (int j=2 jmentions something about square root of a number. This c++ code prints out the following prime numbers: 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97.Ä«ut I don't think that's the way my book wants it to be written.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |