Recent postings Project Euler [C/C++] Project Euler #52 - Permuted Multiples Problem 52 is so famous that I was able to find the answer without even writing a program after just reading the problem.This problem is related to recurring decimals and prime numbers.The problem itself is not difficult, so you can solve it one way or another. It’s a 5% difficulty problem. (Problem 51 had briefly jumped to 15% difficulty.)A number like 125874, when multiplied by 2, becomes 2517.. BOJ [C/C++] BOJ #1038 - Decreasing Number This problem is one I solved back when I was working through a lot of Baekjoon problems. At that time, I used to solve around 20 problems a day, but now the difficulty level of the problems has increased, and they require more thinking, so I can’t solve as many anymore.A “decreasing number” is a number where the digits strictly decrease, such as 31, 7543, or 941. The goal is to list all such num.. Project Euler [C/C++] Project Euler #51 - Prime Digit Replacements Project Euler Problem #51 is the first problem that doesn’t have a difficulty rating of 5%.Compared to problems #1 to #50, this one is a bit more challenging, with a difficulty rating set at 15%.The problem gives a number like 56xx3, where the x represents blank digits. By replacing each x with digits from 0 to 9, we want to find out how many of the resulting numbers are prime.For example, by re.. Project Euler [C/C++] Project Euler #50 - Consecutive Prime Sum This problem itself is not particularly difficult. On the Project Euler website, its difficulty rating is listed as 5%.Personally, I did not find the problem conceptually hard. However, trying to optimize the speed required a lot of careful thought.The goal of the problem is to find the prime number within a given range that can be expressed as the sum of the most consecutive primes, where the s.. BOJ [C/C++] BOJ #1037 - Divisors BOJ #1037 is a problem related to divisors and number properties. The problem provides a list of divisors of a certain positive integer N, except for 1 and N itself. The task is to determine the actual value of N using only the given divisors. The key observation is that the smallest and largest divisors in the given list, when multiplied together, yield N. This is because if all the divisors ar.. Popular postings [C/C++] Project Euler #52 - Permuted Multiples [C/C++] BOJ #1037 - Divisors [C/C++] BOJ #1024 - Sum of Number Sequences. [C/C++] BOJ #1018 - Paint Chess Board Again