Programming (102) 썸네일형 리스트형 [C/C++] Project Euler #54 - Poker Hands I didn’t find this problem particularly difficult, but it certainly is quite tedious.On the Project Euler site, it is rated as a 10% difficulty problem.I’ve never made a poker-style game before, but determining who wins according to poker rules is the most important part of a poker game. Everything else is just the interface.This problem is quite long.Ultimately, the main point is to determine h.. [C/C++] BOJ #1041 - Dice This problem can be easily solved if you can visualize the geometric structure of a cube. Perhaps that’s why its difficulty level is quite low. The difficulty is Silver III. While it is harder than Bronze-level problems, it’s not too difficult.As of now, the correct answer rate is 22.6%, which is quite low, and there are 531 correct submissions.The goal of the problem is to calculate the minimum.. [C/C++] Project Euler #53 - Combinatoric Selections This problem is a simple one. As long as you know the combination formula, it’s easy to solve.When n varies from 1 to 100, you need to calculate the number of times the combination value \(_nC_r\) exceeds one million.Even with a brute-force approach, you only need to compute combinations 10,100 times.If you know that \(_nC_r = nC{n-r}\), you can reduce the calculations by half. Because of its sy.. [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.. [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.. [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.. [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.. [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.. [C/C++] Project Euler #49 - Prime Permutations Project Euler #49 Prime Permutations problem may seem complex, but the actual solving process is straightforward. The difficulty level is 5%, classified as “very easy.”Three four-digit numbers are prime numbers that are permutations of each other.For example, the numbers 1487, 4817, and 8147 consist of the same four digits (1, 4, 7, 8) arranged in different orders, and each of them is a prime nu.. [C/C++] Project Euler #48 - Self Powers This problem is also an easy one.The content is quite short. The difficulty level is 5%.Project Euler problem #48, “Self Powers”, requires calculating the last 10 digits of the sum of each natural number from 1 to n raised to the power of itself. In other words, we need to compute the following sum:\[1^1 + 2^2 + 3^3 + \dots + n^n\]Since the result of this sum can be extremely large, the key as.. 이전 1 2 3 4 5 6 ··· 11 다음
