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🧷 문제 링크
https://www.acmicpc.net/problem/11692
🧭 풀이 시간
45분
👀 체감 난이도
✏️ 문제 설명
1이상 N이하의 수 중, 약수의 합이 짝수인 수의 개수를 구해보자.
🔍 풀이 방법
홀수인 개수를 구해서 전체에서 빼기
약수 중 홀수의 개수가 짝수개
-> 소인수분해 했을 때 2가 아닌 것들로만 개수 구하기
N = 2^m * a1^p1 * a2^p2 * ...
라고 하면, N의 홀수인 약수의 개수는 (p1+1) * (p2+1) * ...
이게 홀수이려면 pk+1 이 모두 홀수여야 함. == pk가 모두 짝수여야 함.
-> 소인수분해의 결과가, 2를 제외한 다른 소수들은 모두 짝수번 곱해져있는 꼴이어야 함.
1부터 출발해서 DFS느낌으로 백트래킹, 수가 크니까 HashSet으로 중복 방지
⏳ 회고
시간 초과가 나야 하는 풀이같은데 억지로 뚫은 느낌...
개어려운데 왜 골드지