[이윤정] 11주차 과제 제출합니다. #19
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yunjeongiya
commented
May 23, 2022
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yunjeongiya
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riroan
reviewed
May 24, 2022
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| //조합 구현 어떻게...? | ||
| for (int i = 1; i <= M; i++) //1번부터 M번까지 학생 | ||
| { | ||
| for (int j = 1; j <= N; j++) | ||
| { | ||
| } | ||
| } |
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비트마스킹 방법으로 리팩터링하는 것은 현재로썬 꽤 어려워보입니다. 백트래킹을 현재 인덱스부터 도는 방법으로 한번 해결해보세요. 만약 아직도 어렵다면 저한테 카톡 또는 디스커션 열어주시면 코드 봐드릴게요.
| cin >> N >> M; | ||
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| //입력의 양이 방대하므로 빠른 입력 사용 권장... | ||
| bool G[100001][100001] = { 0, }; |
| { | ||
| int x = now.first + ways[i][0]; | ||
| int y = now.second + ways[i][1]; | ||
| if (1 <= x && x <= N && 1 <= y && y <= M && !visited[x][y]) |
| Q.push(make_pair(x, y)); | ||
| break; | ||
| case '#': | ||
| cell[x][y] = 'x'; |
kth990303
reviewed
May 30, 2022
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| //����, ���� ���� gcd ���ϸ� �� �Ǵ� ���?? | ||
| //a+d, a+3d -> a+d, 2d -> ? |
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유클리드 호제법 따라서 a랑 a+d의 최대공약수 구하면 a+d%a=d라서 a,d 돌린거랑 같아지는건 확인했었는데
a+d랑 a+3d일 때는
a+3d%a+d = 2d
2d % a+d = ...? 에서 어떻게 증명해야 될 지 모르겠습니다ㅠㅠ!
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제가 볼 때엔 차가 2d인 경우의 최대공약수를 구할 때가 없어보여요! 혹시 언제 차가 2d일 때 최대공약수를 구할 일이 있는지 여쭤봐도 될까요?
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아아아 범위 내 모든 항들에 대해서니까 항상 연속된 항들에 대해서 연산하게 되어서 다 차가 d군요 감사합니다!
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