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bsa0322
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Oct 4, 2021
bsa0322
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p1. 전반적으로 잘 풀어주셨어요!! 3190만 수정한 후, 저 다시 리뷰어로 호출해주세요! 수고하셨습니다~!
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| int min(int a, int b, int c){ | ||
| return b > a ? c > a ? a : c : c > b ? b : c; | ||
| } |
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p2. 삼항연산자는 가독성이 안 좋아서 최대한 피하는 게 좋아요! min 함수 2개 쓰거나, 반복문 활용해서 작성해주시면 좋을 것 같아요!
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| if(i+counsel[i].d < n+1) // 상담 마지막 날 다음 날 dp가 있으면 | ||
| dp[i] = max(dp[i+counsel[i].d]+counsel[i].m, dp[i+1]); | ||
| else // 상담 마지막 날 다음 날 dp가 없으면 | ||
| dp[i] = max(counsel[i].m, dp[i+1]); |
| int find(vector<vector<int>> dp, vector<vector<int>> cost, int n){ | ||
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| int ans [3]; | ||
| for(int i=0; i<3; i++){ |
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p2. 첫 번째 집을 어디로 고정할지를 main에서 정하고, 함수로 보내주면 더 간결하게 작성할 수 있을 것 같아요!
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| while(!su_gr.empty()){ | ||
| dodo.push_back(su_gr.back()); | ||
| su_gr.pop_back(); | ||
| } | ||
| while(!do_gr.empty()){ | ||
| dodo.push_back(do_gr.back()); | ||
| do_gr.pop_back(); | ||
| } |
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p2. 어느 덱으로 넣는지와, 어느 그라운드에서 가져오는지 보내주면 함수화 해봐도 좋을 것 같아요~!
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| char changeDir(char from, char to){ | ||
| if(from == 'E'){ | ||
| if(to == 'L') | ||
| return 'N'; | ||
| else | ||
| return 'S'; | ||
| } | ||
| else if(from == 'W'){ | ||
| if(to == 'L') | ||
| return 'S'; | ||
| else | ||
| return 'N'; | ||
| } | ||
| else if(from == 'S'){ | ||
| if(to == 'L') | ||
| return 'E'; | ||
| else | ||
| return 'W'; | ||
| } | ||
| else{ | ||
| if(to == 'L') | ||
| return 'W'; | ||
| else | ||
| return 'E'; | ||
| } | ||
| } |
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p1. 방향 관련 배열을 사용하면 더 간편하게 작성할 수 있어요!! 예를 들어 좌표로 표현했을 때, 위로 가는 건 (-1,0)을 현재 좌표에 더해주는 거고, 왼쪽으로 가는 건 (0, -1)을 더해주는 거죠! 이런 식의 상,하,좌,우 변화값을 저장한 배열을 만들어서 관리하면 훨씬 좋아요! 한 번 수정해볼까요!!
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