Week09 #9
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Week09 #9
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sforseohn
reviewed
Oct 30, 2024
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| int main() { | ||
| int n; | ||
| cin >> n; | ||
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| vector <int> dir(6); | ||
| vector <int> len(6); | ||
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| vector <pair <int, int>> v(6); | ||
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| vector <int> is_used(5,0); | ||
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| for (int i=0; i<6 ; i++){ | ||
| int d; | ||
| int l; | ||
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| cin >> d >> l; | ||
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| dir[i]=d; | ||
| len[i]=l; | ||
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| v[i]=make_pair(d,l); | ||
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| //방향 나온 빈도 체크하기 | ||
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| is_used[d]++; | ||
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| } | ||
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| int total=1; | ||
| int minus=1; | ||
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| vector <int> two; | ||
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| for(int i=1; i<5 ; i++){ | ||
| //한번만 나온 방향의 경우, 즉 전체 사각형의 크기를 먼저 구한다 | ||
| if(is_used[i]==1){ | ||
| for (int j=0 ; j<6 ; j++){ | ||
| if(v[j].first==i){ | ||
| total*=v[j].second; | ||
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| if(j>=3) two.push_back(j-3); | ||
| else two.push_back (j+3); | ||
| } | ||
| } | ||
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| } | ||
| } | ||
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| minus=len[two[0]]*len[two[1]]; | ||
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| cout << (total-minus)*n; | ||
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| return 0; | ||
| } |
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P2. 현재 메인함수에서 문제의 모든 기능을 구현하고 있죠. 영역을 구하는 부분을 함수로 빼서 재사용성을 높여 볼까요? 😊
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| vector <int> dir(6); | ||
| vector <int> len(6); | ||
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| vector <pair <int, int>> v(6); |
There was a problem hiding this comment.
P2. 두 벡터와 int pair 중 하나만 써도 될 것 같네요! 🥰
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| for(int i=1; i<5 ; i++){ | ||
| //한번만 나온 방향의 경우, 즉 전체 사각형의 크기를 먼저 구한다 |
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방향이 몇 개 나왔는지 세서 케이스를 분리해주셨군요! 좋습니다👍👍👍
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| vector <int> is_used(5,0); | ||
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| for (int i=0; i<6 ; i++){ |
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P2. 6 이 반복되고 있네요! 이렇게 특정 상수가 반복될 때 하드코딩하기보다는 상수 변수를 선언해서 사용하면 가독성도 좋아지고 실수할 일이 줄어듭니다 🥰
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| if(j>=3) two.push_back(j-3); | ||
| else two.push_back (j+3); | ||
| } | ||
| } | ||
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| } | ||
| } | ||
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| minus=len[two[0]]*len[two[1]]; |
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이 부분 로직이 궁금해요! 어떻게 동작하는지 설명해주실 수 있으실까요?
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이름: 김재이
학번: 2376071
제출: 14888, 2477, 15665