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bsa0322
approved these changes
Sep 12, 2021
| } | ||
| else{ | ||
| for(int i = 1; i <= a.length(); i++) | ||
| cout << res[i]; |
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string을 가공하지 않고 너무 잘 풀어주셨어요!!!! 👍 가독성 면에서 조금 코멘트를 드리자면, res 벡터를 push_back 으로 일의 자리 수부터 저장한 후, 끝에서부터 출력하는 형태로 하면 출력을 하나로 합칠 수 있어요!!
| cin >> n >> m; | ||
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| string a; | ||
| for(int i=0; i<n; i++){ |
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n, m 은 다시 쓰지 않으니 while(n--) 으로 입력 받아도 좋을 것 같아요~!
| } | ||
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| //출력 | ||
| cout << cnt << '\n'; |
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cnt 변수 사용도 좋지만, ans 만으로도 크기를 알 수 있는 방법이 있죠!
| // | ||
| // Created by LG on 2021-09-12. | ||
| // | ||
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| //음수를 2로 나눈 몫 | ||
| int divideByTwo(int num){ | ||
| int q = 0; | ||
| while(2*q-num > 0) | ||
| q--; | ||
| return q; | ||
| } |
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p2. 음수를 2로 나눈 몫을 잘 처리해주셨어요! 하지만 반복문의 사용 없이 한 번의 연산 만으로 처리할 수 있는 방법이 있어요. -5/2를 하면 -3이 나와야 하는데, c++에선 -2가 나오죠! 어떻게 하면 좋을까요? 아니면 floor 함수에 대해 알아봐도 좋을 것 같아요.
| //입력 | ||
| for(int i=0; i<n; i++){ | ||
| cin >> name >> type; | ||
| wear[type]++; |
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