[스텍, 큐, 덱] 8월 26일 #2
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Goldchae
reviewed
Sep 1, 2023
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| using namespace std; | ||
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| int main() |
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p2. main함수에서는 입출력만 넣어주시는 편이 좋습니다! 기능을 분리하여 함수로 만들어보세요!
| } | ||
| } | ||
| cout << '>'; | ||
| return 0; |
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| using namespace std; | ||
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| int main() |
chae-jpg
approved these changes
Sep 1, 2023
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| if (sum >= 10) | ||
| { | ||
| carry = true; | ||
| result.push_front(sum - 10); | ||
| } | ||
| else | ||
| { | ||
| carry = false; | ||
| result.push_front(sum); | ||
| } |
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사소하지만,
carry = sum / 10;
result.push_back(sum % 10);
이렇게 하면 조금 더 간결하게 표현할 수 있어요~
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| // 문자열로 입력받기 | ||
| string a, b; | ||
| deque<char> A, B; |
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P3
알튜비튜에서는 변수 이름으로 알파벳 소문자를 사용하고 있습니다! a_deque와 같은 이름으로 변경해 보는 것은 어떨까요?
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| for (int i = 0; i < a.size(); i++) | ||
| { | ||
| A.push_back(a[i]); | ||
| } | ||
| for (int i = 0; i < b.size(); i++) | ||
| { | ||
| B.push_back(b[i]); | ||
| } |
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문자열을 덱 2개에 각각 집어넣은 다음에 덱에서 값을 꺼내 연산하는 식으로 문제를 풀어주셨네요! 덱의 성질을 잘 이용해주셨습니다👍👍 단, 이러면 덱에 값을 푸시하고 팝하는 과정에서 연산이 많이 일어나기 때문에, 효율성 면에서는 성능이 조금 떨어질 수 있다는 점 알아주세요!
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| for (int i = 0; i < size; i++) | ||
| { // 뒷자리부터 차례대로 계산 | ||
| if (A.empty()) | ||
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| m = 0; | ||
| } | ||
| else | ||
| { | ||
| m = A.back() - '0'; // char to int를 위해 '0'을 뺀다 | ||
| A.pop_back(); | ||
| } | ||
| if (B.empty()) | ||
| { | ||
| n = 0; | ||
| } | ||
| else | ||
| { | ||
| n = B.back() - '0'; | ||
| B.pop_back(); | ||
| } | ||
| int sum = m + n; | ||
| if (carry) | ||
| { // 이전 계산에서 carry가 있다면? | ||
| sum++; | ||
| } | ||
| if (sum >= 10) | ||
| { | ||
| carry = true; | ||
| result.push_front(sum - 10); | ||
| } | ||
| else | ||
| { | ||
| carry = false; | ||
| result.push_front(sum); | ||
| } | ||
| } |
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문서영
2017012
1158/4949/10757