[이분탐색] 11월 5일 #12
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[이분탐색] 11월 5일 #12
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이름: 조수아 학번: 2371060 과제 제출 기존 제출: 17266, 3079, 2343, 16401, 10815
avocado8
approved these changes
Nov 10, 2024
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| while (left <= right) { | ||
| int mid = (left + right) / 2; | ||
| if (arr[mid] == target) return 1; | ||
| else if (arr[mid] > target) right = mid - 1; | ||
| else left = mid + 1; | ||
| } |
sforseohn
reviewed
Nov 11, 2024
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| int numPeople, numSnacks; | ||
| cin >> numPeople >> numSnacks; | ||
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| //과자 길이 배열 입력 | ||
| vector<int> snackLengths = getInputArray(numSnacks); |
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P3. 변수명은 스네이크 코드 형식을 따라주세요!
int hello_world; // 변수
int helloWorld() // 함수
const int HELLO_WORLD // 상수
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| // 사용자 입력을 받아 과자 길이 배열을 반환하는 함수 | ||
| vector<int> getInputArray(int size) { | ||
| vector<int> array(size); | ||
| for (int& length : array) { | ||
| cin >> length; | ||
| } | ||
| return array; | ||
| } |
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P2. 함수화 좋습니다! 그런데 알튜비튜에서는 입출력은 메인에서 받고, 주요 로직은 각 함수에서 구현하는 방식으로 제안드리고 있으니 참고해주세요 😊
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| int count = 0; | ||
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| //각 과자 길이를 현재 중앙값(mid)로 나눠 몇 명에게 나눠줄 수 있는지 계산 | ||
| for (int length : snackLengths) { | ||
| count += length / mid; | ||
| } |
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P3. 길이가 mid일 때 나누어줄 수 있는 조각의 수를 구하는 부분도 함수로 빼면 좋겠네요 🤗🤗
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| //과자를 나눠줄 수 있는 최대 길이를 이진 탐색을 통해 찾는 함수 | ||
| int getMaxSnackLength(int numPeople, const vector<int>& snackLengths) { | ||
| int left = 1, right = *max_element(snackLengths.begin(), snackLengths.end()); |
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| int main() { | ||
| ios_base::sync_with_stdio(false); | ||
| cin.tie(NULL); | ||
| cout.tie(NULL); | ||
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| //인원 수와 과자 개수 입력 | ||
| int numPeople, numSnacks; | ||
| cin >> numPeople >> numSnacks; | ||
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| //과자 길이 배열 입력 | ||
| vector<int> snackLengths = getInputArray(numSnacks); | ||
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| //나눠줄 수 있는 최대 과자 길이를 계산하고 출력 | ||
| int result = getMaxSnackLength(numPeople, snackLengths); | ||
| cout << result << "\n"; | ||
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| return 0; | ||
| } |
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이름: 조수아
학번: 2371060
과제 제출
기존 제출: 17266, 3079, 2343, 16401, 10815