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[ 스택, 큐, 덱 ] 8월 30일 #2
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p2. 로직은 좋습니다! 한번 함수형 프로그래밍으로 기능들을 분리해 보세요! 수고하셨습니다!
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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!
for(int i=1;i<=n;i++){ | ||
q.push(i); | ||
} | ||
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cout << "<"; | ||
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while(!q.empty()){ | ||
for(int i=1;i<k;i++){ | ||
q.push(q.front()); | ||
q.pop(); | ||
} | ||
if(q.size()!=1){ | ||
cout << q.front() << ", "; | ||
q.pop(); | ||
} | ||
else{ | ||
cout << q.front() << ">"; | ||
q.pop(); | ||
} | ||
} | ||
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로직 깔끔하고 좋은 것 같습니다!
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using namespace std; | ||
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int main(){ |
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p2. 마찬가지로 함수 분리해주시면 좋을 것 같네요!
cout << "yes" << '\n'; | ||
} | ||
} | ||
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마찬가지로 로직 깔끔하고 좋습니다!
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[스택, 큐, 덱 구현 코드 리뷰 완료]
10757(P3)
안녕하세요 하린님! 코드 리뷰 완료되었습니다👍
깔끔하게 문제를 잘 풀어주셔서 특별히 드릴 말씀은 크게 없었습니다!
다음부터는 간단한 주석을 달아주시면 더 좋을 것 같아요👍
궁금한 점이 있으시다면 리뷰어를 호출해주세요~
while (a_size>0||b_size>0){ | ||
if(a_size>0){ | ||
a_digit=a[--a_size]-'0'; | ||
} | ||
else{ | ||
a_digit=0; | ||
} | ||
if(b_size>0){ | ||
b_digit=b[--b_size]-'0'; | ||
} | ||
else{ | ||
b_digit=0; | ||
} | ||
a_b=a_digit+b_digit+ab_over; | ||
ab_over=a_b/10; | ||
a_b%=10; | ||
ab_char=a_b+'0'; | ||
ab_total+=ab_char; | ||
} |
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덧셈의 원리를 이용해서 덧셈을 잘 구현해주셨네요!👍
인적사항
학번: 2071079
이름: 이하린
###과제제출
기존 제출: 10757, 4949, 1158
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