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| int main(){ |
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와 이걸 직접구현하셨네요! 정말 잘하셨어요!!
근데 사실 삭제&삽입 연산은 front와 back에서 아주 약간 다르게 작동해요!
아무튼 정말 멋져요...!
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| map<char, double> value; | ||
| for (int i = 0; i < n; i++) { | ||
| cin >> t; | ||
| value['A' + i] = t; | ||
| } |
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map도 좋지만, 다른 방법도 있어요! 코드를 잘 살펴보시면 답이 나올 것 같아요~
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| else if(c == '*'){ | ||
| a = st.top(); | ||
| st.pop(); | ||
| b = st.top(); | ||
| st.pop(); | ||
| st.push(a*b); | ||
| } | ||
| else if(c == '+'){ | ||
| a = st.top(); | ||
| st.pop(); | ||
| b = st.top(); | ||
| st.pop(); | ||
| st.push(a+b); | ||
| } | ||
| else if(c == '/'){ | ||
| a = st.top(); | ||
| st.pop(); | ||
| b = st.top(); | ||
| st.pop(); | ||
| st.push(b/a); | ||
| } | ||
| else if(c == '-'){ | ||
| a = st.top(); | ||
| st.pop(); | ||
| b = st.top(); | ||
| st.pop(); | ||
| st.push(b-a); | ||
| } |
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여기 중복이 많네요! 사칙연산 함수를 만드는건 어떨까요?
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| //벡터에 맵 입력 | ||
| vector<pair<string, int>> answer(word.begin(), word.end()); |
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| //스택에 인형 종류 입력 | ||
| for(int i=n-1; i>=0; i--){ | ||
| for(int j=0; j<n; j++){ | ||
| if(board[i][j] != 0) | ||
| doll[j].push(board[i][j]); | ||
| } | ||
| } |
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| if(doll[move-1].empty()) | ||
| continue; |
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