21-Seol-Munhyeok

Seol-Munhyeok/.cph/.리모컨.py_a16f81a5bbad4f26e190f44fb779d889.prob

@@ -0,0 +1 @@
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Seol-Munhyeok/README.md

@@ -12,4 +12,7 @@
 | 16차시 | 2025.07.30 | 트리, 분할 정복, 재귀 | [트리의 순회](https://www.acmicpc.net/problem/2263) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/58 |
 | 17차시 | 2025.08.01 | 구현, 시뮬레이션, 백트래킹 | [청소년 상어](https://www.acmicpc.net/problem/19236) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/61 |
 | 18차시 | 2025.08.05 | 브루트포스, 애드 혹, 홀짝성 | [램프](https://www.acmicpc.net/problem/1034) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/62 |
-| 19차시 | 2025.08.10 | 구현, BFS | [거리두기 확인하기](https://school.programmers.co.kr/learn/courses/30/lessons/81302) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/67 |
+| 19차시 | 2025.08.10 | 구현, BFS | [거리두기 확인하기](https://school.programmers.co.kr/learn/courses/30/lessons/81302) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/67 |
+| 20차시 | 2025.08.14 | DP | [코딩 테스트 공부](https://school.programmers.co.kr/learn/courses/30/lessons/118668) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/73 |
+AlgoLeadMe/AlgoLeadMe-15/pull/67 |
+| 21차시 | 2025.08.18 | 시뮬레이션, 그리디 | [양궁 대회](https://school.programmers.co.kr/learn/courses/30/lessons/92342) | https://github.com/AlgoLeadMe/AlgoLeadMe-15/pull/75 |

Seol-Munhyeok/리모컨.py

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+n = int(input())
+m = int(input())
+lst = list(map(int, input().split())) if m > 0 else []
+cand = [i for i in range(10) if i not in lst]  # 사용 가능한 숫자 버튼
+length = len(str(n))
+
+answer = int(1e9)
+
+def brute(arr, r, depth, value):
+    global n, answer
+    if depth == r:
+        answer = min(answer, r + abs(n - value))
+        return
+    for num in arr:
+        brute(arr, r, depth + 1, value * 10 + num)
+
+def solve(num):
+    global answer
+    # 기본값: 숫자 버튼 안 쓰고 +/-
+    answer = abs(num - 100)
+
+    # [1] 채널이 100인 경우
+    if num == 100:
+        return 0
+
+    # [2] 100 근처 채널 예외 처리
+    if num == 101 or num == 99:
+        return 1
+    if num == 102 or num == 98:
+        return 2
+    if num == 103:
+        return 3
+
+    # [3] 모두 고장
+    if m == 10:
+        return answer
+
+    # [4] 아무것도 안 고장
+    if m == 0:
+        return len(str(num))
+
+    # [5] 모든 자릿수마다 경우의 수 체크
+    for r in range(1, length + 2):
+        brute(cand, r, 0, 0)
+
+    return answer
+
+print(solve(n))

Seol-Munhyeok/양궁대회.py

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+def get_points(a, r):
+    a_point, r_point = 0, 0
+    for i in range(11):
+        if a[i] == r[i] == 0: continue
+        if a[i] >= r[i]:
+            a_point += (10 - i)
+        else:
+            r_point += (10 - i)
+    return (a_point, r_point)
+
+def get_combinations(n, r, cur):
+    # 마지막일 때는 n을 추가
+    if r == 1:
+        return [cur + [n]]
+
+    results = []
+    for i in range(n + 1):
+        results += get_combinations(n - i, r - 1, cur + [i])
+
+    return results
+
+
+def solution(n, info):
+    answer = []
+    max_point = 0
+    cands = get_combinations(n, 11, [])
+    sorted_cands = sorted(cands, key=lambda x: tuple(x[::-1]), reverse=True)
+
+    for cand in sorted_cands:
+        a_point, r_point = get_points(info, cand)
+        if a_point >= r_point: continue
+
+        diff = r_point - a_point
+        if  diff > max_point:
+            max_point = diff
+            answer = cand
+
+    return answer if answer else [-1]

Seol-Munhyeok/코딩 테스트 공부.py

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+def solution(alp, cop, problems):
+    INF = int(1e9)
+    maxA = max(p[0] for p in problems)
+    maxC = max(p[1] for p in problems)
+
+    # 초기의 알고력(코딩력) 자체가 alp_max(cop_max)보다 높은 경우에 대한 예외처리
+    alp, cop = min(alp, maxA), min(cop, maxC)
+
+    # dp[a][c] : 알고력 a, 코딩력 c를 얻는데 걸린 시간
+    dp = [[INF for _ in range(maxC + 1)] for _ in range(maxA + 1)]
+    dp[alp][cop] = 0  # 초기값 설정
+
+    # DP
+    for a in range(alp, maxA + 1):
+        for c in range(cop, maxC + 1): 
+            # 공부하는 경우
+            ta, tc = min(maxA, a+1), min(maxC, c+1)
+            dp[ta][c] = min(dp[ta][c], dp[a][c] + 1)
+            dp[a][tc] = min(dp[a][tc], dp[a][c] + 1)
+
+            # 문제 풀이하는 경우
+            for alp_req, cop_req, alp_rwd, cop_rwd, cost in problems:
+                if a >= alp_req and c >= cop_req:
+                    na, nc = min(maxA, a + alp_rwd), min(maxC, c + cop_rwd)
+                    dp[na][nc] = min(dp[na][nc], dp[a][c] + cost)
+
+    return dp[maxA][maxC]