Add 2023 solutions (#1)

Merge pull request #1 from jfsscsclub/2023-Solutions

2016/README.md

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+# CCC 2016
+Note that these are only accessible if you are a student of the Peel District School Board and are logged into your school Google account. Otherwise, you will not have access.
+## Slide Decks
+- [S2](https://docs.google.com/presentation/d/1c3o4cSQfKFScKp3BfXM0b216HOMg3hURkNv4HBU2Ppc/)

2022/README.md

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+# CCC 2022
+Note that these are only accessible if you are a student of the Peel District School Board and are logged into your school Google account. Otherwise, you will not have access.
+## Slide Decks
+- [S2](https://docs.google.com/presentation/d/1c3o4cSQfKFScKp3BfXM0b216HOMg3hURkNv4HBU2Ppc/)

2023/README.md

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+# CCC 2023
+Note that these are only accessible if you are a student of the Peel District School Board and are logged into your school Google account. Otherwise, you will not have access.
+## Slide Decks
+- [S1/J4, S2, S3](https://docs.google.com/presentation/d/1ZOA_IlWxJPJdOiX4Cxhhy1WOK-nyPMLahiBfGsJwEmM/)

2023/s1.py

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+# CCC '23 J4/S1 - Aritro Saha
+c = int(input())
+
+# Gathering input, True represents a black triangle and False represents a white triangle
+row1 = [bool(int(triangle)) for triangle in input().split()]
+row2 = [bool(int(triangle)) for triangle in input().split()]
+
+# The maximum tape that we need is the number of black triangles * the edges per triangle (3)
+tape = (row1.count(True) + row2.count(True)) * 3
+
+# Checking for left-right adjacency in both rows
+for i in range(c - 1):
+  if row1[i] and row1[i + 1]:
+    tape -= 2  # Subtract 2 because one edge on both adjacent triangles aren't needed
+
+  if row2[i] and row2[i + 1]:
+    tape -= 2
+
+# Checking for top-bottom adjacency
+for i in range(c):
+  # If the triangles are pointing into each other, they don't share an edge
+  # Referring to the diagram, this only happens on even triangle columns (starting at 0)
+  if row1[i] and row2[i] and i % 2 == 0:
+    tape -= 2
+
+print(tape)

2023/s2.py

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+# CCC '23 S2 - Roy Zhang
+N = int(input())
+hills = [int(i) for i in input().split(" ")]
+
+# initialize grid
+grid = []
+for i in range(N):
+    grid.append([0] * N)
+
+# solve
+ans = [float('infinity')] * N
+ans[0] = 0
+for j in range(1, N):
+    for i in range(j - 1, -1, -1):
+        grid[i][j] = grid[i + 1][j - 1] + abs(hills[i] - hills[j])
+        if grid[i][j] < ans[j - i]:
+            ans[j - i] = grid[i][j]
+
+print(" ".join(str(i) for i in ans))

2023/s3.py

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+# CCC '23 S3 - Roy Zhang
+import math
+
+NMRC = input().split(" ")
+N, M, R, C = int(NMRC[0]), int(NMRC[1]), int(NMRC[2]), int(NMRC[3])
+
+# initialize grid
+grid = []
+for i in range(N):
+    grid.append([0] * M)
+
+# impossible case
+if (R == N and C % 2 != 0 and M % 2 == 0) or (C == M and R % 2 != 0 and N % 2 == 0):
+    print("IMPOSSIBLE")
+else:
+    # col palindromes
+    count = 0
+    if C % 2 == 1 and M % 2 == 1:
+        for i in range(N):
+            grid[i][M//2] = 'a'
+        C -= 1
+        count += 1
+
+    # col palindromes
+    for j in range(C):
+        for i in range(N):
+            if j % 2 == 0:
+                col_coord = math.ceil(j / 2)
+            else:
+                col_coord = M - math.ceil(j / 2)
+            grid[i][col_coord] = "a"
+
+    # row palindromes
+    for i in range(R):
+        if i == 0:
+            grid[i] = ['a'] * M
+        else:
+            for j in range(M):
+                if grid[i][j]: continue
+                if grid[i][M - j - 1]:
+                    grid[i][j] = grid[i][M - j - 1]
+                else:
+                    grid[i][j] = 'b'
+
+    # remaining cells
+    for i in range(N):
+        for j in range(M):
+            if grid[i][j]: continue
+            if i == 0:
+                grid[i][j] = 'f'
+                if not grid[i][M - j - 1]:
+                    grid[i][M - j - 1] = 'g'
+            elif i % 2 == 0:
+                grid[i][j] = 'c'
+                if not grid[i][M - j - 1]:
+                    grid[i][M - j - 1] = 'd'
+            else:
+                grid[i][j] = 'd'
+                if not grid[i][M - j - 1]:
+                    grid[i][M - j - 1] = 'c'
+
+    # special cases
+    if C + count == M and R == 0:
+        for i in range(N):
+            grid[i][-1] = "e"
+
+    elif C + count == M and R != 0:
+        for i in range((N - R) // 2):
+            grid[i][0] = "e"
+            grid[N - i - 1][0] = 'e'
+
+    # print ans
+    for row in grid:
+        print("".join(str(i) for i in row))

README.md

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 A repository of CCC questions and solutions that were reviewed during John Fraser CS Club's lessons.
 
 ## Years
+- [2023](/2023/)
 - [2022](/2022/)
 - [2021](/2021/)
 - [2020](/2020/)