YoonYn9915 / 4월 2주차 / 3문제

YoonYn9915/dp/2025-04-10-[백준]-#10844-쉬운 계단 수.py

@@ -0,0 +1,19 @@
+
+N = int(input())
+
+dp = [[0] * 10 for _ in range(N + 1)]
+
+for i in range(1, 10):
+    dp[1][i] = 1
+
+for i in range(2, N + 1):
+    for j in range(10):
+        if j == 0:
+            dp[i][j] = dp[i - 1][1]
+        elif 1 <= j <= 8:
+            dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j + 1]
+        else:
+            dp[i][j] = dp[i - 1][8]
+
+
+print(sum(dp[N]) % 1000000000)

YoonYn9915/dp/2025-04-12-[백준]-#1562-계단 수.py

@@ -0,0 +1,26 @@
+import sys
+
+input = sys.stdin.readline
+
+N = int(input())
+dp = [[[0 for _ in range(1 << 10)] for _ in range(10)] for _ in range(N)]
+mod = 1000000000
+res = 0
+
+for k in range(1, 10):
+    dp[0][k][1 << k] = 1
+
+for i in range(1, N):
+    for k in range(10):
+        for bit in range(1024):
+            if k - 1 >= 0:
+                dp[i][k][bit | (1 << k)] += dp[i - 1][k - 1][bit]
+            if k + 1 <= 9:
+                dp[i][k][bit | (1 << k)] += dp[i - 1][k + 1][bit]
+            dp[i][k][bit | (1 << k)] %= mod
+
+for k in range(10):
+    res += dp[N - 1][k][1023]
+    res %= mod
+
+print(res)

YoonYn9915/recursion/2025-04-10-[백준]-#1780-종이의 개수.py

@@ -0,0 +1,45 @@
+
+import sys
+
+def recursion(paper, row_start, col_start, step):
+    # 종이가 같은지 검사
+    result = check_paper_same(paper, row_start, col_start, step)
+
+    # 같지 않다면 9분할한뒤 각각을 재귀적으로 검사
+    if result == 2:
+        for i in range(3):
+            for j in range(3):
+                row = row_start + (i * step // 3)
+                col = col_start + (j * step // 3)
+                recursion(paper, row, col, step // 3)
+    else:
+        # 종이의 칸이 다 같은 경우
+        answer[result] += 1
+
+
+
+def check_paper_same(paper, row_start, col_start, step):
+    base = paper[row_start][col_start]
+    for i in range(row_start, row_start + step):
+        for j in range(col_start, col_start + step):
+            if paper[i][j] != base:
+                return 2
+    return base
+
+
+inp = sys.stdin.readline
+
+n = int(inp())
+
+paper = []
+
+# 각각 -1, 0, 1로만 이루어진 종이의 개수 저장 튜플
+global answer
+answer = {-1: 0, 0: 0, 1: 0}
+
+for i in range(n):
+    paper.append(list(map(int, inp().split())))
+
+recursion(paper, 0, 0, n)
+
+print('\n'.join(str(value) for value in answer.values()))