Problem Solving
[BOJ] (구현) 20327번 배열돌리기 6
주씨.
2022. 4. 3. 23:14
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별다른 스킬이 필요한 문제는 아니었지만, 배열을 돌리는 다양한 상황을 연습할 수 있었다.
나중에 참고할 때가 있을 것 같아 코드를 업로드 해본다.
def oper1(arr, l):
n = len(arr)
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
tmp = reverseTopDown(tmp)
for i in range(s):
for j in range(s):
arr[x+i][y+j] = tmp[i][j]
return arr
def oper2(arr, l):
n = len(arr)
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
tmp = reverseLeftRight(tmp)
for i in range(s):
for j in range(s):
arr[x+i][y+j] = tmp[i][j]
return arr
def oper3(arr, l):
n = len(arr)
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
tmp = rotateRight(tmp)
for i in range(s):
for j in range(s):
arr[x+i][y+j] = tmp[i][j]
return arr
def oper4(arr, l):
n = len(arr)
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
tmp = rotateLeft(tmp)
for i in range(s):
for j in range(s):
arr[x+i][y+j] = tmp[i][j]
return arr
def oper5(arr, l):
n = len(arr)
new_arr = [[0]*n for _ in range(n)]
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
for i in range(s):
for j in range(s):
new_arr[n-s-x+i][y+j] = tmp[i][j]
return new_arr
def oper6(arr, l):
n = len(arr)
new_arr = [[0]*n for _ in range(n)]
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
for i in range(s):
for j in range(s):
new_arr[x+i][n-s-y+j] = tmp[i][j]
return new_arr
def oper7(arr, l):
n = len(arr)
new_arr = [[0]*n for _ in range(n)]
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
for i in range(s):
for j in range(s):
new_arr[i+y][n-s-x+j] = tmp[i][j]
return new_arr
def oper8(arr, l):
n = len(arr)
new_arr = [[0]*n for _ in range(n)]
s = 2**l
for x in range(0, n, s):
for y in range(0, n, s):
tmp = []
for i in range(s):
row = []
for j in range(s):
row.append(arr[x+i][y+j])
tmp.append(row)
for i in range(s):
for j in range(s):
new_arr[i+n-s-y][j+x] = tmp[i][j]
return new_arr
def reverseTopDown(arr):
n = len(arr)
m = len(arr[0])
tmp = [[0]*m for _ in range(n)]
for i in range(n):
for j in range(m):
tmp[i][j] = arr[n-i-1][j]
return tmp
def reverseLeftRight(arr):
n = len(arr)
m = len(arr[0])
tmp = [[0]*m for _ in range(n)]
for i in range(n):
for j in range(m):
tmp[i][j] = arr[i][m-1-j]
return tmp
def rotateRight(arr):
n, m = len(arr), len(arr[0]) # 행, 열
result = [[0] * n for _ in range(m)]
for i in range(n):
for j in range(m):
result[j][n-1-i] = arr[i][j]
return result
def rotateLeft(arr):
n, m = len(arr), len(arr[0]) # 행, 열
result = [[0] * n for _ in range(m)]
for i in range(n):
for j in range(m):
result[m-1-j][i] = arr[i][j]
return result
def main():
N, R = map(int, input().split())
arr = [list(map(int, input().split())) for _ in range(2**N)]
for _ in range(R):
k, l = map(int, input().split())
if k==1:
arr = oper1(arr, l)
elif k==2:
arr = oper2(arr, l)
elif k==3:
arr = oper3(arr, l)
elif k==4:
arr = oper4(arr, l)
elif k==5:
arr = oper5(arr, l)
elif k==6:
arr = oper6(arr, l)
elif k==7:
arr = oper7(arr, l)
elif k==8:
arr = oper8(arr, l)
for a in arr:
print(*a)
main()
https://www.acmicpc.net/problem/20327
20327번: 배열 돌리기 6
크기가 2N×2N인 배열이 있을 때, 배열에 연산을 R번 적용하려고 한다. 연산은 8가지가 있고, 연산에는 단계 ℓ (0 ≤ ℓ < N)이 있다. 단계 ℓ은 배열을 부분 배열로 나눌때 사용하는 값이며, 부분
www.acmicpc.net