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cf:B. Almost Ternary Matrix【對稱 + 找規律 + 構造 + 我是構造垃圾】
2022-07-05 18:58:00 【白速龍王的回眸】
分析
趨於對稱,由於是2的倍數,所以考慮基本單元2 * 2
1 0 | 0 1
0 1 | 1 0
ABA…
BAB…
…
我們如上定義兩種特殊的2 * 2的A和B
然後就可以構造出來AB交錯的樣子,符合第一第二個實例
ac code
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n, m = list(map(int, input().split()))
# 1 0 | 0 1
# 0 1 | 1 0
# ABA...
# BAB...
# ...
even = [1, 0, 0, 1]
odd = [0, 1, 1, 0]
ans = [[0] * m for _ in range(n)]
for i in range(n):
for j in range(m):
if i % 4 == 0 or i % 4 == 3:
ans[i][j] = even[j % 4]
else:
ans[i][j] = odd[j % 4]
for row in ans:
print(*row)
總結
我是構造垃圾
我是找規律小垃圾
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