結果
| 問題 | No.2082 AND OR XOR |
| コンテスト | |
| ユーザー |
mkawa2
|
| 提出日時 | 2022-09-25 22:56:10 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 124 ms / 2,000 ms |
| コード長 | 2,194 bytes |
| 記録 | |
| コンパイル時間 | 284 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 76,544 KB |
| 最終ジャッジ日時 | 2024-12-22 14:57:19 |
| 合計ジャッジ時間 | 4,193 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 29 |
ソースコード
import sys
# sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = (1 << 63)-1
# inf = (1 << 31)-1
# md = 10**9+7
md = 998244353
def f(a, b, c, x, y):
res = a*(x & y)+b*(x | y)+c*(x ^ y)
return res%4
# return bin(res%4)[2:].zfill(2)
# for a in range(4):
# for b in range(4):
# for c in range(4):
# pDB(f"a={a},b={b},c={c}")
# cur = [[0]*4 for _ in range(4)]
# for x in range(4):
# for y in range(4):
# cur[x][y] = f(a, b, c, x, y)
# p2D(cur)
to = [[-1]*4 for _ in range(4)]
ot = [[set() for _ in range(4)] for _ in range(4)]
n, a, b, c = LI()
for x in range(4):
for y in range(4):
to[x][y] = f(a, b, c, x, y)
ot[to[x][y]][x].add(y)
ot[to[x][y]][y].add(x)
# p2D(to)
# p2D(ot)
dp = [[[0]*4 for _ in range(4)] for _ in range(4)]
for i in range(4):
for s in range(4):
for j in ot[s][i]:
dp[i][j][s] = 1
# p2D(dp)
for _ in range(n-2):
ndp = [[[0]*4 for _ in range(4)] for _ in range(4)]
for i in range(4):
for j in range(4):
for s in range(4):
pre = dp[i][j][s]
if pre == 0: continue
for x in range(4):
ni = to[i][x]
for nj in ot[j][x]:
ndp[ni][nj][s] += pre
ndp[ni][nj][s] %= md
dp = ndp
ans = 0
for i in range(4):
for j in range(4):
for s in range(4):
if to[i][j] == s:
ans += dp[i][j][s]
ans %= md
print(ans)
mkawa2