結果
| 問題 |
No.2152 [Cherry Anniversary 2] 19 Petals of Cherry
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-12-09 20:00:41 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 411 ms / 1,000 ms |
| コード長 | 2,603 bytes |
| コンパイル時間 | 345 ms |
| コンパイル使用メモリ | 82,308 KB |
| 実行使用メモリ | 165,616 KB |
| 最終ジャッジ日時 | 2024-10-14 18:49:45 |
| 合計ジャッジ時間 | 18,629 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 |
|---|---|
| other | AC * 49 |
ソースコード
import sys
#input = lambda :sys.stdin.readline()[:-1]
ni = lambda :int(input())
na = lambda :list(map(int,input().split()))
yes = lambda :print("yes");Yes = lambda :print("Yes");YES = lambda : print("YES")
no = lambda :print("no");No = lambda :print("No");NO = lambda : print("NO")
#######################################################################
n = 19
c = [0] * n
for i in range(n):
que = na()
for j in que[1:]:
c[i] += 1<<j-1
dp = [0] * (1<<n)
dp[0] = 1
for i in range(n):
ndp = [0] * (1<<n)
ndp, dp = dp, ndp
for j in range(1<<n):
if ndp[j]==0:
continue
for k in range(n):
if j >> k & 1:
continue
dp[j+(1<<k)] += ndp[j] * (c[i] >> k & 1)
from fractions import Fraction
#2D matrix
def add(x,y):
return x + y
def mul(x, y):
return x * y
def mat_add(A, B, replace=False):
assert len(A)==len(B) and len(A[0]) == len(B[0])
if not replace:
A = [a.copy() for a in A]
n = len(A)
m = len(A[0])
for i in range(n):
for j in range(m):
A[i][j] = add(A[i][j], B[i][j])
return A
def mat_mul(A,B):
assert len(A[0]) == len(B)
n = len(A)
m = len(B[0])
p = len(A[0])
R = [[0 for j in range(m)]for i in range(n)]
for i in range(n):
for j in range(m):
for k in range(p):
R[i][j] = add(R[i][j],mul(A[i][k],B[k][j]))
return R
def mat_pow(A, x):
assert len(A)==len(A[0])
n = len(A)
R = [[0 for j in range(n)]for i in range(n)]
while x > 0:
if x&1:
R = mat_mul(R, A)
A = mat_mul(A,A)
x >>= 1
return R
def determinant(A, replace=False):
if not replace:
A = [a.copy() for a in A]
n = len(A)
res = 1
for i, a_i in enumerate(A):
if a_i[i] == 0:
for j in range(i+1, n):
if A[j][i]:
break
else:
return 0
A[i], A[j] = A[j], A[i]
a_i = A[i]
res = -res
inv = 1/a_i[i]
for j in range(i+1, n):
a_j = A[j]
t = a_j[i] * inv
for k in range(i+1, n):
a_j[k] -= t * a_i[k]
for i in range(n):
res *= A[i][i]
return res
def mat_pri(A):
for i in A:
print(*i)
def fractionize(A):
for i in range(len(A)):
for j in range(len(A[0])):
A[i][j] = Fraction(A[i][j])
A = [[Fraction(c[i]>>j&1) for j in range(n)]for i in range(n)]
D = determinant(A)
C = dp[-1]
print(int(C+D)//2,int(C-D)//2)