結果
| 問題 | No.1900 Don't be Powers of 2 |
| ユーザー |
 Shirotsume
|
| 提出日時 | 2022-02-14 00:38:13 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 7,025 bytes |
| コンパイル時間 | 295 ms |
| コンパイル使用メモリ | 87,180 KB |
| 実行使用メモリ | 79,336 KB |
| 最終ジャッジ日時 | 2023-09-25 14:07:52 |
| 合計ジャッジ時間 | 8,390 ms |
|
ジャッジサーバーID （参考情報） |
judge15 / judge13 |
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テストケース
テストケース表示| 入力 | 結果 | 実行時間 実行使用メモリ |
|---|---|---|
| testcase_00 | WA | - |
| testcase_01 | WA | - |
| testcase_02 | WA | - |
| testcase_03 | RE | - |
| testcase_04 | RE | - |
| testcase_05 | RE | - |
| testcase_06 | RE | - |
| testcase_07 | RE | - |
| testcase_08 | RE | - |
| testcase_09 | RE | - |
| testcase_10 | RE | - |
| testcase_11 | RE | - |
| testcase_12 | RE | - |
| testcase_13 | RE | - |
| testcase_14 | RE | - |
| testcase_15 | RE | - |
| testcase_16 | RE | - |
| testcase_17 | RE | - |
| testcase_18 | RE | - |
| testcase_19 | RE | - |
| testcase_20 | RE | - |
| testcase_21 | RE | - |
| testcase_22 | RE | - |
| testcase_23 | RE | - |
| testcase_24 | RE | - |
| testcase_25 | RE | - |
| testcase_26 | RE | - |
| testcase_27 | RE | - |
| testcase_28 | RE | - |
| testcase_29 | RE | - |
| testcase_30 | RE | - |
| testcase_31 | AC | 76 ms
75,660 KB |
| testcase_32 | AC | 80 ms
75,588 KB |
| testcase_33 | RE | - |
| testcase_34 | RE | - |
| testcase_35 | RE | - |
| testcase_36 | RE | - |
| testcase_37 | RE | - |
| testcase_38 | RE | - |
| testcase_39 | RE | - |
| testcase_40 | RE | - |
| testcase_41 | RE | - |
| testcase_42 | AC | 73 ms
71,336 KB |
| testcase_43 | AC | 73 ms
71,288 KB |
| testcase_44 | AC | 73 ms
71,388 KB |
ソースコード
class maxflow:
"""It solves maximum flow problem.
"""
def __init__(self, n):
"""It creates a graph of n vertices and 0 edges.
Constraints
-----------
> 0 <= n <= 10 ** 8
Complexity
----------
> O(n)
"""
self.n = n
self.g = [[] for _ in range(self.n)]
self.pos = []
def add_edge(self, from_, to, capacity):
cap=capacity
"""It adds an edge oriented from the vertex `from_` to the vertex `to`
with the capacity `cap` and the flow amount 0.
It returns an integer k such that this is the k-th edge that is added.
Constraints
-----------
> 0 <= from_, to < n
> 0 <= cap
Complexity
----------
> O(1) amortized
"""
# assert 0 <= from_ < self.n
# assert 0 <= to < self.n
# assert 0 <= cap
m = len(self.pos)
self.pos.append((from_, len(self.g[from_])))
self.g[from_].append(self.__class__._edge(to, len(self.g[to]), cap))
self.g[to].append(self.__class__._edge(
from_, len(self.g[from_]) - 1, 0))
return m
class edge:
def __init__(self, from_, to, cap, flow):
self.from_ = from_
self.to = to
self.cap = cap
self.flow = flow
def get_edge(self, i):
"""It returns the current internal state of the edges.
The edges are ordered in the same order as added by add_edge.
Complexity
----------
> O(1)
"""
# assert 0 <= i < len(self.pos)
_e = self.g[self.pos[i][0]][self.pos[i][1]]
_re = self.g[_e.to][_e.rev]
return self.__class__.edge(self.pos[i][0], _e.to, _e.cap + _re.cap, _re.cap)
def edges(self):
"""It returns the current internal state of the edges.
The edges are ordered in the same order as added by add_edge.
Complexity
----------
> O(m), where m is the number of added edges.
"""
result = []
for i in range(len(self.pos)):
_e = self.g[self.pos[i][0]][self.pos[i][1]]
_re = self.g[_e.to][_e.rev]
result.append(self.__class__.edge(
self.pos[i][0], _e.to, _e.cap + _re.cap, _re.cap))
return result
def change_edge(self, i, new_cap, new_flow):
"""It changes the capacity and the flow amount of the ii-th edge to new_cap and new_flow, respectively. It doesn't change the capacity or the flow amount of other edges. See Appendix for further details.
Constraints
-----------
> 0 <= newflow <= newcap
Complexity
----------
> O(1)
"""
# assert 0 <= i < len(self.pos)
# assert 0 <= new_flow <= new_cap
_e = self.g[self.pos[i][0]][self.pos[i][1]]
_re = self.g[_e.to][_e.rev]
_e.cap = new_cap - new_flow
_re.cap = new_flow
def _bfs(self, s, t):
self.level = [-1] * self.n
self.level[s] = 0
q = [s]
while q:
nq = []
for v in q:
for e in self.g[v]:
if e.cap and self.level[e.to] == -1:
self.level[e.to] = self.level[v] + 1
if e.to == t:
return True
nq.append(e.to)
q = nq
return False
def _dfs(self, s, t, up):
st = [t]
while st:
v = st[-1]
if v == s:
st.pop()
flow = up
for w in st:
e = self.g[w][self.it[w]]
flow = min(flow, self.g[e.to][e.rev].cap)
for w in st:
e = self.g[w][self.it[w]]
e.cap += flow
self.g[e.to][e.rev].cap -= flow
return flow
while self.it[v] < len(self.g[v]):
e = self.g[v][self.it[v]]
w = e.to
cap = self.g[e.to][e.rev].cap
if cap and self.level[v] > self.level[w]:
st.append(w)
break
self.it[v] += 1
else:
st.pop()
self.level[v] = self.n
return 0
def flow(self, s, t, flow_limit=float('inf')):
"""It augments the flow from s to t as much as possible.
It returns the amount of the flow augmented.
You may call it multiple times.
See Appendix in the document of AC Library for further details.
Constraints
-----------
> s != t
Complexity
----------
> O(min(n^(2/3)m, m^(3/2))) (if all the capacities are 1) or
> O(n^2 m) (general),
where m is the number of added edges.
"""
# assert 0 <= s < self.n
# assert 0 <= t < self.n
# assert s != t
flow = 0
while flow < flow_limit and self._bfs(s, t):
self.it = [0] * self.n
while flow < flow_limit:
f = self._dfs(s, t, flow_limit - flow)
if not f:
break
flow += f
return flow
def min_cut(self, s):
"""It returns a vector of length n,
such that the i-th element is true if and only if there is a directed path from s to i in the residual network.
The returned vector corresponds to a s−t minimum cut after calling flow(s, t) exactly once without flow_limit.
See Appendix in the document of AC Library for further details.
Complexity
----------
> O(n + m), where m is the number of added edges.
"""
visited = [False] * self.n
q = [s]
while q:
nq = []
for p in q:
visited[p] = True
for e in self.g[p]:
if e.cap and not visited[e.to]:
nq.append(e.to)
q = nq
return visited
class _edge:
def __init__(self, to, rev, cap):
self.to = to
self.rev = rev
self.cap = cap
def isSquare(x):
#xは平方数か？
ng = 3 * 10 ** 9 + 1
ok = 0
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if mid * mid <= x:
ok = mid
else:
ng = mid
if ok * ok == x:
return True
else:
return False
N = int(input())
A = list(map(int,input().split()))
if N > 20:
raise Exception
ans = 0
for i in range(2 ** N):
b = []
for j in range(N):
if 1 & (i >> j):
b.append(A[j])
l = len(b)
flag = True
for x in range(l):
for y in range(l):
if isSquare(b[x] ** 2 + b[y] ** 2):
flag = False
if flag:
ans = max(ans, len(b))
print(ans)