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AtCoder Beginner Contest 245
A - Good morning
import sys
import heapq, math, itertools
from collections import defaultdict, deque
from bisect import bisect_left, bisect_right, insort_left, insort_right
inputs = sys.stdin.readline
mod = 10**9+7
inf = float('inf')
#sys.setrecursionlimit(10**7)
def main():
a,b,c,d = map(int, inputs().split())
print('Takahashi' if a*60+b<=c*60+d else 'Aoki')
if __name__ == '__main__':
main()
Aoki君は1秒後に起きているということは、
B - Mex
import sys
import heapq, math, itertools
from collections import defaultdict, deque
from bisect import bisect_left, bisect_right, insort_left, insort_right
inputs = sys.stdin.readline
mod = 10**9+7
inf = float('inf')
#sys.setrecursionlimit(10**7)
def main():
n = int(input())
a = list(map(int, inputs().split()))
for i in range(2001):
if i not in set(a):
print(i)
exit()
if __name__ == '__main__':
main()
Nが高々2000程度なので候補となる値を全探索すればよさそうです
集合型を使うことで高速に判定しましょう。
C - Choose Elements
import sys
import heapq, math, itertools
from collections import defaultdict, deque
from bisect import bisect_left, bisect_right, insort_left, insort_right
from xmlrpc.client import FastMarshaller
inputs = sys.stdin.readline
mod = 10**9+7
inf = float('inf')
#sys.setrecursionlimit(10**7)
def main():
n,k = map(int, inputs().split())
a = list(map(int, inputs().split()))
b = list(map(int, inputs().split()))
lst = [a[0], b[0]]
flg = True
for i in range(1, n):
if not lst:
flg = False
break
else:
nxt = []
for x in lst:
if abs(x-a[i])<=k:
nxt.append(a[i])
if abs(x-b[i])<=k:
nxt.append(b[i])
lst = list(set(nxt))
if not lst:
flg = False
print('Yes' if flg else 'No')
if __name__ == '__main__':
main()
1つ1つシミュレーションしてあげればよさそうです。
候補となる
また
これをケアしないと際限なく候補の数が増えてしてしまいTLEとなります。
D - Polynomial division
import sys
import numpy as np
import heapq, math, itertools
from collections import defaultdict, deque
from bisect import bisect_left, bisect_right, insort_left, insort_right
inputs = sys.stdin.readline
mod = 10**9+7
inf = float('inf')
#sys.setrecursionlimit(10**7)
def main():
n,m = map(int, inputs().split())
a = list(map(int, inputs().split()))
c = list(map(int, inputs().split()))
pa = np.poly1d(a[::-1])
pc = np.poly1d(c[::-1])
p,q = np.polydiv(pc,pa)
p = np.array(p).astype('int')
print(*list(p)[::-1])
if __name__ == '__main__':
main()
numpyは便利です...!
pythonを使う大きなメリットの1つなので、このような学術的(?)なものはnumpyで簡単に解決できないかということを検討するとよい場合があります
E - Wrapping Chocolate
import sys
import heapq, math, itertools
from collections import defaultdict, deque
from bisect import bisect_left, bisect_right, insort_left, insort_right, insort
inputs = sys.stdin.readline
mod = 10**9+7
inf = float('inf')
#sys.setrecursionlimit(10**7)
from typing import Generic, Iterable, Iterator, TypeVar, Union, List
T = TypeVar('T')
class SortedMultiset(Generic[T]):
BUCKET_RATIO = 50
REBUILD_RATIO = 170
def _build(self, a=None) -> None:
"Evenly divide `a` into buckets."
if a is None: a = list(self)
size = self.size = len(a)
bucket_size = int(math.ceil(math.sqrt(size / self.BUCKET_RATIO)))
self.a = [a[size * i // bucket_size : size * (i + 1) // bucket_size] for i in range(bucket_size)]
def __init__(self, a: Iterable[T] = []) -> None:
"Make a new SortedMultiset from iterable. / O(N) if sorted / O(N log N)"
a = list(a)
if not all(a[i] <= a[i + 1] for i in range(len(a) - 1)):
a = sorted(a)
self._build(a)
def __iter__(self) -> Iterator[T]:
for i in self.a:
for j in i: yield j
def __reversed__(self) -> Iterator[T]:
for i in reversed(self.a):
for j in reversed(i): yield j
def __len__(self) -> int:
return self.size
def __repr__(self) -> str:
return "SortedMultiset" + str(self.a)
def __str__(self) -> str:
s = str(list(self))
return "{" + s[1 : len(s) - 1] + "}"
def _find_bucket(self, x: T) -> List[T]:
"Find the bucket which should contain x. self must not be empty."
for a in self.a:
if x <= a[-1]: return a
return a
def __contains__(self, x: T) -> bool:
if self.size == 0: return False
a = self._find_bucket(x)
i = bisect_left(a, x)
return i != len(a) and a[i] == x
def count(self, x: T) -> int:
"Count the number of x."
return self.index_right(x) - self.index(x)
def add(self, x: T) -> None:
"Add an element. / O(√N)"
if self.size == 0:
self.a = [[x]]
self.size = 1
return
a = self._find_bucket(x)
insort(a, x)
self.size += 1
if len(a) > len(self.a) * self.REBUILD_RATIO:
self._build()
def discard(self, x: T) -> bool:
"Remove an element and return True if removed. / O(√N)"
if self.size == 0: return False
a = self._find_bucket(x)
i = bisect_left(a, x)
if i == len(a) or a[i] != x: return False
a.pop(i)
self.size -= 1
if len(a) == 0: self._build()
return True
def lt(self, x: T) -> Union[T, None]:
"Find the largest element < x, or None if it doesn't exist."
for a in reversed(self.a):
if a[0] < x:
return a[bisect_left(a, x) - 1]
def le(self, x: T) -> Union[T, None]:
"Find the largest element <= x, or None if it doesn't exist."
for a in reversed(self.a):
if a[0] <= x:
return a[bisect_right(a, x) - 1]
def gt(self, x: T) -> Union[T, None]:
"Find the smallest element > x, or None if it doesn't exist."
for a in self.a:
if a[-1] > x:
return a[bisect_right(a, x)]
def ge(self, x: T) -> Union[T, None]:
"Find the smallest element >= x, or None if it doesn't exist."
for a in self.a:
if a[-1] >= x:
return a[bisect_left(a, x)]
def __getitem__(self, x: int) -> T:
"Return the x-th element, or IndexError if it doesn't exist."
if x < 0: x += self.size
if x < 0: raise IndexError
for a in self.a:
if x < len(a): return a[x]
x -= len(a)
raise IndexError
def index(self, x: T) -> int:
"Count the number of elements < x."
ans = 0
for a in self.a:
if a[-1] >= x:
return ans + bisect_left(a, x)
ans += len(a)
return ans
def index_right(self, x: T) -> int:
"Count the number of elements <= x."
ans = 0
for a in self.a:
if a[-1] > x:
return ans + bisect_right(a, x)
ans += len(a)
return ans
def main():
n,m = map(int, inputs().split())
a = list(map(int, inputs().split()))
b = list(map(int, inputs().split()))
c = list(map(int, inputs().split()))
d = list(map(int, inputs().split()))
tot = []
for i in range(n):
tot.append((a[i], b[i], 0))
for i in range(m):
tot.append((c[i], d[i], 1))
tot = sorted(tot, reverse=True)
flg = True
cand = SortedMultiset()
for i,j,k in tot:
if k==1:
cand.add(j)
else:
ele = cand.ge(j)
if ele==None:
flg = False
break
else:
cand.discard(ele)
print('Yes' if flg else 'No')
if __name__ == '__main__':
main()
この問題の難しいポイントは、縦と横の長さの両方を比較・管理しなければいけないところにあると思います。そこで縦の長さか横の長さでソートしてしまえば、比較する方向が１つに減らすことができそうです。(同じ長さの場合は箱を優先することに注意しましょう)
よって問題は「チョコの横幅以上の箱の数が足りているか」という問題に帰着するので、MultiSetを使ってACです。
F - Endless Walk
import sys
import heapq, math, itertools
from collections import defaultdict, deque
from bisect import bisect_left, bisect_right, insort_left, insort_right
inputs = sys.stdin.readline
mod = 10**9+7
inf = float('inf')
#sys.setrecursionlimit(10**7)
def main():
n,m = map(int, inputs().split())
g = [[] for _ in range(n)]
frm = [[] for _ in range(n)]
out = [0]*n
for _ in range(m):
u,v = map(lambda x:int(x)-1, inputs().split())
g[u].append(v)
frm[v].append(u)
out[u] += 1
ans = [1]*n
que = deque([])
for i in range(n):
if out[i]==0:
que.append(i)
while que:
v = que.popleft()
ans[v] = 0
for nx in frm[v]:
out[nx] -= 1
if out[nx]==0 and ans[nx]==1:
que.append(nx)
print(sum(ans))
if __name__ == '__main__':
main()
ダブリングみたいな印象の問題でした。閉路へつながることができるノードの個数を考えればよさそうです。
そこで逆に閉路へつながらないノードの個数を数えることを考えます。これはつまりグラフの端の枝の部分となるので、出次数が0のノードをQueueにより全て削除していけば強連結成分が得られることとなり、これに含まれるノード数が答えとなります。
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