[牛客周赛复盘] 牛客周赛 Round 28 20240114
- 总结
- A\B
- 1. 题目描述
- 2. 思路分析
- 3. 代码实现
- 小红的炸砖块
- 1. 题目描述
- 2. 思路分析
- 3. 代码实现
- 小红统计区间(easy)
- 1. 题目描述
- 2. 思路分析
- 3. 代码实现
- 小红的好数组
- 1. 题目描述
- 2. 思路分析
- 3. 代码实现
- 小红统计区间(hard)
- 1. 题目描述
- 2. 思路分析
- 3. 代码实现
- 六、参考链接
总结
- 这周据说难度对标abc,根本扯淡。。
- E比较有难度,D赛中没想出来正数有啥用,直接前缀和莽F交了。
- A 模拟
- B 模拟
- C 模拟
- D 滑窗
- E 数学找规律
- F 前缀和+有序集合
A\B
链接: 小红的新周赛
链接: 小红的字符串
1. 题目描述
2. 思路分析
- 按题意模拟。
3. 代码实现
def solve():print(sum(RI()))
def solve():s, = RS()ans = []for i in range(len(s)-1):ans.append(s[i:i+2])ans.sort()print(*ans,sep='\n')
小红的炸砖块
链接: 小红的炸砖块
1. 题目描述
2. 思路分析
- 用一个长为列宽的数组a记录每列当前最高砖块的行坐标即可。
- 炸x,y的时候可以判断a[y]<=x,即y列的x位置是否有砖块。
3. 代码实现
def solve():m,n,k = RI()a = [0]*nfor _ in range(k):x,y = RI()x -= 1y -= 1if a[y] <= x:a[y] += 1for i in range(m):ans = ['.']*nfor j in range(n):if a[j] <= i:ans[j] = '*'print(*ans,sep='')
小红统计区间(easy)
链接: 小红统计区间(easy)
1. 题目描述
2. 思路分析
- 全是正数,那么可以滑窗,复杂度O(n)
- 维护[j,i]区间,使这个区间和恰好不到k,那么j向左一下就能到k,所以把所有j(窗首)加起来即可。
- 注意由于这个窗可能为空,所以双指针写法比队列方便。
3. 代码实现
# ms
def solve2():n, k = RI()a = RILST()s = ans = 0q = deque()for i, v in enumerate(a):q.append(i)s += vwhile q and s >= k:s -= a[q.popleft()]ans += q[0] if q else i + 1print(ans)def solve():n, k = RI()a = RILST()j = s = ans = 0for i, v in enumerate(a):s += vwhile j <= i and s >= k:s -= a[j]j += 1ans += jprint(ans)
小红的好数组
链接: 小红的好数组
1. 题目描述
2. 思路分析
- 长为三的子段和全是偶数的话,显然可以所有数据全填偶数。而1~k里一共m=k//2个偶数,每个位置填任意偶数是独立的。显然方案数是pow(m,n)。
- 同理,在这些位置中选一些奇数,每个位置有x=k-m个奇数,也是独立事件。只需讨论一下奇数偶数的位置和个数即可。
- 在纸上画了一下,发现合法好数组一定是121 121 121 …(这里12代表奇数偶数) 三个一循环的数,且可以从中任意位置截取。
- 即开始位置共有三种:[1]21121121…;[2]11211211…;[1]12112112…。
- 分别讨论这三种,然后计算1和2的数量即可,这里我只计算2的数量m(因为每个循环节里只有1个2),取个模分类讨论即可。那么1的数量就是n-m。
3. 代码实现
def solve():n, k = RI()a, b = k // 2, (k + 1) // 2 # 1~k的偶数个数,奇数个数# 121 121 121 三个一循环ans = pow(a, n, MOD) # 全填偶数x, y = divmod(n, 3)# 121开头two = x + int(y >= 2)ans += pow(a, two, MOD) * pow(b, n - two, MOD) % MOD# 211开头two = x + int(y >= 1)ans += pow(a, two, MOD) * pow(b, n - two, MOD) % MOD# 112开头two = xans += pow(a, two, MOD) * pow(b, n - two, MOD) % MODprint(ans % MOD)
小红统计区间(hard)
链接: 小红统计区间(hard)
1. 题目描述
2. 思路分析
- 看到字段和,首先想到前缀和。s[i…j] = pre[j+1]-pre[i] (pre前补0)
- 那么对于每个j,其实就是找前边满足<=pre[j+1]-k的数量,这可以用有序集合二分来解决。
3. 代码实现
class CuteSortedList:def __init__(self, iterable=[], _load=200):"""Initialize sorted list instance."""values = sorted(iterable)self._len = _len = len(values)self._load = _loadself._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]self._list_lens = [len(_list) for _list in _lists]self._mins = [_list[0] for _list in _lists]self._fen_tree = []self._rebuild = Truedef _fen_build(self):"""Build a fenwick tree instance."""self._fen_tree[:] = self._list_lens_fen_tree = self._fen_treefor i in range(len(_fen_tree)):if i | i + 1 < len(_fen_tree):_fen_tree[i | i + 1] += _fen_tree[i]self._rebuild = Falsedef _fen_update(self, index, value):"""Update `fen_tree[index] += value`."""if not self._rebuild:_fen_tree = self._fen_treewhile index < len(_fen_tree):_fen_tree[index] += valueindex |= index + 1def _fen_query(self, end):"""Return `sum(_fen_tree[:end])`."""if self._rebuild:self._fen_build()_fen_tree = self._fen_treex = 0while end:x += _fen_tree[end - 1]end &= end - 1return xdef _fen_findkth(self, k):"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""_list_lens = self._list_lensif k < _list_lens[0]:return 0, kif k >= self._len - _list_lens[-1]:return len(_list_lens) - 1, k + _list_lens[-1] - self._lenif self._rebuild:self._fen_build()_fen_tree = self._fen_treeidx = -1for d in reversed(range(len(_fen_tree).bit_length())):right_idx = idx + (1 << d)if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:idx = right_idxk -= _fen_tree[idx]return idx + 1, kdef _delete(self, pos, idx):"""Delete value at the given `(pos, idx)`."""_lists = self._lists_mins = self._mins_list_lens = self._list_lensself._len -= 1self._fen_update(pos, -1)del _lists[pos][idx]_list_lens[pos] -= 1if _list_lens[pos]:_mins[pos] = _lists[pos][0]else:del _lists[pos]del _list_lens[pos]del _mins[pos]self._rebuild = Truedef _loc_left(self, value):"""Return an index pair that corresponds to the first position of `value` in the sorted list."""if not self._len:return 0, 0_lists = self._lists_mins = self._minslo, pos = -1, len(_lists) - 1while lo + 1 < pos:mi = (lo + pos) >> 1if value <= _mins[mi]:pos = mielse:lo = miif pos and value <= _lists[pos - 1][-1]:pos -= 1_list = _lists[pos]lo, idx = -1, len(_list)while lo + 1 < idx:mi = (lo + idx) >> 1if value <= _list[mi]:idx = mielse:lo = mireturn pos, idxdef _loc_right(self, value):"""Return an index pair that corresponds to the last position of `value` in the sorted list."""if not self._len:return 0, 0_lists = self._lists_mins = self._minspos, hi = 0, len(_lists)while pos + 1 < hi:mi = (pos + hi) >> 1if value < _mins[mi]:hi = mielse:pos = mi_list = _lists[pos]lo, idx = -1, len(_list)while lo + 1 < idx:mi = (lo + idx) >> 1if value < _list[mi]:idx = mielse:lo = mireturn pos, idxdef add(self, value):"""Add `value` to sorted list."""_load = self._load_lists = self._lists_mins = self._mins_list_lens = self._list_lensself._len += 1if _lists:pos, idx = self._loc_right(value)self._fen_update(pos, 1)_list = _lists[pos]_list.insert(idx, value)_list_lens[pos] += 1_mins[pos] = _list[0]if _load + _load < len(_list):_lists.insert(pos + 1, _list[_load:])_list_lens.insert(pos + 1, len(_list) - _load)_mins.insert(pos + 1, _list[_load])_list_lens[pos] = _loaddel _list[_load:]self._rebuild = Trueelse:_lists.append([value])_mins.append(value)_list_lens.append(1)self._rebuild = Truedef discard(self, value):"""Remove `value` from sorted list if it is a member."""_lists = self._listsif _lists:pos, idx = self._loc_right(value)if idx and _lists[pos][idx - 1] == value:self._delete(pos, idx - 1)def remove(self, value):"""Remove `value` from sorted list; `value` must be a member."""_len = self._lenself.discard(value)if _len == self._len:raise ValueError('{0!r} not in list'.format(value))def pop(self, index=-1):"""Remove and return value at `index` in sorted list."""pos, idx = self._fen_findkth(self._len + index if index < 0 else index)value = self._lists[pos][idx]self._delete(pos, idx)return valuedef bisect_left(self, value):"""Return the first index to insert `value` in the sorted list."""pos, idx = self._loc_left(value)return self._fen_query(pos) + idxdef bisect_right(self, value):"""Return the last index to insert `value` in the sorted list."""pos, idx = self._loc_right(value)return self._fen_query(pos) + idxdef count(self, value):"""Return number of occurrences of `value` in the sorted list."""return self.bisect_right(value) - self.bisect_left(value)def __len__(self):"""Return the size of the sorted list."""return self._len# def __getitem__(self, index):# """Lookup value at `index` in sorted list."""# pos, idx = self._fen_findkth(self._len + index if index < 0 else index)# return self._lists[pos][idx]def __getitem__(self, index):"""Lookup value at `index` in sorted list."""if isinstance(index, slice):_lists = self._listsstart, stop, step = index.indices(self._len)if step == 1 and start < stop: # 如果是正向的步进1,找到起起止点,然后把中间的拼接起来即可if start == 0 and stop == self._len: # 全部return reduce(iadd, self._lists, [])start_pos, start_idx = self._fen_findkth(start)start_list = _lists[start_pos]stop_idx = start_idx + stop - start# Small slice optimization: start index and stop index are# within the start list.if len(start_list) >= stop_idx:return start_list[start_idx:stop_idx]if stop == self._len:stop_pos = len(_lists) - 1stop_idx = len(_lists[stop_pos])else:stop_pos, stop_idx = self._fen_findkth(stop)prefix = _lists[start_pos][start_idx:]middle = _lists[(start_pos + 1):stop_pos]result = reduce(iadd, middle, prefix)result += _lists[stop_pos][:stop_idx]return resultif step == -1 and start > stop: # 如果是负向的步进1,直接翻转调用自己再翻转即可result = self.__getitem__(slice(stop + 1, start + 1))result.reverse()return resultindices = range(start, stop, step) # 若不是步进1,只好一个一个取return list(self.__getitem__(index) for index in indices)else:pos, idx = self._fen_findkth(self._len + index if index < 0 else index)return self._lists[pos][idx]def __delitem__(self, index):"""Remove value at `index` from sorted list."""pos, idx = self._fen_findkth(self._len + index if index < 0 else index)self._delete(pos, idx)def __contains__(self, value):"""Return true if `value` is an element of the sorted list."""_lists = self._listsif _lists:pos, idx = self._loc_left(value)return idx < len(_lists[pos]) and _lists[pos][idx] == valuereturn Falsedef __iter__(self):"""Return an iterator over the sorted list."""return (value for _list in self._lists for value in _list)def __reversed__(self):"""Return a reverse iterator over the sorted list."""return (value for _list in reversed(self._lists) for value in reversed(_list))def __repr__(self):"""Return string representation of sorted list."""return 'SortedList({0})'.format(list(self))# ms
def solve():n, k = RI()a = RILST()s = ans = 0"""y-x>=kx <= y-k"""sl = CuteSortedList([0])for v in a:s += v# print(s,sl)ans += sl.bisect_right(s-k)sl.add(s)print(ans)
六、参考链接
- 无