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sqrt decomposition; tree transfer; tree diameter;
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old-yan
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,117 @@ | ||
| /* | ||
| 最后修改: | ||
| 20240621 | ||
| 测试环境: | ||
| gcc11.2,c++11 | ||
| clang12.0,C++11 | ||
| msvc14.2,C++14 | ||
| */ | ||
| #ifndef __OY_SQRTDECOMPOSITION__ | ||
| #define __OY_SQRTDECOMPOSITION__ | ||
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| #include <algorithm> | ||
| #include <cmath> | ||
| #include <cstdint> | ||
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| namespace OY { | ||
| struct SqrtDecomposition { | ||
| struct iterator { | ||
| struct _range { | ||
| uint64_t m_quot, m_left, m_right; | ||
| template <typename Ostream> | ||
| friend Ostream &operator<<(Ostream &out, const _range &x) { return out << '(' << x.m_quot << ": " << x.m_left << '~' << x.m_right << ')'; } | ||
| } m_range; | ||
| uint64_t m_val; | ||
| iterator &operator--() { | ||
| if (m_range.m_quot == 1) | ||
| --m_range.m_quot; | ||
| else { | ||
| m_range.m_left = m_range.m_right + 1; | ||
| m_range.m_quot = m_val / m_range.m_left; | ||
| if (m_range.m_quot >= m_range.m_left) | ||
| m_range.m_right = m_range.m_left; | ||
| else | ||
| m_range.m_right = m_val / m_range.m_quot; | ||
| } | ||
| return *this; | ||
| } | ||
| iterator &operator++() { | ||
| if (m_range.m_quot + 1 < m_range.m_left) { | ||
| m_range.m_right = m_range.m_left - 1; | ||
| m_range.m_left = m_val / (m_range.m_quot + 2) + 1; | ||
| m_range.m_quot++; | ||
| } else if (--m_range.m_left) | ||
| m_range.m_quot = m_val / (m_range.m_right = m_range.m_left); | ||
| else | ||
| m_range.m_quot = -1; | ||
| return *this; | ||
| } | ||
| const _range &operator*() const { return m_range; } | ||
| const _range *operator->() const { return &m_range; } | ||
| bool operator!=(const iterator &rhs) const { return m_range.m_quot != rhs.m_range.m_quot; } | ||
| bool operator==(const iterator &rhs) const { return m_range.m_quot == rhs.m_range.m_quot; } | ||
| static iterator null(uint64_t val) { return {{uint64_t(-1)}, val}; } | ||
| }; | ||
| struct reverse_iterator : iterator { | ||
| iterator base() const { return *this; } | ||
| reverse_iterator &operator--() { | ||
| iterator::operator++(); | ||
| return *this; | ||
| } | ||
| reverse_iterator &operator++() { | ||
| iterator::operator--(); | ||
| return *this; | ||
| } | ||
| static reverse_iterator null(uint64_t val) { | ||
| reverse_iterator res; | ||
| res.m_range = {0, val + 1, uint64_t(-1)}, res.m_val = val; | ||
| return res; | ||
| } | ||
| }; | ||
| uint64_t m_val; | ||
| SqrtDecomposition(uint64_t val) : m_val(val) {} | ||
| uint64_t size() const { | ||
| uint64_t r = std::sqrt((long double)m_val); | ||
| return r * 2 - (m_val * 2 + 1 < (r + 1) * r * 2); | ||
| } | ||
| iterator lower_bound(uint64_t quot) const { | ||
| if (!quot) return rend().base(); | ||
| if (quot > m_val) return end(); | ||
| uint64_t right = m_val / quot; | ||
| if (quot < right) { | ||
| uint64_t left = m_val / (quot + 1) + 1; | ||
| return iterator{{quot, left, right}, m_val}; | ||
| } else | ||
| return iterator{{m_val / right, right, right}, m_val}; | ||
| } | ||
| iterator upper_bound(uint64_t quot) const { return lower_bound(quot + 1); } | ||
| iterator find_by_divisor(uint64_t divisor) const { | ||
| if (!divisor) return end(); | ||
| if (divisor > m_val) return rend().base(); | ||
| uint64_t quot = m_val / divisor; | ||
| if (divisor <= quot) return iterator{{quot, divisor, divisor}, m_val}; | ||
| uint64_t right = m_val / quot; | ||
| uint64_t left = m_val / (quot + 1) + 1; | ||
| return iterator{{quot, left, right}, m_val}; | ||
| } | ||
| iterator begin() const { return iterator{{1, m_val / 2 + 1, m_val}, m_val}; } | ||
| iterator end() const { return iterator::null(m_val); } | ||
| reverse_iterator rbegin() const { | ||
| reverse_iterator res; | ||
| res.m_range = {m_val, 1, 1}, res.m_val = m_val; | ||
| return res; | ||
| } | ||
| reverse_iterator rend() const { return reverse_iterator::null(m_val); } | ||
| }; | ||
| template <typename Ostream> | ||
| Ostream &operator<<(Ostream &out, const SqrtDecomposition &x) { | ||
| out << '{'; | ||
| for (auto i : x) { | ||
| if (i.m_quot != 1) out << ", "; | ||
| out << i; | ||
| } | ||
| return out << '}'; | ||
| } | ||
| } | ||
|
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||
| #endif |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,95 @@ | ||
| ### 一、模板类别 | ||
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| 数学：数论分块。 | ||
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| 练习题目： | ||
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| 1. [Enumerate Quotients](https://judge.yosupo.jp/problem/enumerate_quotients)(https://github.com/yosupo06/library-checker-problems/issues/922) | ||
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| ### 二、模板功能 | ||
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| 本模板提供一个 `SqrtDecomposition` 类，按照 `1~n` 每个数字作为除数去整除 `n` 的结果分为了若干个区间。 | ||
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| 本模板支持基于范围的 `for` 循环，也支持反向迭代循环；可以通过整除结果使用 `lower_bound` 和 `upper_bound` 获取迭代器。 | ||
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| ### 三、模板示例 | ||
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| ```c++ | ||
| #include "IO/FastIO.h" | ||
| #include "MATH/SqrtDecomposition.h" | ||
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| int main() { | ||
| uint32_t n = 105; | ||
| OY::SqrtDecomposition sd(n); | ||
| cout << "there are " << sd.size() << " blocks\n"; | ||
| for (auto range : sd) { | ||
| cout << "floor(" << n << " / x) = " << range.m_quot << ", \t" << range.m_left << " <= x <= " << range.m_right << endl; | ||
| } | ||
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| // 也可以反向迭代 | ||
| cout << "\nreversed:\n"; | ||
| for (auto it = sd.rbegin(); it != sd.rend(); ++it) { | ||
| cout << "floor(" << n << " / x) = " << it->m_quot << ", \t" << it->m_left << " <= x <= " << it->m_right << endl; | ||
| } | ||
|
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| // 可以二分找到一个迭代器 | ||
| cout << "\nlower_bound:\n"; | ||
| auto it = sd.lower_bound(13); | ||
| cout << "floor(" << n << " / x) = " << it->m_quot << ", \t" << it->m_left << " <= x <= " << it->m_right << endl; | ||
| it = sd.upper_bound(13); | ||
| cout << "floor(" << n << " / x) = " << it->m_quot << ", \t" << it->m_left << " <= x <= " << it->m_right << endl; | ||
| } | ||
| ``` | ||
|
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| ``` | ||
| #输出如下 | ||
| there are 19 blocks | ||
| floor(105 / x) = 1, 53 <= x <= 105 | ||
| floor(105 / x) = 2, 36 <= x <= 52 | ||
| floor(105 / x) = 3, 27 <= x <= 35 | ||
| floor(105 / x) = 4, 22 <= x <= 26 | ||
| floor(105 / x) = 5, 18 <= x <= 21 | ||
| floor(105 / x) = 6, 16 <= x <= 17 | ||
| floor(105 / x) = 7, 14 <= x <= 15 | ||
| floor(105 / x) = 8, 12 <= x <= 13 | ||
| floor(105 / x) = 9, 11 <= x <= 11 | ||
| floor(105 / x) = 10, 10 <= x <= 10 | ||
| floor(105 / x) = 11, 9 <= x <= 9 | ||
| floor(105 / x) = 13, 8 <= x <= 8 | ||
| floor(105 / x) = 15, 7 <= x <= 7 | ||
| floor(105 / x) = 17, 6 <= x <= 6 | ||
| floor(105 / x) = 21, 5 <= x <= 5 | ||
| floor(105 / x) = 26, 4 <= x <= 4 | ||
| floor(105 / x) = 35, 3 <= x <= 3 | ||
| floor(105 / x) = 52, 2 <= x <= 2 | ||
| floor(105 / x) = 105, 1 <= x <= 1 | ||
| reversed: | ||
| floor(105 / x) = 105, 1 <= x <= 1 | ||
| floor(105 / x) = 52, 2 <= x <= 2 | ||
| floor(105 / x) = 35, 3 <= x <= 3 | ||
| floor(105 / x) = 26, 4 <= x <= 4 | ||
| floor(105 / x) = 21, 5 <= x <= 5 | ||
| floor(105 / x) = 17, 6 <= x <= 6 | ||
| floor(105 / x) = 15, 7 <= x <= 7 | ||
| floor(105 / x) = 13, 8 <= x <= 8 | ||
| floor(105 / x) = 11, 9 <= x <= 9 | ||
| floor(105 / x) = 10, 10 <= x <= 10 | ||
| floor(105 / x) = 9, 11 <= x <= 11 | ||
| floor(105 / x) = 8, 12 <= x <= 13 | ||
| floor(105 / x) = 7, 14 <= x <= 15 | ||
| floor(105 / x) = 6, 16 <= x <= 17 | ||
| floor(105 / x) = 5, 18 <= x <= 21 | ||
| floor(105 / x) = 4, 22 <= x <= 26 | ||
| floor(105 / x) = 3, 27 <= x <= 35 | ||
| floor(105 / x) = 2, 36 <= x <= 52 | ||
| floor(105 / x) = 1, 53 <= x <= 105 | ||
| lower_bound: | ||
| floor(105 / x) = 13, 8 <= x <= 8 | ||
| floor(105 / x) = 15, 7 <= x <= 7 | ||
| ``` | ||
|
|
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,73 @@ | ||
| #include "IO/FastIO.h" | ||
| #include "MATH/SqrtDecomposition.h" | ||
|
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||
| int main() { | ||
| uint32_t n = 105; | ||
| OY::SqrtDecomposition sd(n); | ||
| cout << "there are " << sd.size() << " blocks\n"; | ||
| for (auto range : sd) { | ||
| cout << "floor(" << n << " / x) = " << range.m_quot << ", \t" << range.m_left << " <= x <= " << range.m_right << endl; | ||
| } | ||
|
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| // 也可以反向迭代 | ||
| cout << "\nreversed:\n"; | ||
| for (auto it = sd.rbegin(); it != sd.rend(); ++it) { | ||
| cout << "floor(" << n << " / x) = " << it->m_quot << ", \t" << it->m_left << " <= x <= " << it->m_right << endl; | ||
| } | ||
|
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| // 可以二分找到一个迭代器 | ||
| cout << "\nlower_bound:\n"; | ||
| auto it = sd.lower_bound(13); | ||
| cout << "floor(" << n << " / x) = " << it->m_quot << ", \t" << it->m_left << " <= x <= " << it->m_right << endl; | ||
| it = sd.upper_bound(13); | ||
| cout << "floor(" << n << " / x) = " << it->m_quot << ", \t" << it->m_left << " <= x <= " << it->m_right << endl; | ||
| } | ||
| /* | ||
| #输出如下 | ||
| there are 19 blocks | ||
| floor(105 / x) = 1, 53 <= x <= 105 | ||
| floor(105 / x) = 2, 36 <= x <= 52 | ||
| floor(105 / x) = 3, 27 <= x <= 35 | ||
| floor(105 / x) = 4, 22 <= x <= 26 | ||
| floor(105 / x) = 5, 18 <= x <= 21 | ||
| floor(105 / x) = 6, 16 <= x <= 17 | ||
| floor(105 / x) = 7, 14 <= x <= 15 | ||
| floor(105 / x) = 8, 12 <= x <= 13 | ||
| floor(105 / x) = 9, 11 <= x <= 11 | ||
| floor(105 / x) = 10, 10 <= x <= 10 | ||
| floor(105 / x) = 11, 9 <= x <= 9 | ||
| floor(105 / x) = 13, 8 <= x <= 8 | ||
| floor(105 / x) = 15, 7 <= x <= 7 | ||
| floor(105 / x) = 17, 6 <= x <= 6 | ||
| floor(105 / x) = 21, 5 <= x <= 5 | ||
| floor(105 / x) = 26, 4 <= x <= 4 | ||
| floor(105 / x) = 35, 3 <= x <= 3 | ||
| floor(105 / x) = 52, 2 <= x <= 2 | ||
| floor(105 / x) = 105, 1 <= x <= 1 | ||
| reversed: | ||
| floor(105 / x) = 105, 1 <= x <= 1 | ||
| floor(105 / x) = 52, 2 <= x <= 2 | ||
| floor(105 / x) = 35, 3 <= x <= 3 | ||
| floor(105 / x) = 26, 4 <= x <= 4 | ||
| floor(105 / x) = 21, 5 <= x <= 5 | ||
| floor(105 / x) = 17, 6 <= x <= 6 | ||
| floor(105 / x) = 15, 7 <= x <= 7 | ||
| floor(105 / x) = 13, 8 <= x <= 8 | ||
| floor(105 / x) = 11, 9 <= x <= 9 | ||
| floor(105 / x) = 10, 10 <= x <= 10 | ||
| floor(105 / x) = 9, 11 <= x <= 11 | ||
| floor(105 / x) = 8, 12 <= x <= 13 | ||
| floor(105 / x) = 7, 14 <= x <= 15 | ||
| floor(105 / x) = 6, 16 <= x <= 17 | ||
| floor(105 / x) = 5, 18 <= x <= 21 | ||
| floor(105 / x) = 4, 22 <= x <= 26 | ||
| floor(105 / x) = 3, 27 <= x <= 35 | ||
| floor(105 / x) = 2, 36 <= x <= 52 | ||
| floor(105 / x) = 1, 53 <= x <= 105 | ||
| lower_bound: | ||
| floor(105 / x) = 13, 8 <= x <= 8 | ||
| floor(105 / x) = 15, 7 <= x <= 7 | ||
| */ |
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