1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217
//! $\\sum\_{i=1}^n \\lfloor m/i\\rfloor$ および $\\sum\_{i=1}^n (m\\bmod i)$.
use std::fmt::Debug;
use std::ops::{
Bound::{Excluded, Included, Unbounded},
RangeBounds,
};
/// $\\sum\_{i=1}^n \\lfloor m/i\\rfloor$ および $\\sum\_{i=1}^n (m\\bmod i)$.
///
/// # Idea
/// $\\lfloor m/\\bullet\\rfloor$ の値は $O(\\sqrt{m})$ 通りである。
/// $i\\in[q\_l, q\_r]$ において $\\lfloor m/i\\rfloor$ の値が共通であるとき、
/// $\\sum\_{i=q\_l}^{q\_r} \\lfloor m/i\\rfloor$ の値は簡単に求められる。
/// また、この範囲で $(m\\bmod i)$ は等差数列を成すことから、
/// $\\sum\_{i=q\_l}^{q\_r} (m\\bmod i)$ も簡単に求められる。
/// 前計算でこれらの累積和を求めておき、差分計算によってクエリ処理を行う。
///
/// # Notes
/// 考察を進めれば $\\sum\_{i=1}^n \\lfloor\\frac{m}{ai+b}\\rfloor$ を求めることも可能?
///
/// # Complexity
/// $O(\\sqrt{m})$ preprocess, $O(1)$ query time.
///
/// # Examples
/// ```
/// use nekolib::math::HarmonicFloorSum;
///
/// let m = 100;
/// let hs = HarmonicFloorSum::new(m);
/// assert_eq!(hs.quot(1..=m), (1..=m).map(|i| m / i).sum());
/// assert_eq!(hs.rem(1..=m), (1..=m).map(|i| m % i).sum());
///
/// let n = 60;
/// assert_eq!(hs.quot(..=n), (1..=n).map(|i| m / i).sum());
/// ```
#[derive(Clone, Debug)]
pub struct HarmonicFloorSum {
m: usize,
q: Vec<usize>,
qsum: Vec<usize>,
rsum: Vec<usize>,
}
impl HarmonicFloorSum {
/// 前処理を行う。
///
/// # Examples
/// ```
/// use nekolib::math::HarmonicFloorSum;
///
/// let m = 100;
/// let hs = HarmonicFloorSum::new(m);
/// ```
pub fn new(m: usize) -> Self {
let mut q = vec![0];
let mut tmp = vec![];
for i in (1..).take_while(|&i| i * i <= m) {
q.push(i);
if i * i < m {
tmp.push(m / i);
}
}
q.extend(tmp.into_iter().rev());
let mut qsum = vec![0; q.len()];
let mut rsum = vec![0; q.len()];
for i in 1..q.len() {
let ql = q[i - 1] + 1;
let qr = q[i];
let qlen = q[i] - q[i - 1];
qsum[i] = qsum[i - 1] + m / q[i] * qlen;
rsum[i] = rsum[i - 1] + (m % ql + m % qr) * qlen / 2;
}
Self { m, q, qsum, rsum }
}
fn search(&self, n: usize) -> usize {
if n > self.m {
self.q.len()
} else if n * n <= self.m {
n
} else {
self.q.len() - (self.m / n)
}
}
fn quot_internal(&self, n: usize) -> usize {
if n == 0 {
return 0;
}
let i = self.search(n);
if i == self.q.len() {
*self.qsum.last().unwrap()
} else if self.q[i] == n {
self.qsum[i]
} else {
self.qsum[i - 1] + (n - self.q[i - 1]) * (self.m / n)
}
}
/// $\\sum\_{i=s}^e \\lfloor m/i\\rfloor$ を返す。
///
/// $\\sum\_{i=s}^{\\infty} \\lfloor m/i\\rfloor = \\sum\_{i=s}^m \\lfloor m/i\\rfloor$
/// なので、上限を指定しない場合は $m$ までの和を求める。下限を指定しない場合は
/// $1$ からの和を求める。
///
/// # Examples
/// ```
/// use nekolib::math::HarmonicFloorSum;
///
/// let m = 100;
/// let hs = HarmonicFloorSum::new(m);
/// assert_eq!(hs.quot(1..=m), (1..=m).map(|i| m / i).sum());
/// assert_eq!(hs.quot(..), (1..=m).map(|i| m / i).sum());
/// assert_eq!(hs.quot(1..=m), hs.quot(1..=m + 1));
/// ```
pub fn quot(&self, r: impl RangeBounds<usize>) -> usize {
let end = match r.end_bound() {
Included(&e) => self.quot_internal(e),
Excluded(&e) => self.quot_internal(e - 1),
Unbounded => *self.qsum.last().unwrap(),
};
let start = match r.start_bound() {
Included(&s) => self.quot_internal(s - 1),
Excluded(&s) => self.quot_internal(s),
Unbounded => 0,
};
end - start
}
fn rem_internal(&self, n: usize) -> usize {
if n == 0 {
return 0;
}
let i = self.search(n);
if i == self.q.len() {
*self.rsum.last().unwrap() + (n - self.m) * self.m
} else if self.q[i] == n {
self.rsum[i]
} else {
let ql = self.q[i - 1] + 1;
let len = n - self.q[i - 1];
self.rsum[i - 1] + (self.m % n + self.m % ql) * len / 2
}
}
/// $\\sum\_{i=s}^e (m\\bmod i)$ を返す。
///
/// 下限を指定しない場合は $1$ からの和を求める。
///
/// # Panics
/// $\\sum\_{i=s}^{\\infty} (m\\bmod i) = \\infty$ なので、上限が `Unbounded` の場合は
/// panic する。
///
/// # Examples
/// ```
/// use nekolib::math::HarmonicFloorSum;
///
/// let m = 100;
/// let hs = HarmonicFloorSum::new(m);
/// assert_eq!(hs.rem(1..=m), (1..=m).map(|i| m % i).sum());
/// assert_eq!(hs.rem(..=m), (1..=m).map(|i| m % i).sum());
/// assert_ne!(hs.rem(1..=m), hs.rem(1..=m + 1)); // m % (m + 1) = m > 0
/// ```
///
/// ```should_panic
/// use nekolib::math::HarmonicFloorSum;
/// let m = 100;
/// let hs = HarmonicFloorSum::new(m);
/// let infty = hs.rem(1..); // diverges
/// ```
pub fn rem(&self, r: impl RangeBounds<usize>) -> usize {
let end = match r.end_bound() {
Included(&e) => self.rem_internal(e),
Excluded(&e) => self.rem_internal(e - 1),
Unbounded => panic!("the infinite sum does not converge"),
};
let start = match r.start_bound() {
Included(&s) => self.rem_internal(s - 1),
Excluded(&s) => self.rem_internal(s),
Unbounded => 0,
};
end - start
}
}
#[test]
fn test_quot() {
let m = 300;
let hs = HarmonicFloorSum::new(m);
for start in 1..=m + 10 {
let mut sum = 0;
for end in start..=m + 10 {
sum += m / end;
assert_eq!(hs.quot(start..=end), sum);
}
let mut sum = 0;
for end in start..=m + 10 {
assert_eq!(hs.quot(start..end), sum);
sum += m / end;
}
}
}
#[test]
fn test_rem() {
let m = 300;
let hs = HarmonicFloorSum::new(m);
for start in 1..=m + 10 {
let mut sum = 0;
for end in start..=m + 10 {
sum += m % end;
assert_eq!(hs.rem(start..=end), sum);
}
let mut sum = 0;
for end in start..=m + 10 {
assert_eq!(hs.rem(start..end), sum);
sum += m % end;
}
}
}