結果
| 問題 |
No.3333 Consecutive Power Sum (Large)
|
| コンテスト | |
| ユーザー |
akakimidori
|
| 提出日時 | 2025-11-02 22:27:11 |
| 言語 | Rust (1.83.0 + proconio) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 7,723 bytes |
| コンパイル時間 | 13,442 ms |
| コンパイル使用メモリ | 399,980 KB |
| 実行使用メモリ | 60,412 KB |
| 最終ジャッジ日時 | 2025-11-02 22:28:58 |
| 合計ジャッジ時間 | 79,497 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 60 TLE * 3 |
コンパイルメッセージ
warning: unused import: `std::io::Write` --> src/main.rs:22:5 | 22 | use std::io::Write; | ^^^^^^^^^^^^^^ | = note: `#[warn(unused_imports)]` on by default warning: type alias `Map` is never used --> src/main.rs:24:6 | 24 | type Map<K, V> = BTreeMap<K, V>; | ^^^ | = note: `#[warn(dead_code)]` on by default warning: type alias `Set` is never used --> src/main.rs:25:6 | 25 | type Set<T> = BTreeSet<T>; | ^^^ warning: type alias `Deque` is never used --> src/main.rs:26:6 | 26 | type Deque<T> = VecDeque<T>; | ^^^^^
ソースコード
// E=1 の列挙はどうやる?
// 素因数分解できるなら前と同じやつやればいい
// まあ1個くらいなら間に合うだろう
//
// 尺取りは可能?
// E=2は無理
// E=3 はギリギリ間に合うかというくらい
// 間に合いそうなので考えない
//
// E=2はどうする?
// R(R+1)(2R+1)/6 - (L-1)L(2L-1)/6 = N
// L^2+(L+1)^2+..+R^2
// 因数分解できる?
//
// a = R - L + 1
// b = L + R
// として
// a(a^2 + 3b^2 - 1) / 12
// これも素因数分解してaを試してbを固定してみる?
use std::collections::*;
use std::io::Write;
type Map<K, V> = BTreeMap<K, V>;
type Set<T> = BTreeSet<T>;
type Deque<T> = VecDeque<T>;
fn main() {
/*
for l in 1u128..=100 {
for r in l..=100 {
let s = (l..=r).map(|i| i.pow(2)).sum::<u128>();
let n = r - l + 1;
let k = l + r;
let v = n * (n * n + 3 * k * k - 1);
assert_eq!(v, 12 * s, "{l} {r}");
}
}
*/
input!(n: u128);
let mut ans = vec![];
// E=1
let m = 2 * n;
for d in divisor(m) {
// R-L+1 = d
// L + R = m/d
let x = d - 1;
let y = m / d;
if (x + y) % 2 != 0 || x >= y {
continue;
}
let r = (x + y) / 2;
let l = (y - x) / 2;
if l <= r && 1 <= l {
ans.push((1, l, r));
}
}
// E = 2
let m = 12 * n;
for d in divisor(m) {
let q = m / d;
if d.saturating_mul(d) - 1 >= q {
continue;
}
let b = q - d * d + 1;
if b % 3 != 0 {
continue;
}
let b = b / 3;
let sq = kth_root(b, 2);
if sq * sq != b {
continue;
}
let a = d - 1;
let b = sq;
if a & 1 != b & 1 || a >= b {
continue;
}
// a = R - L + 1
// b = L + R
let l = (b - a) / 2;
let r = (a + b) / 2;
ans.push((2, l, r));
}
for e in (3..).take_while(|i| 1u128 << i <= n) {
let mut s = 0;
let mut l = 1u128;
for r in (1u128..).take_while(|i| i.pow(e) <= n) {
s += r.pow(e);
while s > n {
s -= l.pow(e);
l += 1;
}
if s == n {
ans.push((e, l, r));
}
}
}
ans.sort();
println!("{}", ans.len());
for (e, l, r) in ans {
println!("{e} {l} {r}");
}
}
pub fn divisor(n: u128) -> Vec<u128> {
let f = factorize(n);
let mut d = vec![1];
for (p, c) in f {
let l = d.len();
for _ in 0..c {
let k = d.len();
for i in 0..l {
let v = d[k - l + i] * p;
d.push(v);
}
}
}
d
}
// ---------- begin input macro ----------
// reference: https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[macro_export]
macro_rules! input {
(source = $s:expr, $($r:tt)*) => {
let mut iter = $s.split_whitespace();
input_inner!{iter, $($r)*}
};
($($r:tt)*) => {
let s = {
use std::io::Read;
let mut s = String::new();
std::io::stdin().read_to_string(&mut s).unwrap();
s
};
let mut iter = s.split_whitespace();
input_inner!{iter, $($r)*}
};
}
#[macro_export]
macro_rules! input_inner {
($iter:expr) => {};
($iter:expr, ) => {};
($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
let $var = read_value!($iter, $t);
input_inner!{$iter $($r)*}
};
}
#[macro_export]
macro_rules! read_value {
($iter:expr, ( $($t:tt),* )) => {
( $(read_value!($iter, $t)),* )
};
($iter:expr, [ $t:tt ; $len:expr ]) => {
(0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
};
($iter:expr, chars) => {
read_value!($iter, String).chars().collect::<Vec<char>>()
};
($iter:expr, bytes) => {
read_value!($iter, String).bytes().collect::<Vec<u8>>()
};
($iter:expr, usize1) => {
read_value!($iter, usize) - 1
};
($iter:expr, $t:ty) => {
$iter.next().unwrap().parse::<$t>().expect("Parse error")
};
}
// ---------- end input macro ----------
// n = prod_i p_i ^ a_i
// (p_i, a_i) の列を返す
pub fn factorize(n: u128) -> Vec<(u128, u32)> {
assert!(n > 0);
let mut list = vec![];
let mut stack = vec![];
stack.push(n);
while let Some(n) = stack.pop() {
if n <= 1 {
continue;
}
if is_prime_miller(n) {
list.push((n, 1));
continue;
}
if n % 2 == 0 {
list.push((2, 1));
stack.push(n / 2);
continue;
}
'pollard: for i in 1u128.. {
let f = |x: u128| (mul(x, x, n) + i) % n;
let mut x = i;
let mut y = f(x);
loop {
let g = binary_gcd(y - x + n, n);
if g == 0 || g == n {
break;
}
if g != 1 {
stack.push(g);
stack.push(n / g);
break 'pollard;
}
x = f(x);
y = f(f(y));
}
}
}
list.sort();
list.dedup_by(|a, b| (a.0 == b.0).then(|| b.1 += a.1).is_some());
list
}
// ---------- begin miller-rabin ----------
pub fn is_prime_miller(n: u128) -> bool {
if n <= 1 {
return false;
} else if n <= 3 {
return true;
} else if n % 2 == 0 {
return false;
}
let pow = |r: u128, mut m: u128| -> u128 {
let mut t = 1u128;
let mut s = r % n;
let n = n as u128;
while m > 0 {
if m & 1 == 1 {
t = mul(t, s, n);
}
s = mul(s, s, n);
m >>= 1;
}
t
};
let mut d = n - 1;
let mut s = 0;
while d % 2 == 0 {
d /= 2;
s += 1;
}
const B: [u128; 13] = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41];
for &b in B.iter() {
let mut a = pow(b, d);
if a <= 1 {
continue;
}
let mut i = 0;
while i < s && a != n - 1 {
i += 1;
a = mul(a, a, n);
}
if i >= s {
return false;
}
}
true
}
// ---------- end miller-rabin ----------
pub fn mul(a: u128, b: u128, m: u128) -> u128 {
let mut ans = 0;
for i in (0..100).rev() {
ans = ans + ans;
if ans >= m {
ans -= m;
}
if b >> i & 1 == 1{
ans += a;
}
if ans >= m {
ans -= m;
}
}
ans
}
// ---------- begin binary_gcd ----------
pub fn binary_gcd(a: u128, b: u128) -> u128 {
if a == 0 || b == 0 {
return a + b;
}
let x = a.trailing_zeros();
let y = b.trailing_zeros();
let mut a = a >> x;
let mut b = b >> y;
while a != b {
let x = (a ^ b).trailing_zeros();
if a < b {
std::mem::swap(&mut a, &mut b);
}
a = (a - b) >> x;
}
a << x.min(y)
}
// ---------- end binary_gcd ----------
// floor(a^(1/k))
pub fn kth_root(a: u128, k: u64) -> u128 {
assert!(k > 0);
if a == 0 {
return 0;
}
if k >= 128 {
return 1;
}
if k == 1 {
return a;
}
let mut v = ((a as f64).ln() / k as f64).exp().round() as u128;
while v.checked_pow(k as u32).map_or(true, |p| p > a) {
v -= 1;
}
while (v + 1).checked_pow(k as u32).map_or(false, |p| p <= a) {
v += 1;
}
v
}
akakimidori