Problem 50 "Consecutive prime sum"
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
問 50 「連続する素数の和」
素数41は6つの連続する素数の和として表せる:
41 = 2 + 3 + 5 + 7 + 11 + 13.
100未満の素数を連続する素数の和で表したときにこれが最長になる.
同様に, 連続する素数の和で1000未満の素数を表したときに最長になるのは953で21項を持つ.
100万未満の素数を連続する素数の和で表したときに最長になるのはどの素数か?
fn mod_pow(a: u32, exp: u32, m: u32) -> u32 { let (mut a, mut exp, m) = (a as u64, exp as u64, m as u64); if m == 1 { return 0; } if exp == 0 { return 1; } let mut result = 1; a %= m; loop { if exp % 2 == 1 { result *= a; result %= m; } exp >>= 1; if exp == 0 { break; } a *= a; a %= m; } result as u32 } fn gcd(mut a: u32, mut b: u32) -> u32 { assert!(a != 0 && b != 0); while b != 0 { let r = a % b; a = b; b = r; } a } fn pseudo_fermat_test(n: u32) -> bool { gcd(223092870, n) == 1 && mod_pow(223092870, n - 1, n) == 1 } fn rule_out(sieve: &mut Vec<bool>, prime: usize) { for i in (prime * prime..sieve.len()).step_by(prime) { sieve[i] = false; } } fn rule_out_from_previous_position(sieve: &mut Vec<bool>, prime: usize, pp: usize) { use std::cmp::max; let begin = max((((pp - 1) / prime) + 1) * prime, prime * prime); for i in (begin..sieve.len()).step_by(prime) { sieve[i] = false; } } fn extend(sieve: &mut Vec<bool>, primes: &Vec<usize>) { let previous_len = sieve.len(); sieve.extend(vec![true; previous_len]); for &p in primes { rule_out_from_previous_position(sieve, p, previous_len); } } fn main() { let mut sum = 2u32 + 3; let mut sieve = vec![true; 1000]; let mut primes: Vec<usize> = vec![]; let mut cursor = 5usize; let mut ite = [2usize, 4].iter().cycle(); 'sum_fill: loop { while cursor < sieve.len() { if !sieve[cursor] { cursor += ite.next().unwrap(); continue; } &primes.push(cursor); rule_out(&mut sieve, cursor); sum += cursor as u32; if sum >= 1_000_000 { sum -= cursor as u32; break 'sum_fill; } cursor += ite.next().unwrap(); } extend(&mut sieve, &primes); } primes.insert(0, 2); primes.insert(1, 3); for p in primes { sum -= p as u32; if pseudo_fermat_test(sum) { break; } } println!("{}", sum); assert_eq!(sum, 997651); }
struct Sieve { _sieve: Vec<bool>, } impl Sieve { fn rule_out(&mut self, prime: usize) { for i in (prime * prime..self._sieve.len()).step_by(prime) { self._sieve[i] = false; } } fn init(&mut self) { let sqrt = (self._sieve.len() as f64).sqrt() as usize; let mut index = 5usize; for &i in [2usize, 4].iter().cycle() { if index > sqrt { break; } if self._sieve[index] { self.rule_out(index); } index += i; } } fn new(below: u32) -> Self { assert!(below > 4); let sieve = vec![true; below as usize]; let mut s = Self { _sieve: sieve }; s.init(); s } fn is_prime(&self, n: u32) -> bool { assert!(n < self._sieve.len() as u32); if n == 2 || n == 3 { return true; } if n == 0 || n == 1 || n % 2 == 0 || n % 3 == 0 { return false; } self._sieve[n as usize] } fn primes(&self, below: u32) -> Vec<u32> { let mut primes = vec![2u32, 3u32]; let mut index = 5usize; for &i in [2usize, 4].iter().cycle() { if index >= below as usize { break; } if self._sieve[index] { primes.push(index as u32); } index += i; } primes } } fn fill_sum_up_to_million(primes: &[u32]) -> u32 { let mut sum = 0u32; for &p in primes { sum += p; if sum > 1_000_000 { sum -= p; return sum; } } panic!("The prime list was not enough to fill up the sum to be 1 million!"); } fn main() { let sieve = Sieve::new(1_000_000); let primes = sieve.primes(1_000_000); let mut sum = fill_sum_up_to_million(&primes); for p in primes { sum -= p; if sieve.is_prime(sum) { break; } } println!("{}", sum); assert_eq!(sum, 997651); }