// O(A^3)解法 (TLE想定) // mod2048でDP配列を循環させてMLEを回避する手法 fn solve(n: u64, ab: [(u64, u64); 3]) -> u64 { let (a_min, a_max, b_min, b_max) = ab.iter().fold((!0u64, 0u64, !0u64, 0u64), |(a_min, a_max, b_min, b_max), &(a, b)| (a.min(a_min), a.max(a_max), b.min(b_min), b.max(b_max))); assert!(n > 0 && a_min > 0 && a_max < 2048 && b_min > 0); assert!((n / a_min).saturating_mul(b_max) < (1u64 << 61)); let a_lcm = ab .iter() .map(|e| e.0) .reduce(|a, b| a / gcd(a, b) * b) .unwrap(); let mut dp = [i64::MIN; 2048]; dp[0] = 0; let mut v = 0; if n <= a_lcm { for i in 1..=n { for &(a, b) in ab.iter() { v.chmax(dp[(i.wrapping_sub(a) & 2047) as usize] + (b as i64)); } dp[(i & 2047) as usize] = v; } } else { for i in 1..=a_lcm { for &(a, b) in ab.iter() { v.chmax(dp[(i.wrapping_sub(a) & 2047) as usize] + (b as i64)); } dp[(i & 2047) as usize] = v; } let v_lcm = v; for i in a_lcm..=((n % a_lcm) + a_lcm) { for &(a, b) in ab.iter() { v.chmax(dp[(i.wrapping_sub(a) & 2047) as usize] + (b as i64)); } dp[(i & 2047) as usize] = v; } v += v_lcm * ((n / a_lcm - 1) as i64); } v as u64 } fn main() { let mut stdinlock = std::io::stdin().lock(); let mut lines = std::io::BufRead::lines(&mut stdinlock); let n = lines.next().unwrap().unwrap().parse::().unwrap(); let ab = (0..3).map(|_| { let s = lines.next().unwrap().unwrap(); let mut t = s.split_whitespace(); ( t.next().unwrap().parse::().unwrap(), t.next().unwrap().parse::().unwrap(), ) }).collect::>(); println!("{}", solve(n, [ab[0], ab[1], ab[2]])); } fn gcd(mut a: u64, mut b: u64) -> u64 { if a == 0 || b == 0 { return a | b; } let (tza, tzb) = (a.trailing_zeros(), b.trailing_zeros()); a >>= tza; b >>= tzb; let tzt = tza.min(tzb); while a != b { if a > b { a -= b; a >>= a.trailing_zeros(); } else { b -= a; b >>= b.trailing_zeros(); } } a << tzt } trait Change { fn chmax(&mut self, x: Self); fn chmin(&mut self, x: Self); } impl Change for T { fn chmax(&mut self, x: T) { if *self < x { *self = x; } } fn chmin(&mut self, x: T) { if *self > x { *self = x; } } }