// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 macro_rules! input { ($($r:tt)*) => { let stdin = std::io::stdin(); let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock())); let mut next = move || -> String{ bytes.by_ref().map(|r|r.unwrap() as char) .skip_while(|c|c.is_whitespace()) .take_while(|c|!c.is_whitespace()) .collect() }; input_inner!{next, $($r)*} }; } macro_rules! input_inner { ($next:expr) => {}; ($next:expr,) => {}; ($next:expr, $var:ident : $t:tt $($r:tt)*) => { let $var = read_value!($next, $t); input_inner!{$next $($r)*} }; } macro_rules! read_value { ($next:expr, [ $t:tt ; $len:expr ]) => { (0..$len).map(|_| read_value!($next, $t)).collect::>() }; ($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error")); } /** * Segment Tree. This data structure is useful for fast folding on intervals of an array * whose elements are elements of monoid I. Note that constructing this tree requires the identity * element of I and the operation of I. * Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581) * AGC015-E (http://agc015.contest.atcoder.jp/submissions/1461001) * yukicoder No. 833 (https://yukicoder.me/submissions/703521) */ struct SegTree { n: usize, orign: usize, dat: Vec, op: BiOp, e: I, } impl SegTree where BiOp: Fn(I, I) -> I, I: Copy { pub fn new(n_: usize, op: BiOp, e: I) -> Self { let mut n = 1; while n < n_ { n *= 2; } // n is a power of 2 SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e} } /* ary[k] <- v */ pub fn update(&mut self, idx: usize, v: I) { debug_assert!(idx < self.orign); let mut k = idx + self.n - 1; self.dat[k] = v; while k > 0 { k = (k - 1) / 2; self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]); } } /* [a, b) (note: half-inclusive) * http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/ */ #[allow(unused)] pub fn query(&self, mut a: usize, mut b: usize) -> I { debug_assert!(a <= b); debug_assert!(b <= self.orign); let mut left = self.e; let mut right = self.e; a += self.n - 1; b += self.n - 1; while a < b { if (a & 1) == 0 { left = (self.op)(left, self.dat[a]); } if (b & 1) == 0 { right = (self.op)(self.dat[b - 1], right); } a = a / 2; b = (b - 1) / 2; } (self.op)(left, right) } } // Depends on: datastr/SegTree.rs // Verified by: https://yukicoder.me/submissions/717436 impl SegTree where BiOp: Fn(I, I) -> I, I: Copy { // Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp #[allow(unused)] fn max_right bool>( &self, mut l: usize, f: &F, ) -> usize { assert!(f(self.e)); if l == self.orign { return self.orign; } l += self.n - 1; let mut sm = self.e; loop { while l % 2 == 1 { l = (l - 1) / 2; } if !f((self.op)(sm, self.dat[l])) { while l < self.n - 1 { l = 2 * l + 1; let val = (self.op)(sm, self.dat[l]); if f(val) { sm = val; l += 1; } } return std::cmp::min(self.orign, l + 1 - self.n); } sm = (self.op)(sm, self.dat[l]); l += 1; if (l + 1).is_power_of_two() { break; } } self.orign } // Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp #[allow(unused)] fn min_left bool>( &self, mut r: usize, f: &F, ) -> usize { if !f(self.e) { return r + 1; } if r == 0 { return 0; } r += self.n - 1; let mut sm = self.e; loop { r -= 1; while r > 0 && r % 2 == 0 { r = (r - 1) / 2; } if !f((self.op)(self.dat[r], sm)) { while r < self.n - 1 { r = 2 * r + 2; let val = (self.op)(self.dat[r], sm); if f(val) { sm = val; r -= 1; } } return r + 2 - self.n; } sm = (self.op)(self.dat[r], sm); if (r + 1).is_power_of_two() { break; } } 0 } } // https://yukicoder.me/problems/no/1863 (2.5) // 左辺は 2 の倍数で、右辺は 0 か 1 であるため、等式が成立するのであれば両方 0 である。 // l を固定した時、左辺が 0 であるような r の範囲は二分探索で求められる。 // また、その範囲内で右辺が 0 であるような r の個数は、前計算すれば求められる。 fn main() { input! { n: usize, a: [i64; n], b: [usize; n], } let mut acc = vec![0; n + 1]; for i in 0..n { acc[i + 1] = acc[i] ^ b[i]; } let mut bcc = vec![[0; 2]; n + 2]; for i in 0..n + 1 { let mut me = bcc[i]; me[acc[i]] += 1; bcc[i + 1] = me; } let mut st = SegTree::new(n, |x, y| { match (x, y) { (Some(x), Some(y)) => if (x & y) != 0 { Some(x | y) } else { None }, _ => None, } }, Some(0)); for i in 0..n { st.update(i, Some(a[i])); } let mut tot = 0i64; for l in 0..n { let r = st.max_right(l, &|x| x.is_some()); let x = acc[l]; tot += bcc[r + 1][x] - bcc[l + 1][x]; } println!("{}", tot); }