Problem:
Score : 600 points

Problem Statement

Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is A_i.
He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2.
Note that a ball cannot belong to multiple pairs.
Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2^t using some non-negative integer t.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of pairs such that the sum of the integers written on each pair of balls is a power of 2.

Sample Input 1

3
1 2 3

Sample Output 1

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls.
Note that we cannot pair the second ball with itself.

Sample Input 2

5
3 11 14 5 13

Sample Output 2

2

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

// https://atcoder.jp/contests/agc029/tasks/agc029_b
//
#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use std::fmt::*;
use std::io::*;
use std::str::*;

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_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_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>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

fn solve_odd(a: Vec<u64>) -> usize {
    let mut a = a;
    a.sort();
    a.reverse();

    let mut ans = 0;
    let mut mp: HashMap<u64, usize> = HashMap::new();
    for ai in a {
        let mut has = false;
        for w in 1..30 {
            if ai < 1 << w {
                let rem = (1 << w) - ai;
                let mut h = *mp.get(&rem).unwrap_or(&0);
                if h >= 1 {
                    ans += 1;
                    h -= 1;
                    has = true;
                    mp.insert(rem, h);
                    break;
                }
            }
        }
        if !has {
            let w = mp.get(&ai).unwrap_or(&0) + 1;
            mp.insert(ai, w);
        }
    }
    ans
}

fn solve(a: Vec<u64>) -> usize {
    if a.len() <= 1 {
        return 0;
    }
    let mut odds = vec![];
    let mut rem = vec![];
    for ai in a {
        if ai % 2 == 1 {
            odds.push(ai);
        } else {
            rem.push(ai / 2);
        }
    }
    solve_odd(odds) + solve(rem)
}

fn main() {
    input! {
        n: usize,
        a: [u64; n]
    };

    println!("{}", solve(a));
}