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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define IS_SQUARE(i) (((i) & ((i) - 1)) == 0)

// heapsort from https://www.codereading.com/algo_and_ds/algo/heap_sort.html

void max_heap(int* numbers, int root, int bottom);
void max_heap_recursive(int* numbers, int root, int bottom);

void heap_sort(int* numbers, int array_size)
{
    int i, temp;

    for (i = (array_size - 1) / 2; i >= 0; i--)
    {
        max_heap(numbers, i, array_size - 1);
    }

    for (i = array_size - 1; i > 0; i--)
    {
        temp = numbers[0];
        numbers[0] = numbers[i];
        numbers[i] = temp;
        max_heap(numbers, 0, i - 1);
    }
}

void max_heap(int* numbers, int root, int bottom)
{
    int child = (2 * root) + 1;

    int temp = numbers[root];

    while (child <= bottom) {
        if (child < bottom && numbers[child + 1] > numbers[child]) {
            child = child + 1;
        }
        if (temp > numbers[child]) {
            break;
        } else if (temp <= numbers[child]) {
            numbers[(child - 1) / 2] = numbers[child];
            child = (2 * child) + 1;
        }
    }
    numbers[(child - 1) / 2] = temp;
    return;
}

void max_heap_recursive(int* numbers, int root, int bottom)
{
    int l_idx = (root * 2) + 1;
    int r_idx = (root * 2) + 2; 

    int maxChild;

    if (l_idx <= bottom && numbers[l_idx] > numbers[root]) {
        maxChild = l_idx;
    } else {
        maxChild = root;
    }

    if (r_idx <= bottom && numbers[r_idx] > numbers[maxChild]) {
        maxChild = r_idx;
    }

    if (maxChild != root) {
        int temp = numbers[root];
        numbers[root] = numbers[maxChild];
        numbers[maxChild] = temp;
        max_heap_recursive(numbers, maxChild, bottom);
    }

}

int bitscan_sqrt(int value) {
    int mask = 0x7FFFFFFF;

    while(mask) {
        if (value != (value & mask)) {
            return mask + 1;
        }
        mask = mask>>1;
    }

    return 0;
}

int qsort_compare_int(const void *a, const void *b)
{
    // 降順ソート
    return *(int*) b - *(int*) a;
}

int main(int argc, char *argv[])
{
    int n, i, j;
    int a[200000];
    int x, y;
    int count;
    int min_sq;
    int k;

    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    heap_sort(a, n);

    count = 0;
    for (i = n - 1; i > 0; i--) {
        x = a[i];
        if (x == 0) {continue;}
        min_sq = bitscan_sqrt(x)<<1;
        for (j = i - 1; j >= 0; j--) {
            y = a[j];
            if (y == 0) {continue;}
            if (min_sq - x > y) {break;}
            if (IS_SQUARE(x + y)) {
                count++;
                a[j] = 0;
                break;
            }
        }
    }

    printf("%d\n", count);

    return 0;
}