Buggy python submission (verdict: Time Limit Exceeded):
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを作る
    adj = defaultdict(list)
    for a in DISTINCT_A:
        for e in range(MAX_EXP + 1):
            b = pow2[e] - a
            if a <= b:
                break
            if counter[b] == 0:
                continue
            adj[a].append(b)
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        for b in adj[a]:
            cnt = min(counter[a], counter[b])
            ans += cnt
            counter[a] -= cnt
            counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()


Fixed submission:
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを考え
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        b = pow2[a.bit_length()] - a
        if counter[b] == 0:
            continue
        cnt = min(counter[a], counter[b])
        ans += cnt
        counter[a] -= cnt
        counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Buggy python submission (verdict: Time Limit Exceeded):
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを作る
    adj = defaultdict(list)
    for a in DISTINCT_A:
        for e in range(MAX_EXP + 1):
            b = pow2[e] - a
            if counter[b] == 0:
                continue
            if a > b:
                adj[a].append(b)
            elif a < b:
                adj[b].append(a)
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        for b in adj[a]:
            cnt = min(counter[a], counter[b])
            ans += cnt
            counter[a] -= cnt
            counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Fixed submission:
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを考え
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        b = pow2[a.bit_length()] - a
        if counter[b] == 0:
            continue
        cnt = min(counter[a], counter[b])
        ans += cnt
        counter[a] -= cnt
        counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Buggy python submission (verdict: Time Limit Exceeded):
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(set(A))
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを作る
    adj = defaultdict(list)
    for a in DISTINCT_A:
        for e in range(MAX_EXP + 1):
            b = 2**e - a
            if counter[b] == 0:
                continue
            if a > b:
                adj[a].append(b)
            elif a < b:
                adj[b].append(a)
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        for b in adj[a]:
            cnt = min(counter[a], counter[b])
            ans += cnt
            counter[a] -= cnt
            counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = 2**e
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Fixed submission:
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを考え
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        b = pow2[a.bit_length()] - a
        if counter[b] == 0:
            continue
        cnt = min(counter[a], counter[b])
        ans += cnt
        counter[a] -= cnt
        counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Buggy python submission (verdict: Time Limit Exceeded):
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    A.sort(reverse=True)  # 大きい順に並べておく
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを作る
    adj = defaultdict(list)
    for a in A:
        for e in range(MAX_EXP + 1):
            b = 2**e - a
            if counter[b] == 0:
                continue
            if a > b:
                adj[a].append(b)
            elif a < b:
                adj[b].append(a)
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in A:
        for b in adj[a]:
            cnt = min(counter[a], counter[b])
            ans += cnt
            counter[a] -= cnt
            counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = 2**e
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Fixed submission:
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを考え
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        b = pow2[a.bit_length()] - a
        if counter[b] == 0:
            continue
        cnt = min(counter[a], counter[b])
        ans += cnt
        counter[a] -= cnt
        counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Buggy python submission (verdict: Time Limit Exceeded):
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    A.sort(reverse=True)  # 大きい順に並べておく
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを作る
    adj = defaultdict(list)
    for a in A:
        for e in range(MAX_EXP + 1):
            b = 2**e - a
            if counter[b] == 0:
                continue
            if a > b:
                adj[a].append(b)
            elif a < b:
                adj[b].append(a)
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in A:
        for b in adj[a]:
            cnt = min(counter[a], counter[b])
            ans += cnt
            counter[a] -= cnt
            counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = 2**e
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()

Fixed submission:
from collections import defaultdict


def main():
    N = int(input())
    A = list(map(int, input().split()))
    MAX_EXP = 30  # 2**30 < 2 * (10 ** 9) < 2**31
    pow2 = [1] * (MAX_EXP + 1)
    for e in range(1, MAX_EXP + 1):
        pow2[e] = 2 * pow2[e - 1]
    counter = defaultdict(int)
    for a in A:
        counter[a] += 1
    DISTINCT_A = list(counter.keys())
    DISTINCT_A.sort(reverse=True)  # 大きい順に並べておく
    # 和が2のべき乗になるペアの間をedgeで結んだグラフを考え
    # ノードの値が大きい順に貪欲にペアを作っていく
    ans = 0
    for a in DISTINCT_A:
        b = pow2[a.bit_length()] - a
        if counter[b] == 0:
            continue
        cnt = min(counter[a], counter[b])
        ans += cnt
        counter[a] -= cnt
        counter[b] -= cnt
    # 最後に2のべき乗のノードはそれ単体でペアを作れるので、その分を足す
    for e in range(MAX_EXP + 1):
        a = pow2[e]
        ans += counter[a] // 2
    print(ans)


if __name__ == '__main__':
    main()