https://paiza.jp/poh/ec-campaign でテストケース3つともクリアしたコード。http://paiza.jp/poh/ec-campaign/result/609c70c227bb4762a2d3728d461382eb

ec-campaign.py

# -*- coding: utf-8 -*-

# 標準入力から各種値の取得
n, d = raw_input().split(' ')
n, d = int(n), int(d)

items = {}
for i in range(0, n):
    item = int(raw_input())
    if item in items:
        items[item] += 1
    else:
        items[item] = 1

campaigns = [int(raw_input()) for i in range(0, d)]


def searchMaxPair(items, campaign):
    """最大値を探す関数"""
    max_price = campaign - 10
    campaign_items = [p for p in items if p <= max_price]
    max_pair = 0
    for first in campaign_items:
        rest_price = campaign - first
        #重複は予め取り除いているため同じ価格は同じアイテムになってしまう
        rest_items = [p for p in campaign_items if p <= rest_price and p != first]
        if not rest_items:
            continue
        second = max(rest_items)
        pair = first + second
        #キャンペーン価格以上はありえないので即返す
        if campaign == pair:
            return pair
        max_pair = max(max_pair, pair)
    return max_pair


#重複するアイテムは2つまででOK
one_items = [p for p, c in items.items() if c == 1]
two_items = [p for p, c in items.items() if c > 1]
one_items = one_items + two_items
one_items.sort(reverse=True)

#ここから探索処理
for campaign in campaigns:
    #2個あるやつの2倍になってれば即その値を返す。
    if campaign / 2 in two_items:
        print(campaign)
    #最大値を探す
    else:
        print(searchMaxPair(one_items, campaign))