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【单选题】

Hours and Arrival The Museum of Art (MOA) is open daily, from 10:30 a.m. to 5:30 p.m. Mondays are reserved exclusively (专门地) for members and their guests. Entry is by advance timed tickets only and capacity is extremely limited. Tickets must be reserved online and will not be available at the Museum. Free access to MOA through September 27 is available with timed tickets released one week in advance in one-week blocks, every Friday at 10:00 a.m. Reservations are limited to up to two , senior, and student tickets and two children per order. General admission tickets for September 28- -October 31 are also available. Members recive priority access, subject to capacity restrictions, and do not need to book tickets in advance. An allocation(分配)of timed tickets is reserved for members each hour; guest privileges apply. Use the main entrance of the Museum of Art. All the other entrances are for staff only. Please arrive during the 30-minute window of your timed ticket. If your plans change and you are unable to visit during your selected day/time, contact tickets@moa.org for a refund or to change your selection. Your ticket is good for one-time admission only- you may not leave and reenter the Museum. Our checkroom is closed. All bags will be inspected upon arrival. Please wear backpacks (背包) on the front of your chest or carry it to the side. There are several private parking garages located near MOA.1. On Mondays, the Museum ofArt is .

A.
open only to members and their guests
B.
closed to its partime staf members
C.
open toall the publie free of charge
D.
closed to the public for maintenance
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题目标签:分配背包
参考答案:
参考解析:
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举一反三

【单选题】01背包的一维状态转移方程是(i表示物品编号,j表示背包容量)

A.
dp[i] = max(dp[j], dp[j-w[i]]+v[i])
B.
dp[j] = max(dp[i], dp[j-w[i]]+v[i])
C.
dp[j] = max(dp[j], dp[j-w[i]]+v[i])
D.
dp[j] = max(dp[j], dp[j-w[j]]+v[j])

【单选题】溶剂分配法的原理是()

A.
根据物质在两相溶剂中分配系数不同
B.
根据物质的熔点不同
C.
根据物质的沸点不同
D.
根据物质的类型不同

【单选题】在户外最常用的背包上肩的方法是( )。

A.
高处上肩法
B.
提拉上肩法
C.
弓步上肩法
D.
下蹲上肩法

【单选题】完全背包的状态转移方程是()

A.
dp[i] = max(dp[j], dp[j-w[i]]+v[i])
B.
dp[j] = max(dp[i], dp[j-w[i]]+v[i])
C.
dp[j] = max(dp[j], dp[j-w[i]]+v[i])
D.
dp[j] = max(dp[j], dp[j-w[j]]+v[j])