Oracle笔试真题

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只有部分；大家参考；
& ^$ o/ s+ J2 T7 C1 p5 S0 bQuestion Number 20In the implementation of many linear linked structures it is often useful to keep a pointer to the last node in addition to the first, so that a new node can be added at the end without having to loop through the entire structure looking for the last node. In what ways would adding a pointer to the last node improve the efficiency of this dictionary implementation?It would speed up put operations when the dictionary contains an item with the designated key.It would speed up put operations when the dictionary does not contain an item with the designated key.
( t& ~2 k* }8 l5 pneither
: ~' s) d1 h4 U8 U- Z) ^; CA but not B
' q3 D( L! b& m2 b. K4 N4 \- Iboth
B but not A------------
ABC Housekeeping Forces are responsible for maintaining a building that comprises of 100 floors. The maintenance company decides to use a database to schedule work for its employees and also check the status of the work. When an assigned housekeeper does not report for work, an alternate resource is allotted to complete the job.
2 h# g: p* u4 H9 M% f" m& OThe Housekeeping database in its un-normalized form is given below:
3 W3 g7 K5 f2 `+ ^8 vHouseKeeperID   – ID of the housekeeper
; l# p2 L4 q" qHouseKeeperName – Housekeeper's Name
) P$ r. U8 X  F9 `3 p( aSupervisorID    – ID of the Housekeeper's
                  Supervisor
8 f# F) R9 T4 F9 ySupervisorName  - Name of the Housekeeper's
' Y0 Y3 ~" e/ U                  Supervisor
DutyID          – ID of an assigned duty
FloorNo         – Floor to which the duty is
                  assigned
DutyDate        – Date on which the duty has to be
* C* A# ]8 @2 l) A                  done
9 A2 r) \7 x2 P, I, {( b4 jWorkStatus      - Status of the duty
AlternateHID    - ID of the alternate housekeeper's
" ~5 y9 n, y& y                  assigned to complete the job, in
                  case the originally assigned
                  housekeeper is absent.
AlternateWorkSt – Work Status of the Alternate
; i$ T# W# G4 y; N& s% E; e                      Housekeeper's duty.
( w- ^4 t7 E) [( XFrom the following options, select the Housekeeping database in 2NF.
Housekeeper
HouseKeeperID
9 H8 _4 o. u; THouseKeeperName
Supervisor

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SupervisorID
- z  o/ g1 [, I7 p7 P5 ~# rSupervisorName
5 u: v* R8 ^! l  i* d/ _Duty
0 {' k0 k8 M% K' |5 ]' W1 n( QDutyID
" n+ |7 Y; t/ ]: eFloorNo
DutyDate
Transaction
DutyID
' G- K% i) J8 ]FloorNo
-----------
: `  @% Q: \% X' S' J9 Q  ~5 ~6 VHouseKeeperID
6 l7 j" L# {* s5 K; HWorkStatus
AlternateHID
" L) r3 o1 E4 e' U1 jAlternateWorkSt
--------------
/ N9 _5 N& ?* H$ d/ LHousekeeper
HouseKeeperID
$ {2 x$ f$ |9 b5 I/ VHouseKeeperName
SupervisorID
Supervisor
& q) B# X% z5 B5 n. t$ j: k/ b2 SSupervisorID
) B$ R+ K) |7 E% G; }" eSupervisorName
& r* b) G2 q7 k- W0 vDuty
$ ?- P/ Y3 G; z- |1 ?DutyID
. C( W/ }2 R+ f* ]FloorNo
DutyDate
-----------
" X# x- b! y$ `HouseKeeperID
WorkStatus
9 b  c: N1 t, J- CAlternateHID
$ m( g4 Y" T; g2 s! hAlternateWorkSt
4 n+ F, `8 W8 G& @0 a) @------------
( `. B  s1 R9 A/ R7 f) O# vHousekeeper

会计考友

HouseKeeperID
HouseKeeperName
SupervisorID
Supervisor
7 C' s9 \, _" \0 gSupervisorID
; X4 e, P( B3 H- N2 F% S' B8 LSupervisorName
Duty
2 U6 p& d" D7 t' _! a* o% kDutyID
8 @1 N  s! i% h. hFloorNo
DutyDate
. |1 U  W) W- z2 p9 Z1 `) M/ _3 E: B* }Transaction
DutyID
------------
HouseKeeperID
WorkStatus
AlternateHID
2 @( _% o' r, c) t/ _9 a& R9 n, c& ZAlternateWorkSt
------------
* S+ C7 N3 ^' w9 K0 u6 e: o, t$ ^  D% GHousekeeper
7 }( `  ]1 ], r# A8 y) |; iHouseKeeperID
# n' Y1 |5 Y; c' G. `& ^- t9 mHouseKeeperName
Supervisor
% J( a; t4 E4 K9 u7 W0 s- BSupervisorID
" }3 S% v+ {  V% H1 bSupervisorName
! z1 q: d8 f' \7 X% W1 `' y) qDuty
DutyID
: f4 p$ Q# O* A$ x0 n5 kFloorNo
, J* O# @, U# Q/ [5 P7 CDutyDate
! Z/ v8 y  \8 [) C3 K3 kTransaction
DutyID
6 K2 b# G9 c8 @HouseKeeperID
' \; U! m% o& O' Z9 e! i: k- dWorkStatus
AlternateHID
! B& v& i! q# f* W- |" DAlternateWorkSt</p>