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【单选题】
在网络中,为保证192.168.10.0/24 网络中WWW服务器的安全,只允许访问Web服务,现在采用访问控制列表来实现,正确的是( )。
A.
access-list 100 permit tcp any 192.168.10.0 0.0.0.255 eq www
B.
access-list 10 deny tcp any 192.168.10.9 eq www
C.
access-list 100 permit 192.168.10.0 0.0.0.255 eq www
D.
access-list 110 permit ip any 192.168.10.0 0.0.0.255
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参考答案:
举一反三
【单选题】主要以劳动质量来区分劳动差别,进而依此规定工资差别的是()。
A.
职务等级工资制
B.
技术等级工资制
C.
结构工资制
D.
岗位技能工资制
【单选题】The operation of removing an element from the stack is said to( )the stack.
A.
pop
B.
push
C.
store
D.
fetch
【单选题】There are two common types in page replacement algorithm: stack and non-stack strategies. When a real page number increase only stack algorithm can increase the hit rate monotonously. In the following...
A.
FIFO
B.
LRU
C.
PFF
D.
OPT
【单选题】有下列程序: #include<iosteram.h> using namespace std class Stack { public: Stack(unsignedn=10):size(n){rep_=new int [size]top=O} Stack(Stack&s):size(s.size) { rep_=new int[size] fo
A.
4,3,2,1,
B.
4,3,6,7,2,1,
C.
4,3,6,2,1,
D.
1,2,3,4,
【单选题】What is the content of stack pointer (SP)?
A.
address of the current instruction
B.
address of the next instruction
C.
address of the top element of the stack
D.
size of the stack
【单选题】There are two common types in page replacement algorithm : stack and non-stack strategies.When a real page number increase only stack algorithm can increase the hit rate monotonously.In the following ...
A.
FIFO
B.
LRU
C.
OPT
D.
non of above
【单选题】The following is not the basic operation of the stack? ( )
A.
Insert a new element at the top of stack
B.
Adding an element at a random location k.
C.
Determines if stack is empty
D.
Reads the value of stack top
【单选题】有如下程序: #include <iostream.h> using namespace std; class Stack { public: Stack(unsigned n=10):size(n){rep_=new int [size]; top=0;} Stack(Stack&s): size (s.size) { rep_=new int[size]; for (int i=0;i<siz...
A.
4,3,2,1,
B.
4,3,6,7,2,1,
C.
4,3,6,2,1,
D.
1,2,3,4,
【单选题】以下哪种凝血因子不属于蛋白质
A.
因子 I
B.
因子 II
C.
因子 III
D.
因子 IV
E.
因子 X
【单选题】有如下程序: #include <iostream> using namespace std; class Stack { public: Stack(unsigned n= 10):size(n) {rep_=ew int[size]; top=0;} Stack(Stack& s):size(s.size) { rep_=new int[size]; for(int i=0;i<size;i+...
A.
4,3,2,1,
B.
4,3,6,7,2,1,
C.
4,3,6,2,1,
D.
1,2,3,4,
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