单选题75 A progress B goal C result D purposeA AB BC CD D

●Developing the project schedule is often an iterative process. It determines the planned start and finish dates for project activities and milestones. Schedule development can require the review and revision of duration estimates and resource estimates to create an approved project schedule that can serve as a baseline to （75） progress.

（75）

A. analyze

B. track

C. level

D. extend

MIMD systems can be classified into(71)-oriented systems, high-availability systems and response-oriented systems. The goal of(71). oriented multiprocessing is to obtain high(71)(72)minimal computing cost. The techniques employed by multiprocessor operating systems to achieve this goal take advantage of an inherent processing versus input/output balance in the workload to produce(73)and(74)loading of system(75).

A．though

B．through

C．throughout

D．throughput

请补充main函数，该函数的功能是：从键盘输入只有两个整数参与运算的一个四则运算式，然后计算出它的值。注意数字全部为正整数。

例如，输入308*28，结果为8624.000000。

注意：部分源程序给出如下。

请勿改动主函数main和其他函数中的任何内容，仅在 main函数的横线上填入所编写的若干表达式或语句。

试题程序：

include＜stdio, h＞

main ( )

{

int a=0, b=0, flag=l, fg=l;

float result;

char *p, ch;

char str[10];

clrscr ();

printf("\n Input a string \n");

gets (str);

p=str;

while (*p)

{

if (*p＞='0' &&*p＜='9' && flag==l )

a=【 】;

else

if (fg)

{

ch=*p;

flag=2;

fg=0;

}

if (*p＞='0 ' &&*p＜='9' &&flag==2)

b=【 】;

p++;

}

switch(【 】)

{

case '+': result=a+b;break;

case '-': result=a-b;break;

case '*' :result=a*b;break;

case '/' :result= (float) a/b;break;

printf ("\nresult=%f", result);

}

Click the Exhibit button.In the exhibit, you decided to change myHosts addresses.What will happen to the new sessions matching the policy and in-progress sessions that had already matched the policy?（）

A. New sessions will be evaluated. In-progress sessions will be re-evaluated.

B. New sessions will be evaluated. All in-progress sessions will continue.

C. New sessions will be evaluated. All in-progress sessions will be dropped.

D. New sessions will halt until all in-progress sessions are re-evaluated. In-progress sessions will be re-evaluated.

写出下面程序的输出( )。 public class Test { public static void main (String args[ ]) { iht x=1, y=2; System. out. println ("result="+x+y); System. out. println ("result="+(x+y));

A．result=12 result=12

B．result=3 result=3

C．result=3 result=12

D．result=12 result=3

(d) There is considerable evidence to suggest that as a result of implementation problems less than 50% of all

acquisitions achieve their objectives and actually end up reducing shareholder value.

Required:

Provide Ken with a brief report on the most likely sources of integration problems and describe the key

performance indicators he should use to measure progress towards acquisition objectives. (15 marks)

假定a=3，下列程序的运行结果是( )。 #inelude＜iostream.h＞ int fun(int n); void main() { cout＜＜"Please input a number:"; int a,result=0; cin＞＞a; result=fun(a); tout＜＜result＜＜end1; } intfun(int n) { int p=1,result=0; for(int i=1;i＜=n;i++) { p*=i; result+=p; } return result; )

A．4

B．8

C．9

D．20

When the result of an operation becomes larger than the limits of the representation,(75)occurs.

A．overdose

B．orerfnow

C．overdraft

D．overeposure

bool lcd_bz()

{   //  测lcd是否忙碌

   bool  result;

   lcd_rs=0;

   lcd_rw=1;

   lcd_en=1;

   _nop_();

   _nop_();

   _nop_();

   _nop_();

   result=(bool)(P0&0x80);

   lcd_en=0;

   return  result;

}

有以下程序： #include ＜iostream＞ #include ＜cmath＞ using namespace std； class Distance； class point { public： friend class Distance； Point(int a，int B) { x=a； Y=b； } void Print() { cout＜＜"X= "＜＜X＜＜end1； cout＜＜"Y= "＜＜Y＜＜end1； } private： float X，Y； }； class Distance { public： float Dis(Point &p，Point &q)； }； float Distance :: Dis(Point &p，Point &q) { float result； result=sqrt((p.X-q.X)*(p.X-q.X)+(p.Y-q.Y)*(p.Y-q.Y))； cout＜＜result＜＜end1； retUrn result； } int main() { Point p(10，10)，q(10，30)； Distance d； d．Dis(p，q)； return 0； } 运行后的输出结果是( )。

A．10

B．30

C．0

D．20