更多"While （ ）n efficiency test on an en"的相关试题:
[填空题]有如下程序段：
int n=0，sum=0；
while(n++，n＜50)
if(n%2= =0)continue；
sum+=n；
cout＜＜sum；
此程序段执行的输出结果为 [8] ，while语句共执行了 [9] 次。
[单项选择]While the test-oriented approach to teaching is______desirable, it is widely used in China.
A. other than
B. not only
C. nothing but
D. far from
[填空题]有如下程序段：
int n=0，sum=0；
while(n++，n＜50){
if(n%2= =0)continue；
sum+=n；
}
cout＜＜sum；
此程序段执行的输出结果为 [8] ，while语句共执行了 [9] 次。
[填空题]写出下列程序的输出结果______。
main（ ）
int n=0；
while(n++＜=1)；
printf("%d,",n)；
printf("%d",n++)；
[填空题]I dreamed of becoming an (engine) （） while my father wanted me to be a doctor.
[填空题]I dreamed of becoming an (engine) ______ while my father wanted me to be a lawyer.
[单项选择]Intelligence test scores follow an approximately normal distribution, (meaning) that most people score near the middle of the distribution of scores, (and) scores (drop off) fairly rapidly in frequency as one moves in (either) direction from tile center.
A. meaning
B. and
C. drop off
D. either
[填空题]写出下列程序的输出结果 ______。
main（ ）
int=0；
while(n++＜=1)；
printf("%d，"，n)；
printf("%d"，n++)；
[填空题]下列程序的输出结果是 【11】 。
void fun(int *n)
while((*n))
printf("%d",(*n)--);
main（ ）
int a=10;
fun(&a);
[简答题]{{B}} Is a Test of Spoken English Necessary’ {{/B}}
{{I}} You are to write in three parts.
In the first part state what you think is the best way.
In the second part, support your view with one or two reasons.
In the last part, bring what you have written to a natural conclusion or summary.
Marks will be awarded for content, organization, grammar and appropriacy. Failure to follow the instructions may result in a loss of marks.{{/I}}
[单项选择]It is necessary to make a(n) (abstract) while writing a report.
A. summary
B. analysis
C. discussion
D. index
[简答题]完成下面类中成员函数的定义。
class test
private：
int n1；
float f1；
public：
test(int，float f)；
test(test&)；
；
test：：test(______)
n1=n；f1=f；
test：：test(test&t)
n1=t.n1；n=______；
[单项选择]执行下列程序段之后，变量n的值为( )。
public class Test
public static void main(String[ ] args)
int y = 2;
int z = 3;
int n = 4;
n=n+-y* z/n;
System. out. println(n);
A. 3
B. -1
C. -12
D. -3
