更多"Stages of Advertising StrategyA cur"的相关试题:
[单项选择]
Stages of Advertising Strategy
A accepted B acquired C acquainted D acknowledged
[单项选择]
Stages of Advertising Strategy
A ample B much C lots D high
[单项选择]
Stages of Advertising Strategy
A rigid B dull C grim D tough
[单项选择]
Stages of Advertising Strategy
A way B approach C path D key
[单项选择]
Stages of Advertising Strategy
A focus B stress C direct D point
[单项选择]
Stages of Advertising Strategy
A mentioned B described C referred D pointed
[单项选择]
Stages of Advertising Strategy
A aim B target C objective D goal
[单项选择]
Stages of Advertising Strategy
A make B turn C show D face
[单项选择]
Stages of Advertising Strategy
A comprised B composed C consisted D made
[单项选择]
Stages of Advertising Strategy
A ground B fundamental C root D foundation
[单项选择]
Stages of Advertising Strategy
A puts B keeps C turns D takes
[单项选择]
Stages of Advertising Strategy
A protected B differentiated C discriminated D separated
[单项选择]
Stages of Advertising Strategy
A behalf B basis C base D cost
[单项选择]
Stages of Advertising Strategy
A sum B total C amount D volume
[填空题]#include＜stdio.h＞
int add(int a，int b)
{int c；
c=a+b：
return c；
}
void print(int t)
{printf("z=%d/n"，t)；}
main（ ）
{int x，y，z；
x=1；y=2；
z=add(x，y)；
print(z)；
}
程序运行结果是：______
[单项选择]有如下程序
#include
int func(int a,int b)
{ return(a+b);}
void main（ ）
{ int x=2,y=5,z=8,r;
r=func(func(x,y),z);
cout＜＜r;
}
该程序的输出的结果是
A. 12
B. 13
C. 14
D. 15
[单项选择]int fm(int a，int b )
if(b==1)return a；
else return a+fm(a，b-1)；
main（ ）
printf("%d/n"，fm(4，3))；
[单项选择]有以下程序
#include
void fun(int a,int b,int c)
{ a=456,b=567,c=678;}
void main（ ）
{
int x=10,y=20,z=30;
fun(x,y,z);
cout＜＜x＜＜’,’＜＜y＜＜’,’＜＜z＜＜endl;
}
输出结果是
A. 30,20,10
B. 10,20,30
C. 456,567,678
D. 678,567,456
[简答题]
【函数1.2说明】
函数merge(int a[]，int n，int b[]，int m，int
*c)是将两个从小到大有序数组a和b复制合并出一个有序整数序列c，其中形参n和m分别是数组a和b的元素个数。
【函数1.2】
void merge(int a[ ]，int n，int b[ ]，int
m，int *c)
{ int i，j；
for(i=j：0；i＜n
&& j＜m；)
*c++ =a[i] ＜b[j]
a[i++]：b[j++]；
while(i＜n) (2) ；
while(j＜m) (3) ；
}
