更多"The 4,200-acre ranch is kept as one"的相关试题:
[填空题]The 4,200-acre ranch is kept as one large and intact archeological site due to Wilcox’s______.
[单项选择]One of the boys kept laughing, ____ annoyed Jane greatly.
A. whom
B. that
C. what
D. which
[单项选择]下列函数的运行结果是( )。
#include
intadd(inta,intb);
voidmain()
{
externintX,y;
cout< }
intx(20),y(5);
intadd(inta,intb)
{
ints=a+b:
returns;
}
A. 25
B. 30
C. 20
D. 15
[填空题]#include<stdio.h>
main( )
{int i,j;
inta[5]={2,4,6,9,12},b[6]={2,3,4,6,8,9};
for(i=0;i<5;i++)
for(j=0;j<6;j++)
if(*(a+i)==*(b+j)printf("%d",*(a+i));
printf("/n");
}
程序运行结果是:______
[单项选择]下面程序的运行结果是( )。
#include<iostream.h>
intfun(inta[],int n)
int result=1;
for(int i=“i<n;i++)
result’resultxa[i];
return result;
void main( )
inta[3]=3,4,5;
cout<<fun(a,3)<<endl;
A. 12
B. 15
C. 20
D. 60
[填空题]#include<stdio.h>
main( )
int i,j;
inta[5]=2,4,6,9,12,b[6]=2,3,4,6,8,9;
for(i=0;i<5;i++)
for(j=0;j<6;j++)
if(*(a+i)==*(b+j)printf("%d",*(a+i));
printf("/n");
程序运行结果是:______
[填空题]下列程序的输出结果为 【8】 。
include<iostream.h>
voidmain( )inta []=10,20,30,40,*pa=a;
int*&ph=pa;
Pb++;
cout<<*pa<<end1;
[填空题]It is normally dry in Acre between June and December.
[填空题]How long has she kept it
She has kept it______.
[单项选择]下面程序的运行结果是
#nclude<iostream.h>
int fun(inta[],int n)
int result=1;
for(int i=1;i<n;++)
result=result*a[i];
return result;
void main( )
int a[3] =3,4,5;
cout<<fun(a,3)<<endl;
A. 12
B. 15
C. 20
D. 60
[单项选择]下面程序的运行结果是
#include<iostream.h>
int fun(inta[],int n)
int result=1;
for(int i=1:i<n;i++)
result=result*a[i];
remm result;
void main( )
int a[3]=3,4,5;
cout <<fun(a,3) <<end1;
A. 12
B. 15
C. 20
D. 60
[填空题]#include <stdio.h>
main( )
(inta=100,b=200,c=300,d,*p1=&a,*p2=&b,*p3=&c;
d=*p1+*p2:
printf("d=%d/n",d);
p1=&d:
d=a+c;
printf("d=%d/n",*p1);
p1=p2=&c:
a=*p1+*p2:
printf("a=%d/n",a);
c=a+b:
printf("c=%d/n",*p3);
程序运行结果为:
[填空题]有以下程序
#include<stdio.h>
main( )
inta[3][3]=1,2,3),4,5,6,7,8,9;
int b[3]=0,i;
for(i=0;i<3;i++b[i]=a[i][2]+a[2][i];
for(i=0;1<3;i++)printf("%d",b[i]);
printf("/n");
程序运行后的输出结果是______。
[简答题]
程序阅读题:
#include <stdio.h>
main( )
(inta=100,b=200,c=300,d,*p1=&a,*p2=&b,*p3=&c;
d=*p1+*p2:
printf("d=%d/n",d);
p1=&d:
d=a+c;
printf("d=%d/n",*p1);
p1=p2=&c:
a=*p1+*p2:
printf("a=%d/n",a);
c=a+b:
printf("c=%d/n",*p3);
}
程序运行结果为:
[单项选择]
下列程序的输出结果是()。
#include<stdio.h>
main()
inta=4;
printf("%d/n",(a+=a-=a*a))
A. -8
B. 14
C. 0
D. -24