--- Do you like your job?
--- ______.
A. I’m a nurse
B. Yes. I find it very interesting
C. They are very nice
第1题:
阅读以下说明,回答问题1~4,将解答填入对应的解答栏内。
[说明] 假设二叉树采用连接存储结构进行存储,root 指向根接点,p 所指结点为任一给定的结点,编写一个求从根结点到p所指结点之间路径的函数。
void path (root, p)
btree * root, * p;
{
Btree *stack[m0], *s;
int tag[m0], top =0, i, find =0;
s =root;
do
{
while (s ! = NULL)
{
stack [top] = s;
tag[top] =0;
((1))
}
if (top >0)
{
((2))
if (tag[top] = =1)
{
if((3))
{
for (i=1; i< =top; i+ + printf ("%d" ,stack[i]- >data);
find=1;
}
else top - -;
}
if((4))
{
p=p- >right;
((5))
}
}
} while (find || (s! = NULL && top ! =0));
}
第2题:
( ) – Is Miss White __________ English teacher, Maria?
– No, she teaches __________ geography.
A.your;my
B.you;mine
C.you;us
D.your;us
第3题:
A.howdoyoudo!
B.how
C.howdoyoudo
D.hdyd
第4题:
以下能正确计算10!的程序段是 。
A.do {i=1;s=1;; s=s*i; i++; }while(i<=10);
B.do{i=1;s=0 s=s*i; i++; while(i<=10);
C.i=1;s=1; do{s=s*i; i++; }while(i<=10);
D.i=1;s=0; do{s=s*i; i++; }while(i<=10);
第5题:
下面程序段的运行结果是()。 int m[]={5,8,7,6,9,2},i=1; do{ m[i]+=2; } while(m[++i]>5); for(i=0;i<6;i++) printf("%d ",m[i]);
A.7 10 9 8 11 4
B.7 10 9 8 11 2
C.5 10 9 8 11 2
D.5 10 9 8 11 4
第6题:
有如下程序段#include "stdio.h"#include "string.h"#define N 10#define M 10char *find(char(*a)[M],int n){ char *q;int i; q=a[0]; for(i=0;i<n;i++) if(strcmp(a[i],q)<0)q=a[i]; return q;}main(){ char s[N][M]={"tomeetme","you","and","he","china"}; char *p; int n=5; p=find(s,n); puts(p);}则执行后输出的结果为A.he B.and C.you D.tomeetme
第7题:
以下能正确计算1+2+3+…+10的程序段是 。
A.i=1; s=1; do {s=s+i; i++;} while (i<10);
B.do {i=1;s=0; s=s+i; i++;} while (i<=10);
C.do {i=1;s=1; s=s+i; i++;} while (i<=10);
D.i=1,s=0; do {s=s+i; i++;} while (i<=10);
E.i=1; s=1; do {s=s+i; i++;} while (i<=10);
F.i=1,s=0; do {s=s+i; i++;} while (i<10);
第8题:
What's ( )job? Are ( ) British?
A. your, your
B. you, your
C. your, you
第9题:
82、以下语句中循环次数不为10次的语句是()
A.for(i=1;i<10 i>
B.i=1;do{i++;}while(i< >
C.i=10;while(i>;0){--I;}
D.i=1;m:if(i< igotom>
第10题:
以下能正确计算1×2×3×…×10的程序段是 。
A.do {i=1;s=1; s=s*i; i++; } while(i<=10);
B.do {i=1;s=0; s=s*i; i++; } while(i<=10);
C.i=1;s=1; do {s=s*i; i++; } while(i<=10);
D.i=1;s=0; do {s=s*i; i++; } while(i<=10);