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【简答题】
根据短文内容和首字母提示，在空格处填入适当的词使短文完整。（每空一词） Many years ago two men were traveling together. They were w 小题1: along a road in the wood. Suddenly they found a beautiful ax on the g 小题2: . One of them picked it up and said, “Look here , I found an ax.” “Don't say I, but we have found the ax,” said the o 小题3: “We are good friends. We should s 小题4: it between us.” “No,” said the first one, “I found the ax, s 小题5: it is mine.” Soon after, they h 小题6: someone running after them. They looked b 小题7: and saw a strong man. The man was shouting out “Stop, thieves! Stop, thieves !” The f 小题8: traveller said, “What shall we do? He is running after us. We shall be caught by him.” “Don't say we, but I shall be caught. You found the ax, and you say it is yours,” said the other, and left him a 小题9: The first traveler tried to h 小题10: the ax, but he did not know where to hide it. And at last he was caught by the owner of the ax.
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【单选题】若有以下定义语句 struct link { int data; struct link *next; } *p, *q, *r; 指针 p 、 q 和 r 分别指向一个链表中的三个连续的结点。如图所示。 现要将 q 和 r 所指结点的前后位置交换,同时要保持链表的连续,以下错误的程序段是
A.
r->next= q; q->next = r->next; p->next = r;
B.
q->next= r->next; p->next = r; r->next = q;
C.
p->next= r; q->next = r->next; r->next = q;
D.
q->next= r->next; r->next = q; p->next = r;
【单选题】程序段如下所示,下列说法正确的是: #define A 3 #define B(A) ((A+1)*a) int a=4; ...... x=4*(A+B(7));
A.
程序错误，不允许嵌套定义
B.
x=140
C.
程序错误，宏定义不能有参数
D.
x=124
【单选题】有如下所示的C#程序段: int total = 0; for ( int i = 0; i < 4; i++ ){ if ( i==1) continue; if ( i==2) break; total +=i; } 则执行完该程序段后,total的值为()。
A.
0
B.
1
C.
3
D.
6
【单选题】有如下所示的 Java 程序段 : int total = 0; for ( int i = 0; i < 4; i++ ){ if ( i==1) continue; if ( i==2) break; total +=i; } 则执行完该程序段后, total 的值为( )。
A.
0
B.
1
C.
3
D.
6
【单选题】有如下所示的 Java 程序段: int total = 0; for ( int i = 0; i < 4; i++ ){ if ( i==1) { continue;} if ( i==2){ break; } total +=i; } total的值最后是多少
A.
0
B.
1
C.
3
D.
6
【单选题】若有以下定义: struct link { int data; struct link *next; } a,b,c, *head, *p; 其中 , 指针 head 指向变量 a, p 指向变量 c , a.next 指向变量 b ,链表结构如下图所示: a b c ┌──┬──┐ ┌──┬──┐ ┌─...
A.
a.next = c; c.next = b;
B.
head.next = p; p.next = head.next;
C.
head->next = &c; p->next = head->next;
D.
(*head).next = p; (*p).next = &b;
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