问题 写作题

写作.

下面四幅图片描述的是李明和爷爷从养鸟到放鸟的一段经历。请根据图片所提供的信息用英语写一篇短文并发表你的看法。

注意:①短文必须包括所有图片的主要内容,短文的内容要连贯、完整;

②短文单词数:100左右(开头已给出的单词不计入单词总数)。

One Sunday morning, Li Ming and his grandpa ___________________________

答案

___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

One possible Version

One Saturday morning, Li Ming and his grandpa were feeding the bird in the cage. The bird was unhappy though it had enough food and water, Li Ming did not know why.

The next morning on his way to school, Li Ming saw some birds flying in the sky. Then be thought of the lonely bird in the cage, He decided to persuade his grandpa to set the bird free. When he got home, he said to his grandpa, “Birds are friends!” To his delight, his grandpa agreed.

The bird was set free. Grandpa and the children were very happy to see the bird flying away into the sky。

阅读理解

阅读理解。

    It is the duty of every man to work. The life of a lazy (懒惰的) man is of no use to himself and to others. If

the man is too lazy to work, he will usually become a unuseful man. Everyone, when he is young, should learn

some useful work.

    But it is not enough for a man to learn some kind of work. He should put his heart and soul(精力) into his

work, and not waste (浪费) his spare time. "Work while you work and play while you play" is as good rule for

young people as old.

    There is no better to help to diligence(勤奋) than the habit(习惯) of early rising, and this, just like all other

good habits, is most easily formed in youth. There is an English saying "Lost time never returns." This means

that everyboby must be diligent, and make good use of time. One must study hard when one is young so that

one can make great progress, succeed in life and become useful to one's country. So we can say that diligence

is the mother of success.

1. After reading the passage we know that a diligent man will ______. [ ]

A. lose time

B. succeed in life

C. lose his life

D. make little progress

2. If a man wants to be of use to himself and to the people, he should ______. [ ]

A. learn some useful work

B. do something in his spare time

C. put his heart and soul into everything

D. try his best to work without rest.

3. In order to learn to be diligent, it' s very important for the young people to ______.[ ]

A. work while they work and play while they play

B. work all the time without playing

C. form the habit of getting up early

D. learn some good subjects

4. One will be successful(成功的) in life if he ______ when he is young. [ ]

A. spends some time learning

B. is diligent in his study

C. loves his school

D. gets up early

5. Which of the following may be the best title for this passage? [ ]

A. Lazy Boy Can Learn to Be Diligent

B. Young People Should Rise Early

C. Diligence, the Mother of Success

D. Lost Time Never Returns

填空题

阅读以下说明和C函数,填充函数中的空缺。
[说明]
函数Insert _key(*root,key)的功能是将键值key插入到*root指向根结点的二叉查找树中(二叉查找树为空时*root为空指针)。若给定的二叉查找树中已经包含键值为key的结点,则不进行插入操作并返回0;否则申请新结点、存入key的值并将新结点加入树中,返回1。
提示:
二叉查找树又称为二叉排序树,它或者是一棵空树,或者是具有如下性质的二叉树:
若它的左子树非空,则其左子树上所有结点的键值均小于根结点的键值;
若它的右子树非空,则其右子树上所有结点的键值均大于根结点的键值;
左、右子树本身就是二叉查找树。
设二又查找树采用二叉链表存储结构,链表结点类型定义如下:
typedef struct BiTrrode
int key _value; /*结点的键值,为非负整数*/
struct BiTnode *left,*right; /*结点的左、右子树指针*/
BiTnode, *BSTree;
[C函数]
int Insert _key(BsTree *root,int key)

BiTnode *father=NULL,*p=*root,*s;
while(______&&key!=p->key_value)(/*查找键值为]Key的结点*/
father=p;
if(key<p->key_value)p=______; /*进入左子树*/
else p=______; /*进入右子树*/

if (p) return 0; /*二叉查找树中已存在键值为key的结点,无须再插入*/
s=(BiTraode*)malloc(______);/*根据结点类型生成新结点*/
if (!s) return-1;
s->key_value=key; s->left=NULL; s->right=NULL;
if(!father)
______; /*新结点作为二叉查找树的根结点*/
else /*新结点插入二叉查找树的适当位置*/
if(key<father->key_value)father->left=s;
else father->right=s;
return 1;