题目内容

听力原文：GRANT: How do visitors enter the competition on Open Day？
CLARE: Well, firstly, they have to make their way to the 'portal photo booth' on campus.
GRANT: Okay—a bit like a treasure hunt to start with.
CLARE: Yes, and then they get their photo taken stepping through the portal.
GRANT: And they'll have to write down their details—you know, name, phone number, email...No, hang on... let's keep it simple—just name and email address should do.
CLARE: Then after, say, the thirtieth of July, people can visit the university Face book page and vote for their favorite photo.
GRANT: So the photo with the most votes wins.
CLARE: Yes, I think that should generate quite a bit of interest.
GRANT: What about a cut-off date？
CLARE: Of course...maybe, uh...the most popular photo as of 5 pm on the tenth of August will collect the pad.
GRANT: And the winner will be notified by email.
CLARE: And...the winning photo will be enlarged and published in full color on the university Face book page.
Label the flow chart below.
Write NO MORE THAN TWO WORDS AND/OR A NUMBER for each answer.
Step 1 Find 'Portal (28) on campus on Open Day
Step 2 Be photographed stepping through the Portal
Step 3 Give contact (29) (name & email address)
Step 4 Visit the University Face book page and vote
Step 5 The picture with the most votes at 5 pm on (30) wins
(28)

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[说明]
在进行文法分析的时候，通常需要检测一个单词是否在我们的单词列表里。为了提高查找和定位的速度，通常都要画出与单词列表所对应的单词查找树。
程序构造一棵二叉排序树，每个节点存储一个单词，按字典序列，较小的在左子树，较大的在右子树。
函数中使用的预定义符号如下：
typedef struct TreeNode{/二叉排序树节点/
char word;
struct TreeNode left, right;
}BNODE;
[函数]
int getWord(FILE fpt, char word)
/从文件fpt中读取单词到word中，到达文件结束时返回0/
{
char c;
c = fgetc(fpt);
if(c == EOF)return 0;
/跳过单词间的非字母字符/
while(!(tolower(c) ＞= 'a' && tolower(c) ＜= 'z')){
c = fgetc(fpt);
if(c == EOF)return 0;
}
/不区分大小写/
while(tolower(c) ＞= 'a' && tolower(c) ＜= 'z'){
word++ = c;
c = fqetc(fpt);
}
*word = '\0';
return 1;
}
void BTree(BNODE **t, char word)
{
BNODE ptr, p;
int compres;
p = NITLL;
(1) ;
while(ptr){
compres = strcmp(word, (2) );
if(!compres){
return;
}else{
(3) ;
ptr = compres ＞ 0 ？ ptr-＞right : ptr-＞left;
}
}
ptr = (BNODE)malloc(sizeof ptr);
ptr-＞left = ptr-＞right = NULL;
ptr-＞word = (char)malloc(strlen(word) + 1);
strcpy(ptr-＞word, word);
if(p == NULL){
(4) ;
}else if(compres ＞ 0){
p-＞right = ptr;
}else{
p-＞left = ptr;
}
}
int main()
{
FILE fpt;
char word[40];
BNODE *root = NULL;
if((fpt = fopen("text.in", "r")) == NULL){
printf("不能打开文件text.in! \n");
return 1;
}
while(getWord(fpt, word) == 1){
BTree (5) ;
}
fclose(fpt);
return 0;
}

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