妇女。24岁，3个月前曾行引产，术后不久即出现不规则 * * 流血，昨日咳嗽，痰中带血来院

阅读理解。

    When I was seven years old my mom was diagnosed (诊断) with cancer. The doctors told us my mom

might be saved with surgery (手术). But it could also kill her. She chose to have it.

    The day before the surgery I was off school, and my mom planned the best day of my life, everything I

loved at seven and everything that would put a smile on my face. The day began with her waking me up

saying, "Kate, I have a surprise for you. Come and see." The surprise was a doll I had wanted for the longest

time. Throughout the day, she told me everything that she thought I would need to know to grow up and be

a good person; she told me to be the best I could be and that I would always make her proud. We had a picnic

in the park, and it was so cold that we moved to the car. The day was filled with laughter. For the first time

in a long time I could see she was really happy. I would never forget her smile, or the way her eyes shone as

if we were the same age. It was the best day of my life, and I will never forget the conversations we shared. 

    My mother managed to live through the surgery. Now when I look back, I realize that the best day of my

life could be her last and this could be the last day I would remember with her, the last one we shared. I also

understand how unselfish a mother is.

1. Who made the decision to have the surgery in the story? [ ]

A. The doctors.

B. The patient.

C. The writer.

D. The patient's family.

2. What did the writer's mother do the day before the surgery? [ ]

A. She stayed at home to have a good rest.

B. She went to the doctor's for advice.

C. She turned to her family for comfort.

D. She stayed all day with her child.

3. Why was the writer's mother happy that day? [ ]

A. Because she left her child a special memory.

B. Because she realized her child had grown up.

C. Because she was proud of what her child had done.

D. Because she expected the surgery to be successful.

4. The underlined sentence in the last paragraph suggests that ______. [ ]

A. her mother was ill again

B. the surgery was unsuccessful

C. her mother was getting better

D. the surgery was simple

5. Which of the following words best describes the writer's mother?[ ]

A. Hardworking.

B. Famous.

C. Rich.

D. Great.

连词成句。

例: a, is, This, shirt

     This is a shirt.                

1. there, dogs, How, many, are

___________________________________

2. These, are, carrots

___________________________________

3. Can, you, read, a, book

___________________________________

4. I, am, a, boy, not

___________________________________

在企业文化的内容中，一个企业特有的从事生产经营和管理活动的方法论原则属于（）

A.企业精神

B.团体意识

C.价值观念

D.经营哲学

症见头痛经久不愈，痛处固定不痛如锥刺，或有头部外伤史，舌紫，脉细。其治疗最佳方剂是

A．桃红四物汤
B．通窍活血汤
C．血府逐瘀汤
D．复元活血汤
E．失笑散

[说明] 以下[C程序]所完成的功能是在3X3方格中填入数字1～N(N≥10)内的某9个互不相同的整数，使所有相邻两个方格内的两个整数之和为质数。系统输出满足该要求的所有填法。系统的部分输出结果如图3-18所示。

图3-18 系统的部分输出结果 3×3方格从第1行左上角方格开始的序号分别为0、1、2，第2行左边方格开始的序号分别为3、4、 5，第3行左下角方格开始的序号分别为6、7、8。以下[C程序]采用试探法，即从序号为0的方格(左上角)开始，为当前方格寻找一个合理的可填整数，并在当前位置正确填入后，为下一方格寻找可填入的合理整数。如不能为当前方格寻找一个合理的可填整数，就要后退到前一方格，调整前一方格的整数。直至序号为8的方格(右下角)也填入合理的整数时，就找到了一个解，将该解输出，并调整序号为8的方格所填的整数，继续去找下一个解。 为了检查当前方格的填入整数的合理性，C程序引入二维数组checkMatrix，用于存放需要进行合理性检查的相邻方格的序号。 [C程序] #include ＜stdio.h＞ #define N 12 int a [9]; /* 用于存储方格所填入的整数 */ int b[N+1]; int pos; checkMatrix[][3] = {{-1},{0,-1},{1,-1},{0,-1},{1,3,-1},{2,4,-1},{3,-1} {4,6,-1}, 5,7,-1}}; void write(int a[]) { int i, j; for ( i = 0; i ＜ 3; i++) for ( j = 0; j ＜ 3; j++) printf("%3d",a[3*i+j]); printf("\n"); } } int isPrime(int m) { int i;if (m == 2) return 1; if (m == 1 || m % 2 == 0)return 0;for (i = 3; i * i ＜= m; ){ if (m % i == O)return 0; i+ =2;} return 1; } int selectNum(int start) { int j;for (j = start; j ＜= N; j++) if (b[j])return j; return 0; } int check ( ) /* 检查填入pos位置的整数是否合理 */ { int i, j;for (i = 0; (j = (1) ) ＞= 0; i++) if (!isPrime(a[pos] + a[j]))(2) ; (3) ; } extend ()/* 为下一方格找一个尚未使用过的整数 * / { a[ (4) ] = selectNum(1);b[a[pos]] = 0; } void change() /* 为当前方格找下一个尚未使用过的整数(找不到回溯) */ { int j;while (pos ＞= 0 && (j = selectNum( (5) )) == 0 (6) ;if (pos ＜ 0) return;b[a[pos]] = 1;a[pos] = j;b[j] = 0; } find ( ) { int ok = 1;pos = 0; a[pos] = 1; b[a[pos]] = 0;de { if (ok)if ( (7) ) { write (a); change( );}else extend( ); elsechange( ); ok = check(pos);} while (pos ＞=0); } main( ) { int i;for (i = 1; i ＜=N; i++) b[i] = 1;find( ); }