D.Z20-30/100型单体液压支柱的100代表（）。

阅读短文，判断正(T)误(F)。

      An old tiger lives in the forest. He doesn't want to look for food now. He often tells other animals to

bring (带来) him something to eat.

      It's the monkey's turn. He brings him nothing. He says to the tiger, " I can't do that now. There is another

tiger over there. He is stronger than you. He tells me not to bring you anything. I am afraid of him."

      " What? " cries the old tiger. " Show me that tiger. I will talk to him." 

      " Come with me." says the monkey. He takes the tiger to a big river.

      " Now look at the river. Do you see the head, the white teeth and the large green eyes of a tiger? Isn't he

stronger than you? "

      " Yes, he is." cries the old tiger, " I will eat him up! "

      With these words, the tiger jumps into the river.

( ) 1. An old tiger often tells other animals to bring food to him.

( ) 2. The monkey doesn't bring anything to the tiger.

( ) 3. The monkey takes the old tiger to a lake.

( ) 4. The tiger in the river is himself.

( ) 5. The tiger is very clever.

阅读下列诗句，然后回答问题。（8分）

北塘避暑

韩琦（宋）①

尽室林塘涤暑烦，旷然如不在尘寰。

谁人敢议清风价？无乐能过百日闲。

水鸟得鱼长自足，岭云含雨只空还。

酒阑何物醒魂梦？万柄莲香一枕山。

①韩琦是北宋名相，他为官清廉正直，生活俭朴，不尚奢华，尤重保持晚节

小题1:“暑”与“清风”具有双关含义，试加以分析。（4分）

小题2:分析颈联的表现手法。（4分）

2009年18日，我国宣布正式启动和部署国家基本药物制度工作。中药饮片纳入了基本药物目录，保留了中药的传统特色。中药饮片的使用，医生在治疗的过程中形成相对固定的院内制剂。院内制剂经过开发，有的可能会成为一种自主创新的新药。

中药饮片的炮制，国家药品标准没有规定的，必须按照（）

A.县级以上药品监督管理部门制定的炮制规范炮制

B.地方药品标准规定炮制

C.省级药品监督管理部门制定的炮制规范炮制

D.国家中医药管理局制定的炮制规范炮制

目前移动通信系统中用到的多址方式有（）。

A.TDMA

B.FDMA

C.CDMA

D.OFDMA

下列给定程序是建立一个带头结点的单向链表，并用随机函数为各结点赋值。函数fun()的功能是：将单向链表结点(不包括头结点)数据域为偶数的值累加起来，并作为函数值返回。

其累加和通过函数值返回main()函数。例如，若n=5，则应输出8.391667。

请改正程序中的错误，使它能得到正确结果。

[注意] 不要改动main函数，不得增行或删行，也不得更改程序的结构。

[试题源程序]

#include＜stdio.h＞

#include＜stdiib.h＞

typedef struct aa

int data;

struct aa *next;

NODE;

int fun(NODE *h)

int sum=0;

NODE *P;

/**********found**********/

p=h;

while(P-＞next)

if(p-＞data%2==0)

sum+=p-＞data;

/**********found**********/

p=h-＞next;

return sum;

NODE *creatlink(int n)

NODE *h, *p, *s, *q；

int i, x;

h=p=(NODE *)malloc(si zeof(NODE));

for(i=1; i＜=n; i++)

s=(NODE *)malloc(sizeof(NODE));

s-＞data=rand()%16;

s-＞next=p-＞next;

p-＞next=s;

p=p->next;

p-＞next=NULL;

return h;

outlink(NODE *h, FILE *Pf)

NODE *p;

p=h-＞next;

fprintf(Pf, "\n\nTHE LIST:\n\n HEAD");

while(P)

fprintf(Pf, "-＞%d", p-＞data); p=p-＞next;

fprintf(pf, "\n");

outresult(int s, FILE *pf)

fprintf(Pf, "\nThe sum of even numbers : %d\n", s);

main()

NODE *head; int even;

head=creatlink(12);

head-＞data=9000;

outlink(head, stdout);

even=fun(head);

printf("\nThe result :\n"); outresult(even, stdout);