试题与答案

(1)分析判断A和B 各是什么物质,写出各自的化学式。 (2)写出①~⑦各步反

题型:填空题

题目:

(1)分析判断A和B 各是什么物质,写出各自的化学式。  

(2)写出①~⑦各步反应的化学方程式。

  

答案:

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参考答案:令牌环(IEEE 802.5)解析:令牌环网采用的是令牌环(IEEE 802.5)协议。

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第一部分:短文改错(满分10分)

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增加:在缺词处加一个漏字符号(∧), 并在其下面写上该加的词。

删除:把多余的词用斜线(\)划掉。

修改:在错的词下划一横线,并在该词下面写上修改后的词。

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题型:问答题

已知在文件IN34.DAT中存有100个产品销售记录,每个产品销售记录由产品代码dm(字符型4位)、产品名称mc(字符型10位)、单价dj(整型)、数量s1(整型)、金额je(长整型)几部分组成。其中,金额=单价×数量可计算得出。函数ReadDat()的功能是读取这100个销售记录并存入数组sell中。请编制函数SortDat(),其功能要求:按产品代码从大到小进行排列,若产品代码相同,则按金额从大到小进行排列,最终排列结果仍存入结构数组sell中,最后调用函数WriteDat( )把结果输出到文件OUT34.DAT中。
注意:部分源程序已给出。
请勿改动主函数main()、读函数ReadDat()和写函数WriteDat()的内容。
试题程序:
#include<stdio.h>
#include<mem.h>
#include<string.h>
#include<conio.h>
#include<stdlib.h>
#define MAX 100
typedef struct

char dm[5];
char mc[11];
int dj;
int s1;
long je;
PRO;
PRO sell[MAX];
void ReadDat();
void WriteDat();
void SortDat()

void main ()

memset(sell,0,sizeof sell));
ReadDat();
SortDat();
WriteDat();

void ReadDat ()

FILE *fp;
char str[80],ch[11];
int i;
fp=fopen ( "IN34. DAT", "r" );
for (i=0; i<100; i++)

fgets(str,80,fp);
memcpy (sell [i] .dm, str, 4);
memcpy(sell[i] .mc,str+4,10 ;
memcpy (ch, str+14,4);
ch[4]=0;
sell [i] .dj =atoi (ch);
memcpy (ch, str+18,5);
ch[5]=0;
sell [i] . s1=atoi (ch);
sell [i] .je=(long)sell[i].dj*sell[i] .s1;

fclose (fp);

void WriteDat ()

FILE *fp;
int i ;
fp=fopen ( "OUT34. DAT", "w" );
for (i=0; i<100; i++)

fprintf(fp,"%s %s %4d %5d %101d\n", sell Ii] .dm, sell [i] .mc, sell [i] .dj,
sell[i] .s1,sell [i] .je);

fclose (fp);

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