Ａ－Ｄ：①　2.3 ② 5 ③ 7.5 ④ 15 ⑤ 50

⑥ 75 ⑦ 90 ⑧ 105 ⑨ 119 ⑩ 169

Ｅ：①�。粒拢� ②　ＡＣＢ ③�。拢粒� ④�。拢茫痢、荩茫粒�

試題１１

　　從供選擇的答案中選出應(yīng)填入下列英語(yǔ)文句中＿＿內(nèi)的正確答案，把編號(hào)寫(xiě)在崐答卷的對(duì)應(yīng)欄內(nèi)。

Software products may be _A_ into four basic types: application progr崐ams, programming language processors, operating systems, and system util崐ities.

Application programs atr prograns that _B_ useful tasks such as solvi崐ng statistical problems, or keeping your company's books.

Programming language processors are programs that _C_ the use if a co崐mputer language in a computer system. They are tools for the developmen崐t of application programs.

Operation systems are programs that _D_ the system resources and enab崐le you to run application programs.

System utilities are special programs that _E_ the usefulness of or 崐add capabilities to a computer.

供選擇的答案

Ａ－Ｅ：①manage ②perform ③ support ④ reduce

　�、� divided ⑥ enhance ⑦ implemented ⑧ introduce

　 　�、� ranked ⑩ run

試題１２

　　從供選擇的答案中選出應(yīng)填入下列英語(yǔ)文句中＿＿內(nèi)的正確答案，把編號(hào)寫(xiě)在崐答卷的對(duì)應(yīng)欄內(nèi)。

Here is a useful procedure for choosing a program:

1. Study the features of all the programs you might choose _A_. Deci崐de which features you need, which you would _B_, and which you can do yo崐ur jobs wichout.

2. Eliminate the prograns that clearly do not _C_ you needs.

3. Consider how the remaining programs perform the functions you will崐 use most often. This can affect a program's usability more than all th崐e "nice" features that you will _D_ need.

4. Study the remaining programs carefully - with _E_ experince if you崐 can get it - and decide which one is best for you.

供選擇的答案

Ａ：① for ② on ③ in ④ from

Ｂ、Ｃ：① meet ② require ③ help ④ give

⑤ choose ⑥ like

Ｄ、Ｅ：① often ② seldom ③ always ④ rich

⑤ hands-on ⑥ little

試題１５

從供選擇的答案中選出應(yīng)填入下列敘述中─────　內(nèi)的正確答案,把編號(hào)寫(xiě)崐在答案的對(duì)應(yīng)欄內(nèi)。

編譯系統(tǒng)一般可分成──Ａ──,──Ｂ──，──Ｃ──，──Ｄ── 和"崐目標(biāo)代碼生成"等五大部分,其中──Ａ──，──Ｂ──　和"目標(biāo)代碼生成"三部崐分是每個(gè)編譯程序必不可少的,而──Ｃ──，──Ｄ──　則是可有可無(wú)的。許崐多編譯程序?qū)ⅸぉぃ俩ぉぁ【幹瞥梢粋�(gè)子程序,在──Ｂ── 的分析過(guò)程中根據(jù)需崐要調(diào)用──Ａ──,并且把──Ｂ── 和──Ｃ──　二部分結(jié)合起來(lái),邊分析邊崐產(chǎn)生有關(guān)內(nèi)容。此外,這五個(gè)部分在工作過(guò)程中都會(huì)涉及到表格處理和──Ｅ──。

供選擇的答案

Ａ─Ｅ：①　詞法分析　�、凇∽兞糠治觥　、邸�數(shù)據(jù)分析　�、堋≌Z(yǔ)法分析

　　　�、荨∵^(guò)程分析　�、蕖〕鲥e(cuò)處理　�、摺�優(yōu)化處理　�、唷”磉_(dá)式處理

　　　�、帷≈虚g代碼生成

試題１６

　　從供選擇的答案中，選出應(yīng)填入下列有關(guān)方程組的敘述中─────　內(nèi)的正崐確答案，把編號(hào)寫(xiě)在答案的對(duì)應(yīng)欄內(nèi)。

　�。�. 方程組 ａｘ＋ｂｙ＝１　有唯一的充要條件是──Ａ──。

　　　　　　　 ｂｘ＋ａｙ＝２

　　２　如果直線(xiàn)�。幔�＋ｂｙ＋ｃ＝０（ａ，ｂ，ｃ都不為０）不通過(guò)第一象限，崐那么ａ，ｂ，ｃ應(yīng)該滿(mǎn)足條件──Ｂ──。

下午試題

[程序說(shuō)明]

已知某數(shù)的前二項(xiàng)為２和３，其后繼項(xiàng)根據(jù)當(dāng)前最后二項(xiàng)的乘積按下列規(guī)則生成

　（１）若乘積為一位數(shù)，則該乘積即為數(shù)列的后繼項(xiàng)；

�。ǎ玻┤舫朔e為二位數(shù)，則該乘積的十位數(shù)和個(gè)位數(shù)依次作為數(shù)列的兩個(gè)后繼項(xiàng)。

　本程序輸出該數(shù)列的前ｎ項(xiàng)以及它們的和。其中，函數(shù)ｓｕｍ（ｎ，ｐａ）返回

數(shù)列的前ｎ項(xiàng)之和，并將生成的前ｎ項(xiàng)存放于首指針為ｐａ的數(shù)組中。程序中規(guī)定

輸入的ｎ值必須大于２并且不超過(guò)給定的常數(shù)值ＭＡＸＮＵＭ。

　　列如：若輸入ｎ值為１０，則程序輸出如下內(nèi)容：

　　sum(10)=44

2 3 6 1 8 8 6 4 2 4

[程序]

#include <stdio.h>

#define MAXNUM 100

int sum(n,pa)

int n,*pa;

{ int count,total,temp;

*pa=2;

____________________ =3;

total=5; count=2;

while (count++<n)

{ temp=*(pa-1)**pa;

if (temp<10)

{ total+=temp;

*(++pa)=temp;

}

else

{

_______________ =temp/10;

total+=*pa;

if (count<n)

{ count++;pa++;

_______________________=temp%10;

total+=*pa;

}

}

}

___________________________;

}

main ()

{ int n,*p,*q,num[MAXNUM];

do {printf("Input N=? (2<N<%d):",MAXNUM+1);

scanf("%d",&n);

} while ( ___________________ );

printf("\nsum(%d)=%d\n",n,sum(n,num));

for (p=num,q= ________________ ;p<q;p++) printf("%4d",*p);

printf("\n");

}

　　[程序說(shuō)明]

　　本程序?qū)δ畴姶a文(原文)進(jìn)行加密形成密碼文.其加密算法如下:

假定原文為Ｃ1Ｃ2Ｃ3...Ｃn,加密后形成的密文為Ｓ1Ｓ2Ｓ3...Ｓn,首先讀入

正整數(shù)Key(Key>1)作為加密鑰匙,并將密文字符位置按順時(shí)針?lè)较蜻B成一個(gè)環(huán),

如下圖所示:

┏━Sn━━S1━┓

Sn-1 S2

┃ ┃

┃ S3

┃ ┃

┗....━━S4━┛

加密時(shí)從S1位置起順時(shí)針計(jì)數(shù),當(dāng)數(shù)到第Key個(gè)字符位置時(shí),將原文中的字符放入

該密文字符位置中,同時(shí)從環(huán)中除去該字符位置;接著從環(huán)中下一個(gè)字符位置起

繼續(xù)計(jì)數(shù),當(dāng)再次數(shù)到第Key個(gè)字符位置時(shí),將原文中字符C2放入其中,并從環(huán)中

除去該字符位置;依次類(lèi)推,直至n個(gè)原文字符全部放入密文環(huán)中.由此產(chǎn)生的S1S2

...Sn即為原文的密文.

例如,當(dāng)Key=3時(shí),原文: THIS IS AN DECODING SYSTEM 的密文為:

AOTGNHEDI YS D IMIETSNC SS

當(dāng)Key=4時(shí),該原文的密文為:

SSDTYD HTEGIASISCNM E ION

本程序?qū)㈦姶a的原文存放在字符數(shù)組old中,加密鑰匙存放在整數(shù)Key中.函數(shù)decode

用于將原文old加密并返回密文字符數(shù)組的首指針.其中函數(shù)采用一個(gè)雙向循環(huán)鏈表

CODE來(lái)表示密文環(huán).函數(shù)strlen用于計(jì)算一個(gè)字符串中的字符個(gè)數(shù)(不包括字符串結(jié)

尾符'\0').為了簡(jiǎn)單起見(jiàn),程序中假設(shè)內(nèi)存容量足以滿(mǎn)足動(dòng)態(tài)存貯單元分配的要求.

[程序]

#include <stdio.h>

#include <alloc.h>

#define CR 13

typedef struct node

{ char ch;

struct node *forward; /* Link to next node. */

struct node *backward;/* Link to previous node.*/

} CODE;

main()

{ char _________ ,old[256];

int strlen(),key,num=0;

printf("\nPlease input the telegraph: \n")l

while (num<255&&(old[num++]=getch())!=CR);

old[ ______ ]='\0';

do { printf("\nPlease input Key=?(Key>1):");

scanf("%d",&key);

} while (key<=1);

printf("\nThe decode of telegraph:'%s' is:\n'%s'\n",

old,decode(old,key));

}

char *decode(old,key)

char *old; int key;

{ char *new; int length,count,i;

CODE *loop,*p;

length=strlen(old);

loop=( _______ ) malloc(length*sizeof(CODE));

for (i=1;i<length-1;i++)

{ loop[i].forward=&loop[i+1];

loop[i].backward=&loop[i-1];

}

loop[0].backward=&loop[length-1];

loop[0].forward=&loop[1];

loop[length-1].forward=loop;

loop[length-1].backward=&loop[length-2];

for (p=loop,i=0;i<length;i++)

{ for (count=1;count<key;count++)

p= _____ ;

p->ch=*old++;

p->backward->forward= _________ ;

p->forward->backward= _________ ;

p=p->forward;

}

new=(char *)malloc((length+1)*sizeof(char));

for (i=0;i<length;i++) new[i]=loop[i].ch;

new[length]='\0';

return (new);

}

int strlen(s)

char *s;

{ int len=0;

while ( ________ !='\0') len++;

return( len);

}

[程序說(shuō)明]

本程序?qū)⒁粋�(gè)給定漢字的點(diǎn)陣逆時(shí)針旋轉(zhuǎn)90度,并輸出旋轉(zhuǎn)前后的點(diǎn)陣數(shù)據(jù)及字形.

附圖是漢字"轉(zhuǎn)"字的16*16點(diǎn)陣字形,用數(shù)字'0'表示空白位置,用數(shù)字'1'表示非空

白位置,"轉(zhuǎn)"字的第一行機(jī)即可表示成如下的{0,1}序列:

0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0

如果把它看作一個(gè)字的16個(gè)BIT,"轉(zhuǎn)"字的第一行可以用16進(jìn)制數(shù)的1040來(lái)表示.同

理,"轉(zhuǎn)"字的第二行可以表示為1040,第三行可以表示為1048,...等等.依次類(lèi)推,用

16個(gè)雙字節(jié)整型數(shù)即可存放一個(gè)漢字點(diǎn)陣字形."轉(zhuǎn)"字的點(diǎn)陣數(shù)據(jù)及字形如附圖的

左半部分所示.將一個(gè)漢字逆時(shí)針旋轉(zhuǎn)90度,就是把該漢字點(diǎn)陣的最右列作為旋轉(zhuǎn)后

點(diǎn)陣的第一行,次右列作為旋轉(zhuǎn)后點(diǎn)陣的第二行,...,來(lái)形成一個(gè)旋轉(zhuǎn)后的點(diǎn)陣字形.

附圖的右半部分就是將"轉(zhuǎn)"字旋轉(zhuǎn)90度后的點(diǎn)陣數(shù)據(jù)和字形.

程序中,數(shù)組old中存放著"轉(zhuǎn)"字的點(diǎn)陣數(shù)據(jù).函數(shù)turnleft將該點(diǎn)陣數(shù)據(jù)逆時(shí)針旋

轉(zhuǎn)90度,旋轉(zhuǎn)后的點(diǎn)陣數(shù)據(jù)存放在數(shù)組new中.函數(shù)display將旋轉(zhuǎn)前后的點(diǎn)陣數(shù)據(jù)加

以編輯,用字符'.'表示值為0的Bit,用字符'X'表示值為1的Bit,從而將旋轉(zhuǎn)前后的點(diǎn)

陣按行輸出其16進(jìn)制數(shù)據(jù)及字形,如同附圖所表示的那樣.

[附圖]

1040 ...X.....X...... 0000 ................

1040 ...X.....X...... 0200 ......X.........

1048 ...X.....X..X... 1640 ...X.XX..X......

FDFC XXXXXX.XXXXXXX.. 32E0 ..XX..X.XXX.....

2040 ..X......X...... 1259 ...X..X..X.XX..X

2844 ..X.X....X...X.. 1246 ...X..X..X...XX.

4BFE .X..X.XXXXXXXXX. FE44 XXXXXXX..X...X..

7C80 .XXXXX..X....... 13C8 ...X..XXXX..X...

0888 ....X...X...X... 1240 ...X..X..X......

09FC ....X..XXXXXXX.. 0200 ......X.........

1C08 ...XXX......X... 1120 ...X...X..X.....

E810 XX.X.......X.... 17FF ...X.XXXXXXXXXXX

4890 .X..X...X..X.... F120 XXXX...X..X.....

0860 ....X....XX..... 1D10 ...XXX.X...X....

0820 ....X.....X..... 1318 ...X..XX...XX...

0810 ....X......X.... 1010 ...X.......X....

[程序]

#define <stdio.h>

#define EMPTY '.'

#define NONEMPTY 'X'

#define LEFT 0

#define RIGHT 1

main ()

{ static unsigned old[16]=

{0x1040,0x1040,0x1048,0xFDFC,0x2040,0x2844,

0x4BFE,0x7C80,0x0888,0x09FC,0x1C08,0xE810,

0x4890,0x0860,0x0820,0x0810

};

unsigned new[16];

trunleft(old,new);

display(old,new);

}

turnleft(old,new)

unsigned old[],new[];

{ int row,k;

for (row=0;row<16;row++)

for (k=0,______ ;k<16;k++)

new[row]|=((old[k]>> _______ ) &1) << _______ ;

}

display(old,new)

unsigned *old,*new;

{

char out[2][17],letter[2];

int row,col;

letter[0]=EMPTY;

letter[1]=NONEMPTY;

out[LEFT][16]=out[RIGHT][16]=______;

for (row=0;row<16;row++,old++,new++)

{ for (col=0;col<16;++col)

{ out[LEFT][col]=letter[ ( __________) &1];

out[RIGHT][col]=letter[ (__________) &1];

}

printf("\n %4x %s ",*old,&out[LEFT][0]);

printf(" %4x %s",*new,&out[RIGHT][0]);

}

}

new[16]=______;

for (row=0;row<16;row++,old++,new++)

{

for (col=0;col<16;++col)

{ out[LEFT][col]=letter[ ( __________) &1];

out[RIGHT][col]=letter[ (__________) &1];

}

printf("\n %4x %s ",*old,&out[LEFT][0]);

printf(" %4x %s",*new,&out[RIGHT][0]);

}

}