中央電大C語(yǔ)言程序設(shè)計(jì)模擬試題及解答(親測(cè))小抄參考

1、2012春電大c語(yǔ)言程序設(shè)計(jì)模擬試題及解答一、選擇題 1. 由c語(yǔ)言源程序文件編譯而成的目標(biāo)文件的缺省擴(kuò)展名為( c )。 a. cpp b. exe c. obj d. c 2. c語(yǔ)言程序的基本模塊為（ d ）。 a. 表達(dá)式 b. 標(biāo)識(shí)符 c. 語(yǔ)句 d. 函數(shù) 3. 設(shè)x和y均為邏輯值,則x|y為假的條件是( c )。 a它們均為真 b.其中一個(gè)為真 c.它們均為假 d.其中一個(gè)為假 4. x0 | y=5的相反表達(dá)式為（ b ）。a. x=0 | y!=5 b. x0 | y!=5 d. x0 & y=5 5. 枚舉類型中的每個(gè)枚舉常量的值都是一個(gè)（ a ）。 a. 整數(shù) b. 浮點(diǎn)

2、數(shù) c. 字符 d. 邏輯值6. 循環(huán)體至少被執(zhí)行一次的語(yǔ)句為( c )循環(huán)語(yǔ)句。 afor b.while c.do-while d.任一種 7. 循環(huán)語(yǔ)句“for(int i=0; in; i+) s;”中循環(huán)體s被執(zhí)行的次數(shù)為( c )。 a. 1 b. n-1 c. n d. n+1 8. 在下面的while循環(huán)語(yǔ)句中，其循環(huán)體被執(zhí)行的次數(shù)為（ c ）。 int i=0,s=0; while(s3 & x10的相反表達(dá)式為 x=10 。 17若x=5，y=10，則x!=y的邏輯值為 1 。 18. 假定x=5，則執(zhí)行“int a=(! x? 10:20)；”語(yǔ)句后a的值為 20 。19

3、. 在if語(yǔ)句中，每個(gè)else關(guān)鍵字與它前面同層次并且最接近的 if 關(guān)鍵字相配套。 20. 若while循環(huán)的“頭”為“while（i+name等價(jià)的表達(dá)式是 (*p).name 。三、寫出下列每個(gè)程序運(yùn)行后的輸出結(jié)果 1. #include void main() int i, s=0; for(i=1;i+) if(s50) break; if(i%2=0) s+=i; printf(i,s=%d, %dn,i,s); 結(jié)果：i，s=15，56 2. #include void main() 結(jié)果* char ch=*; int i, n=5; while(1) for(i=0;in;i

4、+) printf(%c,ch); printf(n); if(-n=0) break; 3. #includeconst int n=5;void main()結(jié)果 1 1 1 2 2 2 3 6 9 4 24 33 int i,p=1,s=0; for(i=1;in;i+) p=p*i; s=s+p; printf(%5d%5d,i,p); printf(%5dn,s); 4. #include void main() int x=24,y=88; int i=2,p=1; do while(x%i=0 & y%i=0) p*=i; x/=i; y/=i; i+; while(x=i &

5、y=i); printf(%dn,p*x*y); 結(jié)果：p*x*y=8*3*11=264 5. #include void main() int a9=36,25,48,24,55,40,18,66,20; int i, b1, b2; a0=a1? (b1=a0, b2=a1):(b1=a1,b2=a0); for(i=2; i9; i+) if(aib1) b2=b1; b1=ai; else if(aib2) b2=ai; printf(%d %dn,b1,b2); 結(jié)果：18 20 6. #include void main() int a12=76,63,54,62,40,75,80

6、,92,77,84,44,73; int b4=60,70,90,101; int c4=0; int i,j; for(i=0;i=bj) j+; cj+; for(i=0;i4;i+) printf(%d,ci); printf(n); 結(jié)果：3 2 6 17. #include void main() int a=10, b=20;結(jié)果 10 20 40 35 40 20 printf(%d %dn,a,b); int b=a+25; a*=4; printf(%d %dn,a,b); printf(%d %dn,a,b); 8. #include void selectsort(int

7、 a, int n) int i,j,k,x; for(i=1;in;i+) /進(jìn)行n-1次選擇和交換 k=i-1; for(j=i;jn;j+) if(ajak) k=j; x=ai-1; ai-1=ak; ak=x; void main() int i; int a6=20,15,32,47,36,28; selectsort(a,6); for(i=0; i6; i+) printf(%d ,ai); printf(n); 結(jié)果：15 20 28 32 36 47 9#include void main() int a8=4,8,12,16,20,24,28,32; int *p=a;

8、do printf(%d,*p); p+=2; while(pa+8); printf(n); 結(jié)果：4 12 20 28 10. #include void le(int* a, int* b) int x=*a; *a=*b; *b=x; 結(jié)果15 26 26 15 void main() int x=15, y=26; printf(%d %dn,x,y); le(&x,&y); printf(%d %dn,x,y); 11. #include #include struct worker char name15; /姓名 int age; /年齡 float pay; /工資 ; in

9、t equal(struct worker* r1, struct worker* r2) if(strcmp(r1-name, r2-name)=0) return 1; else return 0; void main() struct worker a4=abc,25,420,defa,58,638, ghin,49,560,jklt,36,375; struct worker x=defa; int i; for(i=0; i=4) printf(沒(méi)有查找所需要的記錄!n); else printf(%s %d %6.2fn,,ai.age,ai.pay); 結(jié)果：def

10、a 58 638.00四、寫出下列每個(gè)函數(shù)的功能 1. int se(int n) /n為大于等于1的整數(shù) int x,s=0; printf(輸入%d個(gè)整數(shù): ,n); scanf(%d,&x); if(n=1) return x; s=x; while(-n) scanf(%d,&x); s+=x; return s; 函數(shù)功能：求出鍵盤上輸入的n個(gè)整數(shù)之和并返回。2. float fh() float x,y=0,n=0;scanf(%f,%x);while(x! =-1.0) n+; y+=x;scanf(%f,%x);if(n-0) return y; else return y/n

11、; 函數(shù)功能：求出從鍵盤上輸出的一批常數(shù)的平均值，以-1.0作為結(jié)束輸入的標(biāo)志。 3. #include void wa(int a, int n) int i,k,j,x; for(i=0;in-1;i+) k=i; for(j=i+1;jak) k=j; x=ai; ai=ak; ak=x; 函數(shù)功能：采用選擇排序的方法對(duì)數(shù)組a中的n個(gè)整數(shù)按照從大到小的次序重新排列。 4. #include int* lj(int a, int n) int i,k=0; for(i=1;iak) k=i; return &ak; 函數(shù)功能：求出數(shù)組a中n個(gè)元素的最大值元素，返回該元素的地址。 5. vo

12、id qc(struct strnode* f) while(f) printf(%sn,f-name);f=f-next; 假定結(jié)構(gòu)類型strnode的定義如下: stuct strnode char name15; /字符串域 struct strnode *next; /指針域 ; 函數(shù)功能：遍歷f單鏈表，即從頭結(jié)點(diǎn)開(kāi)始依次輸出f單鏈表中每個(gè)結(jié)點(diǎn)的值。 6. int output(struct intnode *f) int sum=0; if(!f) return -9999; /用返回特定值-9999表示空表 while(f) sum+=f-data; f=f-next; retur

13、n sum; 假定struct intnode的類型定義為： struct intnode int data; /結(jié)點(diǎn)值域 struct intnode* next; /結(jié)點(diǎn)指針域 ; 函數(shù)功能：對(duì)于以表頭指針為f的鏈表，求出并返回所有結(jié)點(diǎn)中data域的值之和。 7. int wr4(file* fptr) char ch; int c=0; fseek(fptr,0,seek_set); while(1) ch=fgetc(fptr); if(ch!=eof) c+; else break; return c; 函數(shù)功能：求出一個(gè)以fptr為文件流的所對(duì)應(yīng)文件的長(zhǎng)度，即所存字符的個(gè)數(shù)。五、根

14、據(jù)下列每個(gè)題目要求編寫程序1. 編寫一個(gè)函數(shù)fun()，首先從鍵盤上輸入一個(gè)4行4列的一個(gè)實(shí)數(shù)矩陣到一個(gè)二維數(shù)組a44中，接著求出主對(duì)角線上元素之和，最后返回求和結(jié)果。 #include double fun() double a44; double s=0; int i,j; printf(輸入一個(gè)4*4的數(shù)值矩陣:n); for(i=0;i4;i+) for(j=0;j4;j+) scanf(%lf,&aij); for(i=0;i4;i+) s+=aii;i return s; void main() printf(主對(duì)角線上元素之和是%lfn,fun(); 2. 編寫一個(gè)主函數(shù)，計(jì)算1

15、+3+32+.+310的值并輸出，假定分別用i,p,s作為循環(huán)變量、累乘變量和累加變量的標(biāo)識(shí)符。#include void main() int i; /用i作為循環(huán)變量 int p=1; /用p作為累乘變量 int s=1; /用s作為累加循環(huán)變量 for(i=1;i=10;i+) p*=3; s+=p; printf(%dn,s); 3. 編寫一個(gè)主函數(shù)，已知6a40,15b30，求出滿足不定方程2a+5b=120的全部整數(shù)組解。如(13,20)就是其中的一組解。#include void main() int a,b; for(a=6;a=40; a+) for(b=15;b=30;b+

16、) if(2*a+5*b=126) printf(%d, %d)n,a,b); if we dont do that it will go on and go on. we have to stop it; we need the courage to do it.his comments came hours after fifa vice-president jeffrey webb - also in london for the fas celebrations - said he wanted to meet ivory coast international toure to di

17、scuss his complaint.cska general director roman babaev says the matter has been exaggerated by the ivorian and the british media.blatter, 77, said: it has been decided by the fifa congress that it is a nonsense for racism to be dealt with with fines. you can always find money from somebody to pay th

18、em.it is a nonsense to have matches played without spectators because it is against the spirit of football and against the visiting team. it is all nonsense.we can do something better to fight racism and discrimination.this is one of the villains we have today in our game. but it is only with harsh

19、sanctions that racism and discrimination can be washed out of football.the (lack of) air up there watch mcayman islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 150th anniversary celebrations and will attend citys premier league match at chelsea

20、 on sunday.i am going to be at the match tomorrow and i have asked to meet yaya toure, he told bbc sport.for me its about how he felt and i would like to speak to him first to find out what his experience was.uefa hasopened disciplinary proceedings against cskafor the racist behaviour of their fans

21、duringcitys 2-1 win.michel platini, president of european footballs governing body, has also ordered an immediate investigation into the referees actions.cska said they were surprised and disappointed by toures complaint. in a statement the russian side added: we found no racist insults from fans of

22、 cska.baumgartner the disappointing news: mission aborted.the supersonic descent could happen as early as sunda.the weather plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud co

23、ver. the balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (

24、5.6 miles/9.17 kilometers) and into the stratosphere. as he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.the balloon will slowly drift to the edge of space at 120,000 feet ( then, i would assume, he will slowly step out onto something resembling an olympic divi

25、ng platform.below, the earth becomes the concrete bottom of a swimming pool that he wants to land on, but not too hard. still, hell be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. it will be like he is diving into the shallow end.skydiver preps for the big jumpwhen he jumps, he is expected to reach the speed of sound - 690 mph (1,110 kph) - in less than 40 seconds.