C語言程序設(shè)計(jì)教材答案

1、、選擇題第1章C語言概述123456DCACAC填空題1. 一個(gè)主函數(shù)多個(gè)2. mai n()3. c obj exe三、程序填空題mai n()四、運(yùn)行結(jié)果AAAAAAAAASSSSSSSSSSSSSS五、編程題void main()printf(n nprintf( H*請(qǐng)完成以下功能)；n ；printf( H* 1、進(jìn)入程序2、切換用戶n"；prin tf(H* 3、更新程序4、刪除程序n ；printf( H* 請(qǐng)選擇(1-4):n )printf( nn n)；第2章數(shù)據(jù)類型、運(yùn)算符、表達(dá)式與基本輸入輸出、選擇題123456789101112131415BDDBACCAC

2、CBCDBAAACA221二、填空題1. 3,12. 不定值3. 14. 05. 1, 0, 0, 06. 07. 168. 8, 8, 99. 1運(yùn)行結(jié)果題a=32767,b=32768<1 >7.000000<2>6<3>3.000000<4>2<5>6.000000<!>0.0000003. <1>0.000000<2>0<3>0<4>2<5>10<6>5第3章順序程序設(shè)計(jì)、選擇題12345678910BABAADABDCD二、填空題1. dec

3、:-l,oct:37777777777,hex:ffffffff,u nsig ned:42949672952. dec: 1,oct: 1,hex: 1,u nsig ned: 13. dec:65,oct:41,AXCI 上 A4. 3.141593,3.1425. 3.500000,3.56. 31.415920,3.141592e+0017. c=K8. a=l,b= ,c=29. a=12,b=34510. c=A三、編程題1.#in clude<stdio. h>void mai n()float r,h,v;scanf(H%f%f n,&r, &h);

4、v=3.14*T*T*h;prin tf("v=%fn",v);2.#i nclude<stdio.h>void mai n()int n, s=0;sca nf("%d",&n);s=( n+1 )*(n+2)* n/3;n" ,s);3.#i nclude<stdio.h>void mai n()int a,b ,n=0;scamf("%d%d", &a, & b);n=a/2+b/4;prin tf(" n=%dn", n);4.#i nclude&l

5、t;stdio.h>void mai n()printf( nOnH);O 0nH);OnH);printf( nOOOOOOOOOOOOOOn n); printf( nOprintf( nO O0nn);printf( nOOOOOOOOOOOOOOn H); printf( n 第4章分支結(jié)構(gòu)程序設(shè)計(jì)習(xí)題解、選擇題12345BDABD二、填空題1. 缺省 時(shí)，else總是與上面最近的、沒有匹配過的if匹配2. if(a<=b) x=l; printf( <4#x=%dn ”,x);else y=2; printf ( “ * * * y=%dn '?,y); 3

6、. 根據(jù)條件判定值，從兩種選擇對(duì)象中，選取一個(gè)作為整個(gè)運(yùn)算的結(jié)果4. (1)整個(gè)條件表達(dá)式的結(jié)果就取自運(yùn)算分量b的結(jié)果(2)整個(gè)條件表達(dá)式的結(jié)果就取自運(yùn)算分量c的結(jié)果三、程序填空題char grade;gradedefault四、運(yùn)行結(jié)果題1.1,0,12.x=63.32334.*&五、編程題1.#in elude <math.h>#i nclude <stdio.h>void mai n()float a,b,c,s,sl;n ；sca nf("%f%f%f", &a,&b, &c); if(a+b)>c&am

7、p;&(a+c)>b&&( b+c)>a) s=(a+b+c)*0.5; sl=s*(s ? a)*(s ? b)*(s ? c); s=sqrt(sl);printf("nArea of the triangle is %fn",s);else printf("It is not triangle!n n);2.#include <stdio.h>#define PI 3.14159void main()int k;float r,c,a; printfC'input r,kn n);scanf(H%f%d

8、 H,&r,&k);switch(k)case 1: a=PI*T*T ； printf("area=%fn",a);break;case 2: c=2*PI*r;printf("circle=%fn",c);break;case 3: a=PI*r*r;c=2*PI*r;printf( narea=%f circle=%fn",a,c);break;3.#include <stdio.h>void main()int year; scanf("%d",&year);if(year%400

9、=0llyear%4=0&&year% 100! =0) printf("%d is a leap yearn",year); elseprintf("%d is not a leap yearn ",year);4.#include<stdio.h>void main()float score; scanf("%f",&score);if(score>=90)printf( nA n);else f(score>=80)printfCB ”；)else f(score>=70)p

10、rintf( nCH);else f(score>=60)printf( nDn);elseprintf( nEn);5.#i nclude<stdio.h>void mai n()int n um,a,b,c,d,p;sca nf("%d",&n um);if(n um<=9999&&n um>999)P=4;else f(n um>99)P=3 ；else if(n um>9)P=2 ；else if(n um>0)P=l ；printf (” 位數(shù)是：%dnn,p);a=n um/1000;b=n

11、 um/100-a*10;c=n um/10-a* 100-b* 10;d=n um%10;switch(p)case 4:printf("%d %d %d %dn",d,c,b,a);break;case 3:printf("%d %d %d n n,d,c,b);break;case 2:printf("%d %dn",d,c);break;case l:pri ntf("%d n",d);第5章循環(huán)結(jié)構(gòu)程序設(shè)計(jì)習(xí)題解、選擇題12345ACCBB填空題1. (l)while (2)do-while (3)for2. do

12、-while 語句3. for語句4. (1)先判斷條件是否成立(為真)，若成立(為真)再執(zhí)行循環(huán)語句(2)先執(zhí)行循環(huán)語句，再判斷條件是否成立(為真)，若成立則繼續(xù)執(zhí)行循環(huán)，若不成立，則結(jié)束循環(huán)。do-while至少執(zhí)行一次循環(huán)語句5. 表達(dá)式1 ；while (表達(dá)式2)表達(dá)式 3 ；6. (1)結(jié)束本次循環(huán)，后進(jìn)行循環(huán)的條件判定，即不終止整個(gè)循環(huán)的執(zhí)行過程( 2)終止整個(gè)循環(huán)的執(zhí)行，不再進(jìn)行條件判定7. 一個(gè)循環(huán)體內(nèi)包括另一個(gè)完整的循環(huán)結(jié)構(gòu)8. 死循環(huán)，不能確定次數(shù)三、 程序填空題1. 18 或 192. c=getchar()四、運(yùn)行結(jié)果題1. 0 111222. x=8五、編程題1.

13、#include<stdio.h> void main()int i, s=0;for(i=l;i<=101;i+=2)if(i-l)%4=0)s+=i;elses-=i;printf( “d",s);2.#include<stdio.h>#include<math.h>void main()long int i,x,y,z;for(i=l;i<100000;i+)x=sqrt(i+100);/*x 為加上 100后開方后的結(jié)果 */ y=sqrt(i+268);/*y 為再加上 268 后開方后的結(jié) 果 */ if(x*x=i+100

14、&&y*y=i+268)printf( nn%ldn H,i);3.#include<stdio.h>void main()int n,i ；printf("nplease input a number:n");scanf("%d",&n);printf("%d=",n);for(i=2;i<=n;i+)while(n!=i)if(n%i=0)printf( n%d* n,i); n=n/i;elsebreak;printf( n%dn,n);4.#include<stdio.h>

15、void main()char c;charactersnn);int Ietters=0,space=0,digit=0,others=0; printf( npleaseinput somewhile(c=getchar() !='n*)if(c>='a '&&c<= ,z,llc>= ,A '&&cv=Z) letters+;else if(c=*') space+;else if(c>= !0'&&cv=9) digit+;else others+;printf(&q

16、uot;all in all:char=%d space=%d digit=%d others=%dn'*,letters, space,digit,others); 5.#include<stdio.h>#include<math.h>void main()int i,n,j,k,m,l=0;for(i=4;i<=2000;i+=2)for(n=2; n<i ;n+)for(j=2,k=l;j<=(int)sqrt(n);j+)/* 判斷 n 是否素?cái)?shù) */if(n%j=O) k=0;if(k)/*若n是素?cái)?shù)，則繼續(xù)判斷 i? n是否素?cái)?shù)*/

17、for(j=2,m= 1 ;j<=(int)sqrt(i-n)；j+)/* 判斷 i-n 是否素?cái)?shù) */if(i-n)%j=O) m=0;if(m)printf( n%d=%d+%d n,i,n,i-n);1+； if (l%5=0) printf( nnn);break;if(n=i) printf (”n”，i);/*n等于i, i不遵循哥德巴赫猜想*/6.#include<stdio.h>void main()int day,xl,x2;day=9;x2=l; while(day>0) xl=(x2+l)*2;/* 第一天的桃子數(shù)是第 2 天桃子數(shù)加 1 后的 2

18、 倍*/ x2=xl;day;printf( Hthe total is %dn H,xl);7.#include<stdio.h>void main()int i,j,k;for(i=0;i<=3 ;i+)for(j=0;j<=2-i;j+) printfC'");for(k=0;kv=2*i;k+) printf(； p”rin*t”f( )nnn);for(i=0;i<=2; i+)for(j=0;j<=i;j+) printfC'") ； for(k=0;kv=4- 2*i;k+) printf(” *” );pr

19、intf( nnn);&#i nclude<stdio.h> void mai n() double p=l,t=l,v;int i;for(i=l;i<64;i+)p=p*2;/* p是當(dāng)前格子中的麥子粒數(shù)*/t=t+p;/* t是當(dāng)前麥子總粒數(shù)*/v=t/l .42e &printf("total=%en",t); printf("volume=%en",v);第6章數(shù)組與指針、選擇題12345678910CDB 1CCADAAD :111213141516CCCCAAC二、讀程序題1. ABC2.4793.194.

20、65.s=s+1.0/n;或 s=s+l/(float)n;6.-17.48.LANG三、程序填空1.i=strle n(s);i>=j;i一一 si+l=si;!='0'i+2.(ch=sti)ch="word=0n+四、編程題1.#i nclude <stdio.h>#define NUM 10void mai n()int i,cou nt=O;float aNUM ,sum=0,aver;for(i=0;i<NUM;i+)scanf("%f n,&ai); sum+=ai; aver=sum/NUM; printf(&

21、quot;The average is : %.2fn",aver); for(i=0;i<NUM;i+) if(ai>aver)count+;printf("the number that is over average is : %dn*',count); 2.#include <stdio.h>#define NUM 10void main()int i;float aNUM,max,min;scanf( ”f ” ,&a0);max=aO;min=aO;for(i=l;ivNUM;i+)scanf( ”f ” ,&ai

22、); if(ai>max) max=ai;else if(ai<min) min=ai; printf( nmax=%dnmin=%dn n,max,min);3.#include<stdio.h>#define NUM 20void main()int aNUM,i ； for(i=0;i<NUM;i+) scanf( H%dn,&ai); printf( nthe result is:n*');for(i=0;i<NUM;i+)if(i%2=0&&ai %2=0)printf( n%dn n,ai);4.#include&

23、lt;stdio.h>void main()int i,j, sum 1 =0, sum2=0, sum3=0;float aver;for(i=0;i<3;i+)for(j=0;j<3;j+)scanf(H%d n,&ai);for(i=0;i<3;i+)for(j=0;j<3;j+)printf( n%6dH,aij); printf( nnn);for(i=0;i<3;i+) suml+=aij+a2-ii;printf ( ”輸出主對(duì)角線與次對(duì)角線元素的和是:%dnn,suml);for(i=0;i<3;i+)sum2+=ali;pri

24、ntfC 輸出第二行元素的和是 :%dnn,sum2); for(i=0;i<3;i+) sum3+=ai 2;aver=(float)sum3/3;printfC 輸出第三列元素的平均值是 :%.2fn n,aver);5.#include<stdio.h>void main()int i,col=0,co2=0,co3=0,co4=0;char s1000,c;gets(s);for(i=0;(c=ai) != 、 0： i+)if(c>=*a'&&cv='z'llc>='A '&& c&

25、lt;='Z') col+;else if(c=* *)co2+;else if(c>='0 '&&cv=9)co3+;elseco4+;printf ( ”統(tǒng)計(jì)字符串中英文字母、空格符、數(shù)字字符及其它字符的個(gè)數(shù)分別 是 : %d,%d,%d,%dn",col,co2,co3,co4);6.#include <stdio.h>#define N 10void main()intaNN,i,j;for(i=0;i<N;i+)ai0=l; aii=l;for(i=2;i<N;i+)for(j=l;j<i;

26、j+) aiU=ai-lU+ai-lj-l;for(i=0;i<N;i+)for(j=0;j<=i;j+)printf( %”d” ,aij);printf( nnn);7.#i nclude <stdio.h>void mai n()int i,j ;for(i=0;i<5;i+)for(j=0;j<i;j+)prin tf("");for(j=0;j<5;j+)printfC* ");pri ntf("n");第7章函數(shù)與變量、選擇題123456789101112DCBBDADDDAAA二、填空題1

27、.42. double avg(double a,double b)3. 2114. i+11二、編程題1 #in clude <stdio.h>double cyli nder (double r, double h)double result;result =3.1415926 * r * r * h;return result;void mai n()double height, radius, volume; double cyli nder (double r, double h); printf ("E nter radius and height:"

28、;); sca nf ("%lf%lf", &radius, &height); volume = cyli nder (radius, height);printf ("Volume = %.3fn", volume);return 0;2#include <stdio.h>char* fun(char* s, char c)char* t = s;char* f = s;for(;*f;f+)if(*f=c)continue;*t 二 *f?t+;*t = 0;return s;void main()char input1

29、000;char c;Hprintf ( “請(qǐng)輸入一個(gè)字符串： H); gets(input); printf ( ”請(qǐng)輸入一個(gè)字符 : ” ) ； scanf("%c",&c); printf ( ”結(jié)果 : ” ) ； puts(fun(input,c);3#include <stdio.h>long fun(int g)switch(g)case 0: return 0;case 1: case 2: return 1;return (fun(g-1 )+fun(g-2);void main()long fib; int n;printf(&quo

30、t;Input n:");scanf("%d n,&n); printf("n=%dn",n); fib=fun(n); printf ( ”fib=%dnn",fib);4#include <conio.h>#include <stdio.h>int fun (int x)if(x< 10)return !(x * x - x) % 10);elsereturn !(x * x - x) % 100);void main()int x,y;clrscr();Hprintf( nPlease enter a

31、 integer numbers:"); scanf("%d",&x); if(x>100)printf( ndata error !n"); exit(O);y=fun(x);if(y)printf( nn” ,x);elseprintf( n%d NO!n H,x);5#include <stdio.h>#include <conio.h>float fun (float h)long num;h=h+0.005;h=h*100;num=h;h=num;h=h/100;return h;void main()fl

32、oat a;clrscr();printf( nEnter a:；)sca nf("%f', &a);printf( nThe original data is:");printf (”f nnn,a);printf( nThe ?suit: %fn",fun(a);6#in clude vconi o.h>#i nclude <stdio.h>#defi ne N 81void fun (char *s)int i,le ngth;char tmp;len gth=strle n( s);if(le ngth=O)return;

33、tmp=s0;for(i=0;i<le ngth-1 ;i+)si=si+l;s length ? l=tmp;void mai n()char aN;clrscr();pri ntf( nEn ter a str in g:");gets(a);printf( nThe original string is:");puts(a);fun( a);pri ntf("The stri ng after modified:");puts(a);第8章編譯與預(yù)處理、選擇題1234567DBA無答案ADB二、填空題(5)1. c2. 93. 704. 10

34、00 105.9911三、運(yùn)行結(jié)果題1. a=62.a=3a=32. x= 933. a=14,b=15c=04. 錯(cuò)誤四、編程題1.#i nclude<stdio.h>#in elude <math.h>#define BICIR(a,b,c) sqit(a+b+c)/2)*(a+b+c)/2? a)*(a+b+c)/2 ? b)*(a+b+c)/2 ? c)void mai n()float a,b,c;scanf(n%f%f%f n,&a,&b,&c);printf( n%f',BICIR(a,b,c);2.#i nclude<

35、;stdio.h>#in clude <math.h>#defi ne min( a,b,c) a>b?(b>c?c:b):(a>c?c:a)void mai n()float a,b,c;scanf(n%f%f%f n,&a,&b,&c);printf( n%fn,min(a,b,c);3.#i nclude<stdio.h>#define sum(n) (l+(n)*(n)/2void mai n()int n ,sum;prin tf("e nter your n umberA n');sca nf

36、("%d",&n);sum=sum (n);prin tf("the sum is %dn",sum);第9章結(jié)構(gòu)體與其他數(shù)據(jù)類型、選擇題1234567cDDABCC二、填空題1. emoclew2. 13431三、編程題1.#include <stdio.h>#include <string.h>#define N 16typedef structchar num10;int s;STREC;STREC fun(STREC *a,char *b)STREC k;int i,flag=O;for(i=0;i<N;i+

37、)if(strcmp(ai .num,b)=0)k=ai;flag=l ； if(flag=l) return k;elsestrcpy(k.num," n);k.s=-l;return k;void main()STRECsN= ” GA05” ,85, ” GA03” ,76, ” GA02” ,69, ” GA04” ,85, ” GA01” ,91, ” GA07“ ,72, ”GA 08” ,64, “ GA06” ,87, ” GA015” ,85, ” GA013” ,91, ” GA012” ,64, ” GA014” ,9nG1, ” GA011” ,91 , A017 n,64, HGA018 n,64, HGA016 n,72;STREC h;char m10;int i;FILE *out;printf( nThe original data:n");for(i=0;i<N;i+)if(i)%4=0) printf( nnn);nHprintf("%s %3d n,si.num,si.s); printf( HnnEnter