C语言课后习题解答.doc

第七章 函数调用7-1写两个函数，分别求两个整数的最大公约数和最小公倍数，#include void main()int hcf(int,int);int lcd(int,int,int); int u,v,h,l;scanf(%d,%d,&u,&v);h=hcf(u,v);printf(H.C.F=%dn,h);l=lcd(u,v,h);printf(L.C.D=%dn,l);int hcf(int u,int v)int t,r;if (vu)t=u;u=v;v=t;while (r=u%v)!=0)u=v;v=r;return(v);int lcd(int u,int v,int h)return(u*v/h);7-2求方程ax*x+bx+c=0的根，用3个函数分别求当b*b-4ac大于0，等于0、小于0时的根并输出结果。从主函数输入abc的值。#include #include float x1,x2,disc,p,q;void main()void greater_than_zero(float,float);void equal_to_zero(float,float);void smaller_than_zero(float,float);float a,b,c;printf(input a,b,c:);scanf(%f,%f,%f,&a,&b,&c);printf(equation: %5.2f*x*x+%5.2f*x+%5.2f=0n,a,b,c);disc=b*b-4*a*c;printf(root:n);if (disc0)greater_than_zero(a,b);printf(x1=%fttx2=%fn,x1,x2);else if (disc=0)equal_to_zero(a,b);printf(x1=%fttx2=%fn,x1,x2);elsesmaller_than_zero(a,b);printf(x1=%f+%fitx2=%f-%fin,p,q,p,q);void greater_than_zero(float a,float b)x1=(-b+sqrt(disc)/(2*a);x2=(-b-sqrt(disc)/(2*a);void equal_to_zero(float a,float b)x1=x2=(-b)/(2*a);void smaller_than_zero(float a,float b)p=-b/(2*a);q=sqrt(-disc)/(2*a);7-3写一个判断素数的函数，在主函数输入一个整数，输出是否素数的信息。#include void main()int prime(int);int n;printf(ninput an integer:);scanf(%d,&n);if (prime(n)printf(n %d is a prime.n,n);elseprintf(n %d is not a prime.n,n);int prime(int n)int flag=1,i;for (i=2;in/2 & flag=1;i+)if (n%i=0)flag=0;return(flag);7-4写一个函数，使给定的一个3x3的二维整形数组转置，即行列转换。#include #define N 3int arrayNN;void main() void convert(int array3);int i,j;printf(input array:n);for (i=0;iN;i+)for (j=0;jN;j+)scanf(%d,&arrayij);printf(noriginal array :n);for (i=0;iN;i+)for (j=0;jN;j+)printf(%5d,arrayij);printf(n);printf(convert array:n);for (i=0;iN;i+)for (j=0;jN;j+)printf(%5d,arrayij);printf(n);void convert(int array3)int i,j,t;for (i=0;iN;i+)for (j=i+1;jN;j+)t=arrayij;arrayij=arrayji;arrayji=t;7-5写一个函数，是输入的字符串按反序存放，在主函数中输入和输出字符窜。#include #include char str100;void main()void inverse(char str);printf(input string:);scanf(%s,str);inverse(str);printf(inverse string:%sn,str);void inverse(char str)char t;int i,j;for (i=0,j=strlen(str);i(strlen(str)/2);i+,j-)t=stri;stri=strj-1;strj-1=t;7-6写一个函数，将两个字符窜连接。#include void main()void concatenate(char string1,char string2,char string);char s1100,s2100,s100;printf(input string1:);scanf(%s,s1);printf(input string2:);scanf(%s,s2);concatenate(s1,s2,s);printf(nThe new string is %sn,s);void concatenate(char string1,char string2,char string)int i,j;for (i=0;string1i!=0;i+)stringi=string1i;for(j=0;string2j!=0;j+)stringi+j=string2j;stringi+j=0;7-7写一个函数，将字符窜中元音字母复制到另一个字符窜，然后输出。#include void main()void cpy(char ,char );char str80,c80;printf(input string:);gets(str);cpy(str,c);printf(The vowel letters are:%sn,c);void cpy(char s,char c) int i,j;for (i=0,j=0;si!=0;i+)if (si=a|si=A|si=e|si=E|si=i|si=I|si=o|si=O|si=u|si=U)cj=si;j+;cj=0;7-13输入10个学生5门课的成绩，分别用函数实现下列功能。1 计算每个学生平均分。2 计算每门课的平均分。3 找出所有50个分数中最高的分数所对应的学生和课程。4 计算平均方差。#include #define N 10#define M 5float scoreNM;float a_stuN,a_courM;int r,c;void main() int i,j;float h;float s_var(void);float highest();void input_stu(void);void aver_stu(void);void aver_cour(void);input_stu();aver_stu();aver_cour();printf(n NO. cour1 cour2 cour3 cour4 cour5 avern);for(i=0;iN;i+)printf(n NO %2d ,i+1);for(j=0;jM;j+)printf(%8.2f,scoreij);printf(%8.2fn,a_stui);printf(naverage:);for (j=0;jM;j+)printf(%8.2f,a_courj);printf(n);h=highest();printf(highest:%7.2f NO. %2d course %2dn,h,r,c);printf(variance %8.2fn,s_var();void input_stu(void)int i,j;for (i=0;iN;i+)printf(ninput score of student%2d:n,i+1);for (j=0;jM;j+)scanf(%f,&scoreij);void aver_stu(void)int i,j;float s;for (i=0;iN;i+)for (j=0,s=0;jM;j+)s+=scoreij;a_stui=s/5.0;void aver_cour(void)int i,j;float s;for (j=0;jM;j+)s=0;for (i=0;iN;i+)s+=scoreij;a_courj=s/(float)N;float highest()float high;int i,j;high=score00;for (i=0;iN;i+)for (j=0;jhigh)high=scoreij;r=i+1;c=j+1;return(high);float s_var(void)int i;float sumx,sumxn;sumx=0.0;sumxn=0.0;for (i=0;iN;i+)sumx+=a_stui*a_stui;sumxn+=a_stui;return(sumx/N-(sumxn/N)*(sumxn/N);7-14写几个函数：1 输入10个职工的姓名和职工号2 按职工号由小到大的顺序排序，姓名顺序也随之调整。3 要求输入一个职工号，用折半查找法找出该职工的姓名，从主函数输入要查找的职工号，输出该职工姓名。#include #include #define N 10void main()void input(int ,char name8);void sort(int ,char name8);void search(int ,int ,char name8);int numN,number,flag=1,c;char nameN8;input(num,name);sort(num,name);while (flag=1)printf(ninput number to look for:);scanf(%d,&number);search(number,num,name);printf(continue ot not(Y/N)?);getchar();c=getchar();if (c=N|c=n)flag=0;void input(int num,char nameN8)int i;for (i=0;iN;i+)printf(input NO.: );scanf(%d,&numi);printf(input name: );getchar();gets(namei);void sort(int num,char nameN8) int i,j,min,templ;char temp28;for (i=0;iN-1;i+)min=i;for (j=i;jnumj) min=j;templ=numi;strcpy(temp2,namei);numi=nummin;strcpy (namei,namemin);nummin=templ;strcpy(namemin,temp2);printf(n result:n);for (i=0;iN;i+)printf(n %5d%10s,numi,namei);void search(int n,int num,char nameN8)int top,bott,mid,loca,sign;top=0;bott=N-1;loca=0;sign=1;if (nnumN-1)loca=-1;while(sign=1) & (top=bott)mid=(bott+top)/2;if (n=nummid)loca=mid;printf(NO. %d , his name is %s.n,n,nameloca);sign=-1;else if (nnummid)bott=mid-1;elsetop=mid+1;if (sign=1 | loca=-1)printf(%d not been found.n,n);第八章 指针8-11 输入3个整数，按由小到大的顺序输出#includevoid main()void swap(int*,int*);int a,b,c; int *p1,*p2,*p3;p1=&a;p2=&b;p3=&c;scanf(%d,%d,%d,p1,p2,p3);if(*p1*p2) swap(p1,p2);if(*p1*p3) swap(p1,p3);if(*p2*p3) swap(p2,p3);printf(%d,%d,%dn,a,b,c);void swap(int*p1,int*p2)int p;p=*p1;*p1=*p2;*p2=p;8-2输入3个整数按有小到大的顺序输出。#include #include void main()void swap(char *,char *);char str120,str220,str320;printf(input three line:n);gets(str1);gets(str2);gets(str3);if(strcmp(str1,str2)0) swap(str1,str2);if(strcmp(str1,str3)0) swap(str1,str3);if(strcmp(str2,str3)0) swap(str2,str3);printf(Now,the order is:n);printf(%sn%sn%sn,str1,str2,str3);void swap(char *p1,char *p2)char p20;strcpy(p,p1);strcpy(p1,p2);strcpy(p2,p);8-3输入10个整数，将其中最小的数与第一个数对换，把最大的数与最后一个数对换。写3个函数：1输入10个数；2 进行处理；3 输出10个数#include void main() void input(int *); void max_min_value(int *); void output(int *);int number10;input(number); max_min_value(number); output(number); void input(int *number) int i;printf(input 10 numbers:);for (i=0;i10;i+)scanf(%d,&numberi);void max_min_value(int *number) int *max,*min,*p,temp;max=min=number;for (p=number+1;p*max) max=p; else if (*p*min) min=p; temp=number0;number0=*min;*min=temp; if(max=number) max=min;temp=number9;number9=*max;*max=temp; void output(int *number) int *p;printf(Now,they are: );for (p=number;pnumber+10;p+)printf(%d ,*p);printf(n);8-7有一个字符窜，包含n个字符。写一个函数，将此字符窜中从m个字符开始的全部字符复制成为另一个字符窜。#include #include void main()void copystr(char *,char *,int); int m;char str120,str220;printf(input string:);gets(str1);printf(which character that begin to copy?);scanf(%d,&m);if (strlen(str1)m)printf(input error!);elsecopystr(str1,str2,m);printf(result:%sn,str2);void copystr(char *p1,char *p2,int m) int n;n=0;while (nm-1)n+;p1+;while (*p1!=0)*p2=*p1;p1+;p2+;*p2=0;8-9写一个函数，使给定的一个3x3的二维整形数组转置，即行列转换。#include void main()void move(int *pointer);int a33,*p,i;printf(input matrix:n);for (i=0;i3;i+)scanf(%d %d %d,&ai0,&ai1,&ai2);p=&a00;move(p);printf(Now,matrix:n);for (i=0;i3;i+)printf(%d %d %dn,ai0,ai1,ai2);void move(int *pointer)int i,j,t;for (i=0;i3;i+)for (j=i;j3;j+)t=*(pointer+3*i+j);*(pointer+3*i+j)=*(pointer+3*j+i);*(pointer+3*j+i)=t;8-13将n个数按输入顺序的逆序排列，用函数实现。#include void main()void sort (char *p,int m);int i,n;char *p,num20;printf(input n:);scanf(%d,&n);printf(please input these numbers:n);for (i=0;in;i+)scanf(%d,&numi);p=&num0;sort(p,n);printf(Now,the sequence is:n);for (i=0;in;i+)printf(%d ,numi);printf(n);void sort (char *p,int m)int i;char temp, *p1,*p2;for (i=0;im/2;i+)p1=p+i;p2=p+(m-1-i);temp=*p1;*p1=*p2;*p2=temp;第九章 用户建立的数据类型9-1定义一个结构体变量（包扩年、月、日）。计算该日在本年中的第几天，注意闰年问题。#include struct int year;int month;int day;date;void main()int days;printf(input year,month,day:);scanf(%d,%d,%d,&date. year,&date.month,&date.day);switch(date.month) case 1: days=date.day; break;case 2: days=date.day+31; break;case 3: days=date.day+59; break;case 4: days=date.day+90; break;case 5: days=date.day+120; break;case 6: days=date.day+151; break;case 7: days=date.day+181; break;case 8: days=date.day+212; break;case 9: days=date.day+243; break;case 10: days=date.day+273; break;case 11: days=date.day+304; break;case 12: days=date.day+334; break;if (date.year %4= 0 & date.year % 100 != 0|date.year % 400 = 0) & date.month =3) days+=1;printf(%d/%d is the %dth day in %d.n,date.month,date.day,days,date.year);9-2写一个函数days，实现第一题的计算，由主函数将年、月、日传递给days函数，计算后将日子数传回主函数输出。clude struct int year;int month;int day;date;void main()int i,days;int day_tab13=0,31,28,31,30,31,30,31,31,30,31,30,31;printf(input year,month,day:);scanf(%d,%d,%d,&date. year,&date.month,&date.day);days=0;for(i=1;i=3)days=days+1;printf(%d/%d is the %dth day in %d.n,date.month,date.day,days,date.year);9-3编写一个函数print，打印一个学生的成绩数组，该数组中有5个学生的数据记录，每个记录包括,score3,用主函数输入这些记录，用print函数输出这些记录。#include #define N 5struct student char num6;char name8;int score4;stuN;void main()void print(struct student stu6);int i,j;for (i=0;iN;i+)printf(ninput score of student %d:n,i+1);printf(NO.: );scanf(%s,stui.num);printf(name: );scanf(%s,);for (j=0;j3;j+)printf(score %d:,j+1);scanf(%d,&stui.scorej);printf(n);print(stu);void print(struct student stu6)int i,j;printf(n NO. name score1 score2 score3n);for (i=0;iN;i+)printf(%5s%10s,stui.num,);for (j=0;j3;j+)printf(%9d,stui.scorej);printf(n);9-4在第3题的基础上，编写一个函数input，用来输入5个学生的数据记录。#include #define N 5struct studentchar num6;char name8;int score4; stuN;void main()void input(struct student stu);void print(struct student stu);input(stu);print(stu);void input(struct student stu)int i,j;for (i=0;iN;i+)printf(input scores of student %d:n,i+1);printf(NO.: );scanf(%s,stui.num);printf(name: );scanf(%s,);for (j=0;j3;j+)printf(score