1、C语言程序设计第3版何钦铭颜晖第9章结构第 9 章 结构【练习 9-1 】定义一个能够表示复数的结构类型,一个复数包括实数与虚数两个部分。解答:struct complexfloat real;float imaginary;【练习 9-2】人的出生日期由年、月、日组成,请在例 9-1 中的通讯录结构中增加一个成员:出生日期,用嵌套定义的方式重新定义该结构类型。解答:struct dateint year;int month;int day;struct studentint num;char name10;struct date birthday;int computer,english,m
2、ath;double average;【练习 9-3 】例 9-1 中,如果要计算的是三门课程的课程平均成绩,应该如何改写程序?解答:#includestruct studentint num;char name10;int computer,english,math;double average;int main(void)int i, n;double math_sum,english_sum,computer_sum;struct student s1;printf(Input n:);scanf(%d, &n);printf(Input the students number, nam
3、e and course scores:n);math_sum=english_sum=computer_sum=0;for(i=1;i=n;i+)printf(No.%d:,i);scanf(%d%s%d%d%d,&s1.num,s1.name,&s1.math,&s1.english,&pu ter);math_sum+=s1.math;english_sum+=s1.english;computer_sum+=puter;printf(math_ave:%.2lfnenglish_ave:%.2lfncomputer_ave:%.2lfn, math_sum/n,english_sum/
4、n,computer_sum/n);return 0;【练习 9-4】定义一个包含 5 名学生信息的结构数组,并对该结构数组的所有元素进行初始化。解答:struct studentint num;char name10;int computer, english, math;struct student s5=30101, 张一 ,69,75,84 ,30132, 李二 ,80,8 5,90,40231, 王三 ,71,91,74,40754, 赵四 ,65,76,87,50426, 刘五 ,81,92,73;【练习 9-5】参考例 9-2 ,输入并保存 10 个学生的成绩信息,分别输出平均成
5、绩最高和最低的学生信息。解答:#includestruct studentint num;char name10;int computer,english,math;double average;int main(void)int i,n,max,min;struct student students50;printf(Input n:); scanf(%d,&n);for(i=0;in;i+)printf(Input the info of No.%d:n,i+1);printf(number:);scanf(%d,&studentsi.num);printf(name:);scanf(%s
6、,&studentsi.name);printf(math score:);scanf(%d,&studentsi.math);printf(english score:);scanf(%d,&studentsi.english);printf(computer score:);scanf(%d,&puter);studentsi.average=(studentsi.math+studentsi.english+stu puter)/3.0;max=min=0;for(i=1;in;i+)if(studentsmax.averagestudentsi.average) min=i;print
7、f(max_average_infortmation:n);printf(number:%d,name:%s,math:%d,english:%d,computer:%d,average: %.2lfn,studentsmax.num,studentsmax.name,studentsmax.math,stud entsmax.english,puter,studentsmax.average);printf(min_average_information:n);printf(number:%d,name:%s,math:%d,english:%d,computer:%d,average: %
8、.2lfn,studentsmin.num,studentsmin.name,studentsmin.math,stud entsmin.english, puter, studentsmin.average);return 0;【练习 9-6 】定义一个 struct student 类型的结构指针,用其实现一个学生信数组形式。解答:int update_score(struct student s,int n,int num,int course,int int i,pos;for(i=0;in;i+)if(si.num=num)break;if(i”称为 _指向 _运算符。struct
9、sk int a; float b; data, *p=&data;A (*p).data.a BC p-data.a D5对于以下结构定义, (*p)-str+struct int len; char *str; *p;A指针 str 上 BC str 指向的内容上 D二、填空题1“ . ”称为 成员 _( 分量 )_运算符,“2完成下列程序,该程序计算 10 名学生的平均成绩。#include #include struct student int num;char name20;int score;struct student stud10;int main(void)int i , s
10、um = 0 ;for(i = 0; i 10; i+)scanf(%d%s%d , studi.num,_studi.name_, studi.score);sum += studi.score;printf(aver = %d n, sum/10);return 0;3下列程序读入时间数值,将其加 1 秒后输出,时间格式为: hh: mm: ss,即小时:分钟:秒,当小时等于 24 小时,置为 0 。#includestruct int hour, minute, second; time;int main(void)scanf(%d: %d: %d,_&time.hour, &time.
11、minute, &time.second_); time.second+;if( _time.second_ = 60)_time.minute+ _;time.second = 0;if(time.minute = 60)time.hour+;time.minute = 0;if( _time.hour = 24_ )time.hour = 0;printf (%d: %d: %d n, time.hour, time.minute, time.second );return 0;4写出下面程序的运行结果 1 2 A B 。struct s1char c1, c2;int n;struct
12、s2int n;struct s1 m; m = 1, A , B , 2 ;int main(void)printf( “ %d n” , m.n, m.m.n, m.m.c1, m.m.c2);return 0;5写出下面程序的运行结果 23, wang, 98.5, wang 。struct abcint a;float b;char *c; ;int main(void)struct abc x = 23,98.5,wang;struct abc *px = &x;printf(%d, %s, %.1f, %s n, x.a, x.c, (*px).b, px-c );return 0
13、;三、程序设计题1.时间换算:用结构类型表示时间内容(时间以时、分、秒表示) ,输入一个时间数值,再输入一个秒数 n(n60) ,以 h: m: s 的格式输出该时间再过 n 秒后的时间值(超过 24 点就从 0 开始计时) 。试编写相应程序。解答:#includestruct timeint hour;int minute;int second;int main(void)int n;struct time tim;printf(Enter time:);scanf(%d%d%d,&tim.hour,&tim.minute,&tim.second);printf(Enter seconds:
14、);scanf(%d,&n);tim.hour=(tim.hour*60*60+tim.minute*60+tim.second+n)/3600%24;tim.minute=(tim.minute*60+tim.second+n)/60%60;tim.second=(n+tim.second)%60;printf(%d:%d:%d,tim.hour,tim.minute,tim.second);return 0;2.计算两个复数之积:编写程序,利用结构变量求解两个复数之积 : (3+4i) (5+6i ) 。提示:求解 (a1+a2i) (b1+b2i) ,乘积的实部为: a1 b1 - a2
15、 b2,虚部为:a1 b2 + a2 b1。解答:#includestruct complexint real,im;struct complex cmult(struct complex a,struct complex b);int main(void)struct complex a=3,4,b=5,6,c;c=cmult(a,b);printf(%d+%di)(%d+%di)=%d+%din,a.real,a.im,b.real,b.im,c.rea l,c.im);return 0;struct complex cmult(struct complex a,struct comple
16、x b)struct complex c;c.real=a.real*b.real-a.im*b.im;c.im=a.real*b.im+a.im*b.real;return c;3.平面向量加法:输入两个二维平面向量 V1=( x1, y1 )和 V2=( x2, y2)的分量,计算并输出两个向量的和向量。试编写相应程序。解答:#includestruct vectorfloat x,y;struct vector vec(struct vector v1,struct vector v2);int main(void)struct vector v1,v2,v;printf(x1 and
17、y1:);scanf(%f%f,&v1.x,&v1.y);printf(x2 and y2:);scanf(%f%f,&v2.x,&v2.y);v=vec(v1,v2);printf(v=(%f,%f),v.x,v.y);return 0;struct vector vec(struct vector v1,struct vector v2)struct vector v;v.x=v1.x+v2.x;v.y=v1.y+v2.y;return v;4.查找书籍:从键盘输入 10 本书的名称和定价并存入结构数组中,从中查找定价最高和最低的书的名称和定价,并输出。试编写相应程序。解答:#includ
18、e#define NUMBER 10struct bookchar name10;float price;int main(void)int i,maxl,minl,n;struct book testNUMBER;for(i=0;iNUMBER;i+)scanf(%s%f,testi.name,&testi.price);maxl=minl=0;for(i=1;in;i+)if(testmaxl.pricetesti.price) minl=i;printf(Max Price:%f,%sn, testmaxl.price,testmaxl.name);printf(Min Price:%f
19、,%sn, testminl.price,testminl.name);return 0;5通讯录排序:建立一个通讯录,通讯录的结构包括:姓名、生日、电话号码;其中生日又包括三项:年、月、日。编写程序,定一个嵌套的结构类型,输入n(n 10)个联系人信息, 再按照他们的年龄从大到小依次输出其信息。 试编写相应程序。解答:#includestruct dateint year;int month;int day;struct friends_listchar name10;struct date birthday;char phone15;void sort(struct friends_lis
20、t s,int n);int main(void)int i,n;struct friends_list friends10;printf(Input n:);scanf(%d,&n);printf(Input %d friends name ,birthday and phone:,n);for(i=0;in;i+)scanf(%s%d%d%d%s,friendsi.name,&friendsi.birthday.year, &friendsi.birthday.month,&friendsi.birthday.day,&friendsi.phone );sort(friends,n);fo
21、r(i=0;in;i+)printf(%s%d/%d/%d%sn,friendsi.name,friendsi.birthday.y ear,friendsi.birthday.month,friendsi.birthday.day,friendsi.phon e);return 0;void sort(struct friends_list s,int n)int i,j,index;struct friends_list temp;for(i=0;in-1;i+)index=i;for(j=i+1;jsj.birthday.year)index=j;else if(sindex.birth
22、day.year=sj.birthday.year)&(sindex.birthday.monthsj.birthday.month)index=j;else if(sindex.birthday.year=sj.birthday.year)&(sindex.birthday.month=sj.birthday.month)&(sindex.birthday. daysj.birthday.day)index=j;temp=si;si=sindex;sindex=temp;6按等级统计学生成绩:输入 10 个学生的学号、姓名和成绩,输出学生的成绩等级和不及格人数。每个学生的记录包括学号、姓名、
23、成绩和等级,要求定义和调用函数 set_grade() 根据学生成绩设置其等级, 并统计不及格人数, 等级设置: 85 100 为 A, 70 84 为 B, 60 69 为 C, 0 59 为 D。 试编写相应程序。解答:#include#define N 10struct studentsint num;char name20;int score;char grade;int set_grade(struct students *p);int main(void)struct students studentN;int i,count;printf(Input 10 students nu
24、mber,name and score: n);for(i=0;iN;i+)printf(No.%d:,i+1);scanf(%d%s%d,&studenti.num,studenti.name,&studenti.sc ore);count=set_grade(student);printf(The count (60): %dn,count);printf(The student grade:n);for(i=0;iN;i+)printf(%d %s %cn,studenti.num,studenti.name,studenti .grade);return 0;int set_grade(struct students *p)int i,n=0;for(i=0;iscore=85)p-grade=A;else if(p-score=70&p-scoregrade=B;else if(p-score=60&p-scoregrade=C;elsep-grade=D; n+;return n;
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