C语言程序设计(第3版)何钦铭 颜 晖 第9章结构.doc

如有侵权，请联系网站删除，仅供学习与交流C语言程序设计(第3版)何钦铭 颜 晖 第9章 结构【精品文档】第 10 页第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 complex float 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,math;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：#include<stdio.h>struct student int 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 student's number, name 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,&puter); math_sum+=s1.math; english_sum+=s1.english; computer_sum+=puter; printf("math_ave:%.2lfnenglish_ave:%.2lfncomputer_ave:%.2lfn",math_sum/n,english_sum/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 student int num;char name10; int computer, english, math; struct student s5=30101, "张一",69,75,84 ,30132, "李二",80,85,90,40231, "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，分别输出平均成绩最高和最低的学生信息。解答：#include<stdio.h>struct student int 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;i<n;i+) printf("Input the info of No.%d:n",i+1); printf("number:"); scanf("%d",&studentsi.num); printf("name:"); scanf("%s",&studentsi.name); printf("math score:"); scanf("%d",&studentsi.math); printf("english score:"); scanf("%d",&studentsi.english); printf("computer score:"); scanf("%d",&studentsi.computer);studentsi.average=(studentsi.math+studentsi.english+studentsi.computer)/3.0; max=min=0; for(i=1;i<n;i+) if(studentsmax.average<studentsi.average) max=i; if(studentsmin.average>studentsi.average) min=i; printf("max_average_infortmation:n");printf("number:%d,name:%s,math:%d,english:%d,computer:%d,average:%.2lfn",studentsmax.num,studentsmax.name,studentsmax.math,studentsmax.english,studentsmax.computer,studentsmax.average); printf("min_average_information:n");printf("number:%d,name:%s,math:%d,english:%d,computer:%d,average:%.2lfn",studentsmin.num,studentsmin.name,studentsmin.math,studentsmin.english, studentsmin.computer, 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息的输入和输出。解答：struct student /* 学生信息结构定义 */int num; /* 学号 */char name10; /* 姓名 */int computer, english, math ; /* 三门课程成绩 */double average; /* 个人平均成绩 */s, *p;p = &s;scanf("%d%s%d%d%d", &p->num, p->name, &p->math, &p->english, &p->computer);【练习9-7】改写例 9-3 中的函数 update_score( )，将第一个形参改为结构数组形式。解答：int update_score(struct student s,int n,int num,int course,int score) int i,pos; for(i=0;i<n;i+) if(si.num=num) break; if(i<n) switch(course) case 1:si.math=score; break; case 2:si.english=score; break; case 3:si.computer=score; break; pos = i; else pos = -1; return pos;习题9一、选择题1下面定义结构变量的语句中错误的是 D 。Astruct student int num; char name20; s;Bstruct int num; char name20; s;Cstruct student int num; char name20; ; struct student s;Dstruct student int num; char name20; ; student s;2如果有定义语句：struct int x, y; s2 = 1, 3 , 2, 7 ;则语句：printf(“%dn”, s0.y/s1.x ); 输出结果为 B 。A0 B1 C2 D3 3根据下面的定义，能打印出字母 M 的语句是 C 。struct personchar name10; int age; c10 = “John”, 17, “Paul”, 19, “Mary”, 18, “Adam”, 16 ; Aprintf(“%c”, c3.name); Bprintf(“%c”, c3.name1);Cprintf(“%c”, c2.name0); Dprintf(“%c”, c2.name1);4设有如下定义，则对 data 中的 a 成员的正确引用是 B 。struct sk int a; float b; data, *p=&data;A(*p).data.a B(*p).aCp->data.a Dp.data.a5对于以下结构定义，(*p)->str+中的+加在 D 。struct int len; char *str; *p;A 指针 str 上 B指针 p 上Cstr 指向的内容上 D语法错误二、填空题1“.”称为 成员_(分量)_运算符，“->”称为_指向_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 <stdio.h>#include <string.h>struct student int num;char name20; int score;struct student stud10;int main(void)int i , sum = 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。#include<stdio.h>struct int hour, minute, second; time;int main(void)scanf("%d: %d: %d",_&time.hour, &time.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 s1 char c1, c2; int n;struct s2 int n; struct s1 m; m = 1, A, B, 2 ;int main(void)printf(“%dt%dt%ct%cn”, 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;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(n<60)，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：#include<stdio.h>struct time int 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:"); 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×b2，虚部为：a1×b2 + a2×b1。解答：#include<stdio.h>struct complex int 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.real,c.im); return 0; struct complex cmult(struct complex a,struct complex 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：#include<stdio.h>struct vector float x,y;struct vector vec(struct vector v1,struct vector v2);int main(void) struct vector v1,v2,v; printf("x1 and 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：#include<stdio.h>#define NUMBER 10struct book char name10; float price;int main(void) int i,maxl,minl,n; struct book testNUMBER; for(i=0;i<NUMBER;i+) scanf("%s%f",testi.name,&testi.price); maxl=minl=0; for(i=1;i<n;i+) if(testmaxl.price<testi.price) maxl=i; if(testminl.price>testi.price) minl=i; printf("Max Price:%f,%sn", testmaxl.price,testmaxl.name); printf("Min Price:%f,%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(n10)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：#include<stdio.h>struct date int year; int month; int day; struct friends_list char name10; struct date birthday; char phone15; void sort(struct friends_list 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;i<n;i+)scanf("%s%d%d%d%s",friendsi.name,&friendsi.birthday.year,&friendsi.birthday.month,&friendsi.birthday.day,&friendsi.phone); sort(friends,n); for(i=0;i<n;i+)printf("%s%d/%d/%d%sn",friendsi.name,friendsi.birthday.year,friendsi.birthday.month,friendsi.birthday.day,friendsi.phone); return 0; void sort(struct friends_list s,int n) int i,j,index; struct friends_list temp; for(i=0;i<n-1;i+) index=i; for(j=i+1;j<n;j+) if(sindex.birthday.year>sj.birthday.year) index=j; else if(sindex.birthday.year=sj.birthday.year)&&(sindex.birthday.month>sj.birthday.month) index=j; else if(sindex.birthday.year=sj.birthday.year)&&(sindex.birthday.month=sj.birthday.month)&&(sindex.birthday.day>sj.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、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：85100 为 A，7084 为 B，6069 为 C，059 为 D。试编写相应程序。解答：#include<stdio.h>#define N 10struct students int 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' number,name and score: n"); for(i=0;i<N;i+) printf("No.%d:",i+1);scanf("%d%s%d",&studenti.num,studenti.name,&studenti.score); count=set_grade(student); printf("The count (<60): %dn",count); printf("The student grade:n"); for(i=0;i<N;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;i<N;i+,p+) if(p->score>=85) p->grade='A' else if(p->score>=70&&p->score<85) p->grade='B' else if(p->score>=60&&p->score<70) p->grade='C' else p->grade='D' n+; return n;