C语言程序设计第四版第九章答案谭浩强

1、第九章91 定义一个结构体变量（包括年、 月、日）。计算该日在本年中是第几天，注意闰年问题。解： Structint year;int month;int day;date;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.da

2、y+59;break;case 4: days=date.day+90;break;case 5: days=date.day+120;break;case 6: days=date.day+31;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;

3、if(date.year%4=0&&date.year%100!=0|date.year%400=0)&&date.month>=3)days+=1;printf(n%d/%d“ is the %dth day in%d.” ,date.month,data.day,days,date,year);9.2 写一个函数days, 实现上面的计算。由主函数将年、月、日传递给days函数，计算后将日数传回主函数输出。解： struct y_m_dint year:int month;int day;date;intdays(struct y_m_d date1)

4、int sum;switch(data.month)case 1:sum=date1.day;case 2:sum=date1.day+31;case 3:sum=date1.day+59;case 4:sum=date1.day+90;case 5:sum=date1.day+120;case 6:sum=date1.day+151;break;break;break;break;break;break;case 7:sum=date1.day+181;break;case 8:sum=date1.day+212;break;case 9:sum=date1.day+243;breakcas

5、e 10:sum=date1.day+243;breakcase 11:sum=date1.day+243;breakcase 12:sum=date1.day+243;break;9.3 编写一个函数print, 打印一个学生的成绩数，该数组中有录包括 num 、name 、 sore3 ，用主函数输入这些记录，用解：5 个学生的数据记录，每个记print函数输出这些记录。#define N 5struct studentchar num6;char name8;int score4;stuN;main()int I,j ;for(I=0;I<N;I+)printf(“n” ,I+1)

6、;printf(“ no.:” );scanf(“ %s” ,stui.num);printf(“ name: ” );scanf(“ %s” ,);for(j=0;j<3;j+)printf(“ score%d: ” j+1);scanf(“ %d” ,&stui.scorej);printf(n“” );print(stu);print(struct student stu6)int I,j;printf(“ %5s%10s” ,stui.num,);for(j=0;j<3;j+)printf(“ %9d” ,stui.scorej)

7、;print(“n” );9.4 在上题的基础上，编写一个函数input,用来输入5 个学生的数据记录。解：#define N 5struct studentchar num6;char name8;int score4stuN;input(struct student stu)int I,j;for(I=0;I<N;I+)printf(“ input scores of studentn” ,I+1);printf(“ NO.: ” );scanf(“ %s” ,stui.num);printf(“ name:” );scanf(“ %s” , );for(j=0;j

8、<3;j+)printf(“ score%d: ” ,j+);scanf(“ %d” , &stui.scorej);printf(n“” );9.5 有 10 个学生，每个学生的数据包括学号、姓名、3 门课的成绩，从键盘输入10 个学生的数据，要求打印出3 门课的总平均成绩，以及最高分的学生的数据（包括学号、姓名、3门课成绩）解： #define N 10struct studentchar num6char name8int score4float avr;stuN;main()int I,j,max,maxi,sum;float average;for(I=0;I<N

9、;I+)printf(“n” ,I+1);printf(“ NO.: ” );scanf(“ %s” ,stui.num);printf(“ name” );scanf(“ %s” ,);for(j=0;j<3;j+)printf(“ score %d:” ,j+1);scanf(“ %d” , &stui.scorej);average=0;max=0;maxi=0;for(i=0;i<3;i+)sum=0;for(j=0;j<3;j+)sum+=stui.scorej;stui.avr=sum/3.0;average+=stui.avr;if(s

10、um>max)max=sum;maxi=I;average/=N;printf(“ NO. name score1 score2 score3 averagen” );for(I=0;I<N;I+)printf(“ %5s%10s” ,stui.num, );for(j=0;j<3;j+)printf(printf(“ %9d” ,stui.scorej);“ %8.n2f” ,stui.avr);printf(printf(“ average=%6.2fn” ,average);“ The highest score is:%s,score total:%

11、d.” ,max);9.6 编写一个函数 new,对 n 个字符开辟连续的存储空间，此函数应返回一个指针，指向字符串开始的空间。 New(n) 表示分配 n 个字节的内存空间。解： new 函数如下：#define NULL 0#define NEWSIZE 1000char newbufNEWSIZE;char *newp=newbuf;char *new(int n)if (newp+n<=newbuf+ NEWSIZE) newp= newp+n; return(newp-n);elsereturn(NULL);9.7 写一函数free ，将上题用new 函

12、数占用的空间释放。Free(p) 表示将 p 指向的单元以后的内存段释放。解：#define Null o#define NEWSIZE 1000char newbufNEWSIZE;char *newp=newbuf;free(char *p)if(p>=newbuf)&&(p<newbuf+NEWSIZE)newp=p;9.8 已有 a、 b 亮光链表，每个链表中的结点包括学好、成绩。要求把两个链表合并，按学号升序排列。解：#include<stdio.h>#define NULL 0#define LENsizeof(struct student)

13、strut studentlong num;int scor;struct student *next;struct student listA,listB;int n,sum=0;main()struct student *creat(void);struct student *insert(struct student *,struct student *);void print(struct student *);stuct student *ahead , *bhead,*abh;printf(“n” );ahead=creat();sum=sum+|n;abh=insert(ahea

14、d,bhead);print(abh);struct student *creat(void)struct student *p1,*p2,*head;n=0;p1=p2=(struct student *)malloc(LEN);printf(“ input number&scores of student: n” );printf(“ if number Is 0,stop inputing.n” );scanf(“ %ld,%d ” ,&p1->num,&p1->score);head=NULL;while(p1->num!=0)n=n+1;if

15、(n=1)head=p1;else p2->next =p1;p2=p1;p1=(struct student *)malloc(LEN);scanf(“ %ld,%d ” ,&p1->num,&p1->score);p2->next=NULL;return(head);struct student *insert(struct student *ah,struct student *bh)struct student *pa1 , *pa2,*pb1,*pb2;pa2=pa1=ah;pb2=pb1=bh;dowhile(pb1->num>p

16、a1->num)&&(pa1->next!=NULL)pa2=pa1;pa1=pa1->next;if(pb->num<=pa1->num)if(ah=pa1)ah=pb1;else pa2->next=pb1;pb1=pb1->next;pb2->next=pa1;pa2=pb2;pb2=pb1;while(pa1->next!=NULL)|(pa1=NULL&&pb1!=NULL);if(pb1->num>pa1->num)&&(pa1->next=NULl)

17、ap1->next=pb1;return(ah);void print(struct student *head)struct student *p;printf(“ %ld%dn” ,p->num,p->score);p=p->next;while(p!=NULL);9.9 13 个人围成一圈，从第1 个人开始顺序报号1、 2、 3。凡报到“3”者退出圈子。找出最后留在圈子中的人原来的序号。解：#define N 13struct personint number;int nextop;linkN+1;main()int I,count,h;for(I=1;I<

18、=N;I+)if(I=N)linki.nextp=1;elselinki.nextp=I+1;linki.number=I;printf(n“” );count=0;h=N;printf(“ sequence that person2 leave the circle:n” );while(count<N-1)I=0;while(I!=3)h=linkh.nextp;if(linkh.number)I+;printf(“ %4d” ,linkh.number);linkh.number=0;count+;printf(nThe“ last one is ” );for(I=1;ii<

19、;=N;I+)if(linki.number)printf(“ %3d” ,lini.number);9.10 有两个链表a 和b,设结点中包含学号、姓名。从1 链表中删去与b 链表中有相同学号的那些结点。解：#define LA 4#define LB 5#define NULL 0struct studentchar nump6;char name8;struct student *next;ALA,bLB;main()struct student aLA= struct studentbLB=“ 101 ” , ” Wang” , “ 103 ” , ” Zhang ” ,“102”,

20、”LI ”,“ 104 ” , ” Ma” ,“ 105 ” , ” zhang ” ,“ 106 ” , ”“ 105 ” , ” Chen” ,“ 107 ” , ” Gu “ 108 ” , ” Lui ” ; int I,j;struct student *p, *p1,*p2,*pt,*head1,*head2;head1=a;head2=b;printf(“ list an ”: );for(p1=head1,i=1;p1<a=LA;i+)p=p1;p1->next=a+I;printf(“ %8s%8sn” ,p1->num,p1->name);p1=p1

21、->next;p->next=NULL;printf(“n” );for(p2=head2,I=1;p2<b+LB;I+)p=p2;p2->next=b+I;printf( “ %8s%8sn” ,p2->num,p2->name);p2=pa->next;p->next=NULL;printf(n“” );p1=head1;while(p1!=NULL)p2=head2;while(p2!=NULL&&strcmp(p1->num,p2->num)!=0)p2=p2->next;if(strcmp(p1->

22、;num,p2->num=0)if(p1=head1)head1=p1->next;elsep->next=p1->next;p=p1;p1=p1->next;p1=hedad1;printf“n” ;while(p1!=NULL)printf(“ %7s %7sn” ,p1->num,p1->name);p1=p1->next;9.11 建立一个链表，每个结点包括：学号、姓名、性别、年龄。输入一个年龄，如果链表中的结点所包含的年龄等于此年龄，则将此结点删去。解： #define NULL 0#define LEN sizeof(struct s

23、tudent)struct studentchar num6;char name8;char sex2;int age;stuct student *next;stu10;main()struct student *p,*pt,*head;int I,length,iage,flag=1;int find=0;while(flag=1)printf( “ input length of list（ <10 ） : ” );scanf(“ %d” ,&length);if(length<10)flag=0;for(I=0;I<lenth;I+)p=(struct stu

24、dent *)malloc(LEN);if(I=0)head=pt=p;elsept->next=p;pt=p;ptintf(“ NO:” );scanf(“ %s”->num);,pprntf(“ name: ” );scanf(“ %s”->name);,pprintf(“ sex: ” );scanf(“ %s”->sex);,pprintf(“ age: ” );scanf(“ %s”->age);,pp->next=NULL;p=head;printf(“n” );while(p!=NULL)printf(“ %4s%8s%6s%6dn” ,p-&

25、gt;num, p->name, p->sex, p->age);p=p->next;printf(“ Input age:” );scanf(“ %d” ,&iage);pt=head;p=pt;if(pt->age=iage)p=pt->next;head=pt=p;find=1;elsept=pt->next;while(pt!=NULL)if(pt->age=iage)p->next=pt->next;find=1;else p=pt;pt=pt->next;if(!find)printf(“ Not found%d.” ,iage);p=head;printf(“n” );while(p!=NULL)pri