完整版C语言程序设计第四版第九章答案谭浩强.docx
《完整版C语言程序设计第四版第九章答案谭浩强.docx》由会员分享,可在线阅读,更多相关《完整版C语言程序设计第四版第九章答案谭浩强.docx(16页珍藏版)》请在冰豆网上搜索。
![完整版C语言程序设计第四版第九章答案谭浩强.docx](https://file1.bdocx.com/fileroot1/2022-12/16/eaad0420-5fb5-4910-9513-499d05b28da1/eaad0420-5fb5-4910-9513-499d05b28da11.gif)
完整版C语言程序设计第四版第九章答案谭浩强
第九章
9.1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:
Struct
{intyear;
intmonth;
intday;
}date;
main()
{intdays;
printf(“Inputyear,month,day:
”);
scanf(“%d,%D,%d”,&date.year,&date.month,&date.day);
switch(date.month)
{case1:
days=date.day;break;
case2:
days=date.day+31;break;
case3:
days=date.day+59;break;
case4:
days=date.day+90;break;
case5:
days=date.day+120;break;
case6:
days=date.day+31;break;
case7:
days=date.day+181;break;
case8:
days=date.day+212;break;
case9:
days=date.day+243;break;
case10:
days=date.day+273;break;
case11:
days=date.day+304;break;
case12:
days=date.day+334;break;
}
if((date.year%4==0&&date.year%100!
=0||date.year%400==0)&&date.month>=3)days+=1;
printf(“\n%d/%disthe%dthdayin%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:
structy_m_d
{intyear:
intmonth;
intday;
}date;
intdays(structy_m_ddate1)
{intsum;
switch(data.month)
{case1:
sum=date1.day;break;
case2:
sum=date1.day+31;break;
case3:
sum=date1.day+59;break;
case4:
sum=date1.day+90;break;
case5:
sum=date1.day+120;break;
case6:
sum=date1.day+151;break;
case7:
sum=date1.day+181;break;
case8:
sum=date1.day+212;break;
case9:
sum=date1.day+243;break
case10:
sum=date1.day+243;break
case11:
sum=date1.day+243;break
case12:
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5个学生的数据记录,每个记录包括num、name、sore[3],用主函数输入这些记录,用print函数输出这些记录。
解:
#defineN5
structstudent
{charnum[6];
charname[8];
intscore[4];
}stu[N];
main()
{intI,j;
for(I=0;I{printf(“\Inputscoreofstudent%d:
\n”,I+1);
printf(“no.:
”);
scanf(“%s”,stu[i].num);
printf(“name:
”);
scanf(“%s”,stu[i].name);
for(j=0;j<3;j++)
{printf(“score%d:
”j+1);
scanf(“%d”,&stu[i].score[j]);
}
printf(“\n”);
}
print(stu);
}
print(structstudentstu[6])
{intI,j;
printf(“%5s%10s”,stu[i].num,stu[i].name);
for(j=0;j<3;j++)
printf(“%9d”,stu[i].score[j]);
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:
#defineN5
structstudent
{charnum[6];
charname[8];
intscore[4]
}stu[N];
input(structstudentstu[])
{intI,j;
for(I=0;I{printf(“inputscoresofstudent%d:
\n”,I+1);
printf(“NO.:
”);
scanf(“%s”,stu[i].num);
printf(“name:
”);
scanf(“%s”,stu[i].name);
for(j=0;j<3;j++)
{printf(“score%d:
”,j++);
scanf(“%d”,&stu[i].score[j]);}
}
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:
#defineN10
structstudent
{charnum[6]
charname[8]
intscore[4]
floatavr;
}stu[N];
main()
{intI,j,max,maxi,sum;
floataverage;
for(I=0;I{printf(“\nInputscoresofstudent%d:
\n”,I+1);
printf(“NO.:
”);
scanf(“%s”,stu[i].num);
printf(“name”);
scanf(“%s”,stu[i].name);
for(j=0;j<3;j++)
{printf(“score%d:
”,j+1);
scanf(“%d”,&stu[i].score[j]);
}
}
average=0;
max=0;
maxi=0;
for(i=0;i<3;i++)
{sum=0;
for(j=0;j<3;j++)
sum+=stu[i].score[j];
stu[i].avr=sum/3.0;
average+=stu[i].avr;
if(sum>max)
{max=sum;
maxi=I;
}
}
average/=N;
printf(“NO.namescore1score2score3average\n”);
for(I=0;I{printf(“%5s%10s”,stu[i].num,stu[i].name);
for(j=0;j<3;j++)
printf(“%9d”,stu[i].score[j]);
printf(“%8.2f\n”,stu[i].avr);
}
printf(“average=%6.2f\n”,average);
printf(“Thehighestscoreis:
%s,scoretotal:
%d.”stu[maxi].name,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:
#defineNULL0
#defineNEWSIZE1000
charnewbuf[NEWSIZE];
char*newp=newbuf;
char*new(intn)
{if(newp+n<=newbuf+NEWSIZE)
{newp=newp+n;
return(newp-n);
}
else
return(NULL);
}
9.7写一函数free,将上题用new函数占用的空间释放。
Free(p)表示将p指向的单元以后的内存段释放。
解:
#defineNullo
#defineNEWSIZE1000
charnewbuf[NEWSIZE];
char*newp=newbuf;
free(char*p)
{if((p>=newbuf)&&(pnewp=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
#defineNULL0
#defineLENsizeof(structstudent)
strutstudent
{longnum;
intscor;
structstudent*next
};
structstudentlistA,listB;
intn,sum=0;
main()
{structstudent*creat(void);
structstudent*insert(structstudent*,structstudent*);
voidprint(structstudent*);
stuctstudent*ahead,*bhead,*abh;
printf(“\ninputlista:
\n”);
ahead=creat();
sum=sum+|n;
abh=insert(ahead,bhead);
print(abh);
}
structstudent*creat(void)
{structstudent*p1,*p2,*head;
n=0;
p1=p2=(structstudent*)malloc(LEN);
printf(“inputnumber&scoresofstudent:
\n”);
printf(“ifnumberIs0,stopinputing.\n”);
scanf(“%ld,%d”,&p1->num,&p1->score);
head=NULL;
while(p1->num!
=0)
{n=n+1;
if(n==1)head=p1;
elsep2->next=p1;
p2=p1;
p1=(structstudent*)malloc(LEN);
scanf(“%ld,,%d”,&p1->num,&p1->score);
}
p2->next=NULL;
return(head);
}
structstudent*insert(structstudent*ah,structstudent*bh)
{structstudent*pa1,*pa2,*pb1,*pb2;
pa2=pa1=ah;
pb2=pb1=bh;
do
{while((pb1->num>pa1->num)&&(pa1->next!
=NULL))
{pa2=pa1;
pa1=pa1->next;
}
if(pb->num<=pa1->num)
{if(ah=pa1)
ah=pb1;
elsepa2->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))
ap1->next=pb1;
return(ah);
}
voidprint(structstudent*head)
{structstudent*p;
printf(“%ld%d\n”,p->num,p->score);
p=p->next;
while(p!
=NULL);
}
9.913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:
#defineN13
structperson
{intnumber;
intnextop;
}link[N+1];
main()
{intI,count,h;
for(I=1;I<=N;I++)
{if(I==N)
link[i].nextp=1;
else
link[i].nextp=I+1;
link[i].number=I;
}
printf(“\n”);
count=0;
h=N;
printf(“sequencethatperson2leavethecircle:
\n”);
while(count{I=0;
while(I!
=3)
{h=link[h].nextp;
if(link[h].number)
I++;
}
printf(“%4d”,link[h].number);
link[h].number=0;
count++;
}
printf(“\nThelastoneis”);
for(I=1;ii<=N;I++)
if(link[i].number)
printf(“%3d”,lin[i].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:
#defineLA4
#defineLB5
#defineNULL0
structstudent
{charnump[6];
charname[8];
structstudent*next;
}A[LA],b[LB];
main()
{structstudenta[LA]={{“101”,”Wang”},{“102”,”LI”},{“105”,”zhang”},{“106”,”Wei”}};
structstudentb[LB]={{“103”,”Zhang”},{“104”,”Ma”},{“105”,”Chen”},{“107”,”Guo”},
{“108”,”Lui”}};
intI,j;
structstudent*p,*p1,*p2,*pt,*head1,*head2;
head1=a;
head2=b;
printf(“lista:
\n”);
for(p1=head1,i=1;p1{p=p1;
p1->next=a+I;
printf(“%8s%8s\n”,p1->num,p1->name);
p1=p1->next;
}
p->next=NULL;
printf(“\nlistb:
\n”);
for(p2=head2,I=1;p2
{p=p2;
p2->next=b+I;
printf(“%8s%8s\n”,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->num,p2->num==0))
if(p1==head1)
head1=p1->next;
else
p->next=p1->next;
p=p1;
p1=p1->next;
}
p1=hedad1;
printf{“\nresult:
\n”};
while(p1!
=NULL)
{printf(“%7s%7s\n”,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:
#defineNULL0
#defineLENsizeof(structstudent)
structstudent
{charnum[6];
charname[8];
charsex[2];
intage;
stuctstudent*next;
}stu[10];
main()
{structstudent*p,*pt,*head;
intI,length,iage,flag=1;
intfind=0;
while(flag==1)
{printf(“inputlengthoflist(<10):
”);
scanf(“%d”,&length);
if(length<10)
flag=0;
}
for(I=0;I{p=(structstudent*)malloc(LEN);
if(I==0)
head=pt=p;
else
pt->next=p;
pt=p;
ptintf(“NO:
”);
scanf(“%s”,p->num);
prntf(“name:
”);
scanf(“%s”,p->name);
printf(“sex:
”);
scanf(“%s”,p->sex);
printf(“age:
”);
scanf(“%s”,p->age);
}
p->next=NULL;
p=head;
printf(“\nNO.namesexage\n”);
while(p!
=NULL)
{printf(“%4s%8s%6s%6d\n”,p->num,p->name,p->sex,p->age);
p=p->next;
}
printf(“Inputage:
”);
scanf(“%d”,&iage);
pt=head;
p=pt;
if(pt->age==iage)
{p=pt->next;
head=pt=p;
find=1;
}
else
pt=pt->next;
while(pt!
=NULL)
{if(pt->age==iage)
{p->next=pt->next;
find=1;
}
elsep=pt;
pt=pt->next;
}
if(!
find)
printf(“Notfound%d.”,iage);
p=head;
printf(“\nNO.namesexage\n”);
while(p!
=NULL)
{
printf(“%4s%8s”,p->num,p->name);
printf(“%6s%6d”,p->sex,p->age);
p=p->next;
}
}
9.12将一个链表按逆序排列,即将链头当链尾,链尾当链头。
解:
#defineNULL0
structstu
{intnum;
structstu*next;
}
main()
{intlen=1l
structstu*p1,*p2,*head,*new,*newhead;
p1=p2=head=(structstu*)malloc(sizeof(strctstu));
printf(“inputnumber(0:
listend):
”);
scanf(“%d”,&p1->num);
while(p1->num!
=o)
{p1=(structstu*)malloc(sizeof(structstu));
printf(“inputnumber(n:
listend):
”);
scanf(“%d”,&p1->num);
if(p1->num==0)
p2->next=null;
else
{p2=>next=p1;
p2=p1;
len++;
}
}
p1=head;
pritnf(“\ntheoriginallist:
\n”);
do
{printf(“%4d”,p1->num);
if(p1->next!
=NULL)
p1=p1->next;
}
while(p1->next!
=NULL)
{p2=p1;
p1=p1->next;
}
if(I==0)
newhead=new=p1;
else
new=nes->next=p1;
p2->next=NULL;
}
printf(\n\nThenewlistL\n);
p1=newhead;
for(I=0l;I{pritf(“4d,p1->num”);
p1=p1->next,
}
printf(“\n”);
}