C程序设计2-第五讲-链表课件.ppt

假定一台电脑由显示器、CPU、内存、硬盘和机箱5个部件组成，每个部件包括牌子、型号和价格3个属性，从一个文本文件中读取一台电脑的5个部件的信息，然后算出该电脑的总价格，并把5个部件的属性和总价格输出另一个文件中。;structBuJian

{

charPinPai;

charXingHao;

intJiaGe;

};;main()

{

structBuJianxsq,yp,cpu,nc,jx;

intZongJiaGe;

FILE*fp;

fp=fopen(e:\\a.txt,r);

fscanf(fp,%s%s%d,xsq.PinPai,xsq.XingHao,xsq.JiaGe);

fscanf(fp,%s%s%d,yp.PinPai,yp.XingHao,yp.JiaGe);

fscanf(fp,%s%s%d,cpu.PinPai,cpu.XingHao,cpu.JiaGe);

fscanf(fp,%s%s%d,nc.PinPai,nc.XingHao,nc.JiaGe);

fscanf(fp,%s%s%d,jx.PinPai,jx.XingHao,jx.JiaGe);

fclose(fp);; ZongJiaGe=xsq.JiaGe+yp.JiaGe+cpu.JiaGe+nc.JiaGe+jx.JiaGe;

fp=fopen(e:\\b.txt,w);

fprint(fp,%s%s%d\n,xsq.PinPai,xsq.XingHao,xsq.JiaGe);

fprint(fp,%s%s%d\n,yp.PinPai,yp.XingHao,yp.JiaGe);

fprint(fp,%s%s%d\n,cpu.PinPai,cpu.XingHao,cpu.JiaGe);

fprint(fp,%s%s%d\n,nc.PinPai,nc.XingHao,nc.JiaGe);

fprint(fp,%s%s%d\n,jx.PinPai,jx.XingHao,jx.JiaGe);

fprint(fp,%d\n,ZongJiaGe);

fclose(fp);

}

;第二题;structBuJian

{

charPinPai;

charXingHao;

intJiaGe;

};;structDianNao

{

BuJianxsq;

BuJianyp;

BuJiancpu;

BuJiannc;

BuJianjx;

};;main()

{

structDianNaodn[3];

inti;

FILE*fp;

fp=fopen(e:\\a.txt,r);

for(i=0;i3;i++)

{

fscanf(fp,%s%s%d,dn[i].xsq.PinPai,dn[i].xsq.XingHao,dn[i].xsq.JiaGe);

fscanf(fp,%s%s%d,dn[i].yp.PinPai,dn[i].yp.XingHao,dn[i].yp.JiaGe);

fscanf(fp,%s%s%d,dn[i].cpu.PinPai,dn[i].cpu.XingHao,dn[i].cpu.JiaGe);

fscanf(fp,%s%s%d,dn[i].nc.PinPai,dn[i].nc.XingHao,dn[i].nc.JiaGe);

fscanf(fp,%s%s%d,dn[i].jx.PinPai,dn[i].jx.XingHao,dn[i].jx.JiaGe);

}

fclose(fp);;第五讲链表;7;链表结构的定义;例:建立一个简单链表，它由3个学生数据的节点组成。输出各节点中的数据。;10101;对链表的基本操作;(3)插入：在结点ki-1与ki之间插入一个新的结点k’，使表的长度增1，且逻辑关系发生如下变化：

(4)删除操作：删除结点ki，使链表的长度减1，且ki-1、ki和ki+1结点之间的逻辑关系发生如下变化：;建立链表;structstudent

{intnum;

charname[10];

structstudent*next;

};;head=NULL;structstudent*creat(void)

{

structstudent*head=NULL;

structstudent*p,*last;

p=(structstudent*)malloc(sizeof(structstudent));

last=p;