数据结构课设作业，求助

Question

将下面一元多次多项式程序增加一个赋值计算功能，即计算并输出多项式在指定x处的值。另，增加输入 系数或指数为字母时的报错功能。基础C语言

Answer 1

#include
#include
typedef struct node
{

float coef;

int expn;

struct node * next;
}PLOY;

void start()//用户选择界面 { printf("************************************\n");

printf(" 两个一元多项式的相加/相减/相乘\n");

printf("************************************\n");

printf("请选择操作:\n"); printf("0.退出\n");

printf("1.两个一元多项式相加\n");

printf("2.两个一元多项式相乘\n");

printf("3.两个一元多项式相减\n");
}

void insert(PLOY *head,PLOY *inpt)

{ PLOY *pre,*now;

int signal=0;

pre=head;

if(pre->next==NULL)
{
pre->next=inpt;
}

else
{
now=pre->next;

while(signal==0)

{

if(inpt->expnexpn)

{

if(now->next==NULL)

{

now->next=inpt;

signal=1;

}

else

{

pre=now;

now=pre->next;

}

}

else

if(inpt->expn>now->expn)
{

inpt->next=now;

pre->next=inpt;

signal=1;

}

else

{

now->coef=now->coef+inpt->coef;

signal=1;

free(inpt);

if(now->coef==0)

{

pre->next=now->next;

free(now);

}

}

}

}

}

PLOY *creat(char ch)

{

PLOY *head,*inpt;

float x;

int y;

head=(PLOY *)malloc(sizeof(PLOY));

head->next=NULL;

printf("请输入一元多项式%c:(格式是：系数 指数；以0 0 结束！)\n",ch);

scanf("%f %d",&x,&y);

while(x!=0)

{

inpt=(PLOY *)malloc(sizeof(PLOY));

inpt->coef=x;

inpt->expn=y;

inpt->next=NULL;

insert(head,inpt);

printf("请输入一元多项式%c的下一项:(以0 0 结束！)\n",ch);

scanf("%f %d",&x,&y); }

return head; }

PLOY *addPLOY(PLOY *head,PLOY *pre)

{

PLOY *inpt;
int flag=0;

while(flag==0)

{

if(pre->next==NULL)

flag=1;

else

{

pre=pre->next;

inpt=(PLOY *)malloc(sizeof(PLOY));

inpt->coef=pre->coef;

inpt->expn=pre->expn;

inpt->next=NULL;

insert(head,inpt);

}

}

return head;
}

PLOY *minusPLOY(PLOY *head,PLOY *pre)

{ PLOY *inpt;

int flag=0;

while(flag==0)

{

if(pre->next==NULL)

flag=1;

else

{

pre=pre->next;

inpt=(PLOY *)malloc(sizeof(PLOY));

inpt->coef=0-pre->coef;

inpt->expn=pre->expn;

inpt->next=NULL;

insert(head,inpt);

}

}

return head;
}

PLOY *byPLOY(PLOY *head1,PLOY *head2)

{

PLOY *inpt,*res,*pre;

int flag=0;

res=(PLOY *)malloc(sizeof(PLOY));

res->next=NULL;

head1=head1->next;

pre=head2;
while(flag==0)

{

if(pre->next==NULL)

{

pre=head2;

head1=head1->next;

continue;

}

if(head1==NULL)

{

flag=1;

continue;

}

pre=pre->next;

inpt=(PLOY *)malloc(sizeof(PLOY));

inpt->coef=pre->coef*head1->coef;

inpt->expn=pre->expn+head1->expn;

inpt->next=NULL;

insert(res,inpt);

}

return res;
}

void print(PLOY *fun)

{ PLOY *printing;

int flag=0;

printing=fun->next;

if(fun->next==NULL)

{

printf("0\n");

return;

}

while(flag==0)

{

if(printing->coef>0&&fun->next!=printing)

printf("+");

if(printing->coef==1);

else if(printing->coef==-1)

printf("-");

else

printf("%f",printing->coef);

if(printing->expn!=0)
printf("x^%d",printing->expn);

else if((printing->coef==1)||(printing->coef==-1))

printf("1");
if(printing->next==NULL)

flag=1;

else

printing=printing->next;

}

printf("\n");
}

void main()
{

PLOY *f,*g;

int sign=-1;

start();

while(sign!=0)

{

scanf("%d",&sign);

switch(sign)

{

case 0:

break;

case 1:

{

printf("你选择的操作是多项式相加:\n");

f=creat('f');

printf("f(x)=");

print(f);

g=creat('g');

printf("g(x)=");

print(g);

printf("F(x)=f(x)+g(x)=");

f=addPLOY(f,g);

print(f);

sign=-1;

start();

break;

}

case 2:

{

printf("你选择的操作是多项式相乘:\n");

f=creat('f');

printf("f(x)=");

print(f);

g=creat('g');

printf("g(x)=");

print(g);
printf("F(x)=f(x)*g(x)=");

f=byPLOY(f,g);

print(f);

sign=-1;

start();

break;

}

case 3:

{

printf("你选择的操作是多项式相减:\n");

f=creat('f');

printf("f(x)=");

print(f);

g=creat('g');

printf("g(x)=");

print(g);

printf("F(x)=f(x)-g(x)=");

f=minusPLOY(f,g);

print(f);

sign=-1;

start();

break;

}

case 4:

{

sign=-1;

start();

break;

}

default:

{

printf("输入有误!请重新选择操作!\n");

start();

break;

}

}

}

printf("谢谢使用！\n");
}