Question

一元多项式运算一．问题描述设计一个简单的一元稀疏多项式加法运算器.二．基本要求一元稀疏多项式简单计算器的基本功能包括：1

Answer 1

#include

#include

#define A1(a,b,c,d,e) gotoxy(a,b);printf("c%d=",d);scanf("%f",e);

#define A2(a,b,c,d,e) gotoxy(a,b);printf("e%d=",d);scanf("%f",e);

#define B1 do { printf("n想继续吗（求函数值／求导数）(y/N)?")

#define B2 ch2=bioskey(0); printf("[%c]",ch2)

#define B3 if((ch2=='Y')||(ch2=='y')) tc(head1->next,head2->next,r); }

#define B4 while((ch2=='y')||(ch2=='Y'))

struct ma

{ float c;

float e;

struct ma *next; }

main()

{ struct ma *head1,*head2,*p,*q,*r,*rm;

float s1[100][3],s2[100][3],mid1,mid2,mid3;

int i,j,t,num1,num2;

char ch,ch1,ch2,ch11,ch21;

struct ma *add();

struct ma *mul();

void tc();

void output();

loop1: t=0; ch='y';

clrscr();

printf("请输入第一个多项式的项数:");

scanf("%d",&num1);

q=head1=(struct ma*)malloc(sizeof(struct ma));

(*q).next=NULL;

if(num1>0)

{ printf("请输入第一个多项式的系数(c) 和指数(e):");

for(i=1;ic); (*r).e=q->e; q=(*q).next;}

else if(p->ee) { (*r).c=p->c ; (*r).e=p->e; p=(*p).next; }

else { if(ch1=='+') (*r).c=(p->c)+(q->c);

else (*r).c=(p->c)-(q->c);

(*r).e=(p->e); p=(*p).next;q=(*q).next;}

}while((p!=NULL)&&(q!=NULL));

while(q!=NULL)

{ r=(struct ma*)malloc(sizeof(struct ma));

(*s).next=r; s=r;

if(ch1=='+') (*r).c=q->c ;

else (*r).c=-(q->c);(*r).e=q->e; q=(*q).next;}

while(p!=NULL)

{ r=(struct ma*)malloc(sizeof(struct ma));

(*s).next=r; s=r;

(*r).c=p->c ; (*r).e=p->e;p=(*p).next; }

(*r).next=NULL;

s=r=(*head3).next;

while(r->next!=NULL)

if(r->e==(r->next)->e)

{ r->c+=(r->next)->c; s=r->next;

r->next=(r->next)->next; free(s); s=r; }

else { r=r->next; s=s->next; }

s=r=(*head3).next;

return(r);

}

struct ma *mul(struct ma *head1,struct ma *head2)

{ struct ma *head3,*head4,*p,*q,*r,*s,*r1,*s1;

struct ma *w[100];

int t=0;

head3=r=s=(struct ma*)malloc(sizeof(struct ma));

head4=r1=s1=(struct ma*)malloc(sizeof(struct ma));

p=head1->next; q=head2->next;

while(p!=NULL)

{ s=(struct ma*)malloc(sizeof(struct ma));

s->c=(p->c)*(q->c);

s->e=(p->e)+(q->e);

r->next=s;

r=s;

p=p->next;}

s->next=NULL; s=r=head3->next; w[0]=head3;

p=head1->next; q=q->next;

while(q!=NULL)

{ while(p!=NULL)

{ s1=(struct ma*)malloc(sizeof(struct ma));

s1->c=(p->c)*(q->c);s1->e=(p->e)+(q->e);

r1->next=s1;

r1=s1;

p=p->next;

}

s1->next=NULL;r1=s1=head4->next;

t++;

w[t]=(struct ma*)malloc(sizeof(struct ma));

w[t]->next=add(w[t-1],head4,'+');

while(head3->next!=NULL) {r=r->next;free(s);head3->next=s=r;}

s=r=head3;

while(head4->next!=NULL) {r1=r1->next;free(s1);head4->next=s1=r1;}

s1=r1=head4;

p=head1->next; q=q->next;

}

return(w[t]->next);}

void tc(struct ma *p,struct ma *q, struct ma *r)

{ char ch1,ch2;

float x;

float qzhi(struct ma *head,float x);

void qdao();

printf("n请选择多项式的序号:");

printf("n 1--第一个;2--第二个;3--第三个");

ch1=bioskey(0);printf("[%c]",ch1);

printf("n请输入功能代码:");

printf("n z--求函数值,d--求导函数");

ch2=bioskey(0);printf("[%c]",ch2);

switch(ch1)

{ case '1': { if(ch2=='z')

{ printf("n请输入自变量 x:");

scanf("%f",&x);

printf("多项式的值为:%10.4f",qzhi(p,x));}

if(ch2=='d') qdao(p); break;}

case '2': { if(ch2=='z')

{ printf("n请输入自变量 x:");

scanf("%f",&x);

printf("多项式的值为:%10.4f",qzhi(q,x));}

if(ch2=='d') qdao(q); break;}

case '3': { if(ch2=='z')

{ printf("n请输入自变量 x:");

scanf("%f",&x);

printf("多项式的值为:%10.4f",qzhi(r,x));}

if(ch2=='d') qdao(r); break;} } }

float qzhi(struct ma *head,float x)

{ float value=0;

while(head!=NULL)

{ value+=(head->c)*pow(x,head->e);head=head->next;}

return(value);

}

void qdao(struct ma *head)

{ struct ma *p,*q,*r,*t;

p=head;

t=r=q=(struct ma *)malloc(sizeof(struct ma));

while(p!=NULL)

{ q=(struct ma *)malloc(sizeof(struct ma));

r->next=q;r=q;

q->c=(p->c)*((p->e));q->e=(p->e)-1;

p=p->next;}

q->next=NULL;

printf("n多项式的导数为:");

output(t->next);

p=t;

while(t!=NULL) { p=t->next;free(t);t=p;}

}

void output(struct ma *r)

{ int l=0;

struct ma *t,*s,*head;

t=r;

while(r->next!=NULL)

if(r->e==(r->next)->e)

{ r->c+=(r->next)->c; s=r->next;

r->next=(r->next)->next; free(s); s=r; }

else { r=r->next; s=s->next; }

s=r=t;

head=(struct ma*)malloc(sizeof(struct ma));

head->next=r;

s=head;

while(r!=NULL)

{ if((r->c)==0) {s->next=r->next; free(r);r=s->next;}

else { s=s->next;r=r->next;}}

s=r=head->next;

while(r!=NULL)

{ if(r->c==1) { l++;

if(r->e==1) printf("x");

else if(r->ee);

else if(r->e==0) printf("1");

else printf("x^%.0f",r->e);}

else if(r->c==-1) { l++;

if(r->e==1) printf("-x");

else if(r->ee);

else if(r->e==0) printf("-1");

else printf("-x^%.0f",r->e);}

else { l++;

if(r->e==1) printf("%4.1fx",r->c);

else if(r->ec,r->e);

else if(r->e==0) printf("%4.1f",r->c);

else printf("%4.1fx^%.0f",r->c,r->e); }

if(((*r).next!=NULL)&&((r->next)->c>0)) printf(" +");

r=r->next; }

if(l==0) printf("0");

}