Question

顺序存储结构的实现：例如,已知发f（x） = 8x^6 + 5x^5 - 10x^4 + 32x^2 - x + 10,

Answer 1

#include

#include

#include

#define N sizeof(struct duoxs)

struct duoxs

{

int c;

int e;

struct duoxs *nest;

}*head1,*head2;

int main ( )

{

struct duoxs* get(int n);

void print(struct duoxs *L);

struct duoxs* jiangxu(struct duoxs *L);

struct duoxs *add(struct duoxs *L1,struct duoxs *L2);

struct duoxs *he;

int n1,n2;

printf("输入两个多项式的项数:n");

scanf("%d%d",&n1,&n2);

printf("输入第一个多项式的系数与指数:n");

head1=get(n1);

head1=jiangxu(head1);

printf("输入第二个多项式的项数与指数:n");

head2=get(n2);

head2=jiangxu(head2);

he=add(head1,head2);

printf("两多项式之和为：n");

print(he);

system("pause");

return 0;

}

struct duoxs *get(int n)//输入多项式

{ struct duoxs *p1,*p2,*L;

int i;

for(i=1;ic,&p1->e);

p1->nest=NULL;

}

else

{

p2=(struct duoxs*)malloc(N);

scanf("%d%d",&p2->c,&p2->e);

p1->nest=p2;

p1=p2;

}

}

p1->nest=NULL;

return (L);

}

void print(struct duoxs *L) //输出多项式

{

struct duoxs *p;

int i=0;

p=L;

while(p!=NULL)

{

i++;

if(i==1||p->cc,p->e);

if(i!=1&&p->c>0)

printf("+%dX^%d",p->c,p->e);

p=p->nest;

}

printf("n");

}

struct duoxs* jiangxu(struct duoxs*L)//指数降序排列

{

int i;

struct duoxs *p1,*p2;

p1=L;

while(p1->nest!=NULL)

{

p2=p1->nest;

while(p2!=NULL)

{

if(p1->ee)

{

i=p1->e;

p1->e=p2->e;

p2->e=i;

i=p1->c;

p1->c=p2->c;

p2->c=i;

}

p2=p2->nest;

}

p1=p1->nest;

}

return (L);

}

struct duoxs*add(struct duoxs*L1,struct duoxs*L2)

//多项式相加

{

struct duoxs *L,*p,*p1,*p2,*q;

p1=L1;

p2=L2;

int i=0;

while(p1!=NULL&&p2!=NULL)

{

if(p1->e==p2->e)

{

i++;

if(i==1)L=p=p1;

else{p->nest=p1;p=p->nest;}

p->c+=p2->c;

p1=p1->nest;

q=p2;

p2=p2->nest;

free(q);

}

else if((p1->e)>(p2->e))

{

i++;

if(i==1)L=p=p1;

else{p->nest=p1;p=p->nest;}

p1=p1->nest;

}

else

{

i++;

if(i==1)L=p=p2;

else{p->nest=p2;p=p->nest;}

p2=p2->nest;

}

}

if(p1!=NULL)

p->nest=p1;

if(p2!=NULL)

p->nest=p2;

return (L);

}