呵呵，碰巧到你百度空间看了下，看到了你想知道的这个问题，蒽蒽，下午就给你答案哈~~~
恩，可以了：
#include
#include
typedef struct term
{
double coef;//系数
int expn; //指数
struct term *next;
}term,*polynomial;
//注:下面的head_p是头指针，而非头结点。其存放的是多项式的第一项
int depart_polyn(polynomial &head_p,polynomial &Odd_items,polynomial &Even_items)//奇数项，偶数项
{
int i=1,j=1;
polynomial odd,even,p;
Odd_items=Even_items=NULL;
p=head_p;
while(1)
{
if(p->expn%2)
{
if(i)
{
i=0;//使得奇数项头指针只赋值一次
Odd_items=p;
odd=Odd_items;
}
else
{
odd->next=p;
odd=odd->next;
}
p=p->next;
odd->next=Odd_items;
}
else
{
if(j)
{
j=0;//作用同上
Even_items=p;
even=Even_items;
}
else
{
even->next=p;
even=even->next;
}
p=p->next;
even->next=Even_items;
}
if(p==head_p)
break;
}
return 1;
}
void print_polyn(polynomial head)//这函数只是针对main函数内给出的多项式例子编的，处理的系数均为正数。
{
polynomial p=head;
if( !head)
{
puts("该表为空:");
return;
}
while(1)
{
if(p->next!=head)
printf("%gm^%d + ",p->coef,p->expn);
else
printf("%gm^%dn",p->coef,p->expn);
p=p->next;
if(p==head)
break;
}
putchar('n');
}
int main()
{
int i;
polynomial head_p,head,Odd_items,Even_items,newbase;
head_p=(polynomial)malloc(sizeof(term));
head_p->coef=2.71;
head_p->expn=0;
head=head_p;
//随意创建了一个稀疏多项式，以验证
for(i=1;icoef=2.71*i;
newbase->expn=i*13;
head->next=newbase;
head=head->next;
}
head->next=head_p;//构成循环链表
puts("原稀疏多项式:");
print_polyn(head_p);
if(depart_polyn(head_p,Odd_items,Even_items) )
{
puts("分解后的奇数项稀疏多项式:");
print_polyn(Odd_items);
puts("分解后的偶数项稀疏多项式:");
print_polyn(Even_items);
}
return 0;
}
哪里不懂可追问：