1009. Product of Polynomials (25)C语言

题意：给定两个多项式，然后求两个多项式相乘的结果。

1思路：用一个数组a[1001]，i下标为多项式的Ni，a[i]为该项的系数，同理设置一个数组p[2001]来储存结果，乘法操作过程中，先用一个数组n[20]储存一个多项式的Ni，这样便于查找不为0的项数。

2思路：用链表储存，

#include
int main()
{
	int K1,K2,i,j,n[20],M=0;
	double a[1001]={0},p[2001]={0};
	scanf("%d",&K1);
	for(i=0;i0;i--)printf("%d %.1lf ",n[i],p[n[i]]);
	    printf("%d %.1lf",n[0],p[n[0]]);
	}
	return 0;
 }

链表储存：第0个测试点未通过？？？？？

#include
#include
typedef struct polynomial{
	int n;
	double a;
	struct polynomial *next;
	struct polynomial *pre;
}polynomial;
typedef polynomial* list;
int main()
{
	int K,i;
	list head1,p;
	head1=(list)malloc(sizeof(polynomial));
	head1->next=NULL;
	head1->pre=NULL;
	p=head1;
	scanf("%d",&K);
	for(i=0;in,&q->a);
		q->next=NULL;
		q->pre=p;
		p->next=q;
		p=q;
	}
	list head=(list)malloc(sizeof(polynomial));
	list last=(list)malloc(sizeof(polynomial));
	last->next=NULL;
	last->pre=head;
	head->next=last;
	head->pre=NULL;
	scanf("%d",&K);
	int M=0;
	for(i=0;inext;
		while(q){
			int nn;
			double aa;
			nn=n+q->n;
			aa=a*q->a;
			if(aa!=0){
				p=head->next;
				while(p->next&&p->n>nn)p=p->next;
				if(p->n==nn)
				{
				    p->a+=aa;
				    if(p->a==0){
					    p->pre->next=p->next;
					    p->next->pre=p->pre;
					    M--;
					}
				}
				else{
					list m=(list)malloc(sizeof(polynomial));
					m->n=nn;
					m->a=aa;
					m->pre=p->pre;
					m->next=p;
					p->pre->next=m;
					p->pre=m;
					M++;
				}
			}
			q=q->next;
		}
	}
	p=head->next;
	printf("%d",M);
	if(M){
	printf(" ");
	while(p->next){
		printf("%d %.1lf",p->n,p->a);
		p=p->next;
		if(p->next)printf(" "); 
	}
    }
	return 0;
}