pat1009. Product of Polynomials (25)

1009. Product of Polynomials (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

This time, you are supposed to find A*B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.

 

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input 
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output 
3 3 3.6 2 6.0 1 1.6

提交代码

 

当觉得主要的算法没有问题时，花点时间去关注一些关键变量的变化规律，比如这里的num。

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <queue>
 5 #include <stack>
 6 #include <iostream>
 7 #include <cmath>
 8 using namespace std;
 9 #define exp 1e-9
10 struct node{
11     int e;
12     double c;
13     node *next;
14 };
15 int main(){
16     //freopen("D:\\INPUT.txt","r",stdin);
17     node *a,*b,*head;
18     head=new node();
19     head->next=NULL;
20     int na;
21     scanf("%d",&na);
22     int i;
23     a=new node[na];
24     for(i=0;i<na;i++){
25         scanf("%d %lf",&a[i].e,&a[i].c);
26     }
27     int nb;
28     scanf("%d",&nb);
29     b=new node[nb];
30     for(i=0;i<nb;i++){
31         scanf("%d %lf",&b[i].e,&b[i].c);
32     }
33     int j,num=0;
34     for(i=0;i<na;i++){
35         for(j=0;j<nb;j++){
36             int e=a[i].e+b[j].e;
37             double c=a[i].c*b[j].c;
38             node *q=head,*t=head->next;
39             while(t){
40                 if(t->e<e){//q->e>=e
41                     break;
42                 }
43                 q=t;
44                 t=q->next;
45             }
46             if(q!=head&&q->e==e){
47                 q->c+=c;
48             }
49             else{
50                 t=new node();
51                 t->c=c;
52                 t->e=e;
53                 t->next=q->next;
54                 q->next=t;
55             }
56         }
57     }
58     node *p;
59     p=head->next;
60     while(p){//系数可能为0！！  这里注意!!
61         if(abs(p->c)>exp)
62         num++;
63         p=p->next;
64     }
65     p=head->next;
66     printf("%d",num);
67     while(p){
68         if(abs(p->c)>exp)//系数可能为0！！
69         printf(" %d %.1lf",p->e,p->c);
70         head->next=p->next;
71         delete p;
72         p=head->next;
73     }
74     printf("\n");
75     delete []a;
76     delete []b;
77     delete head;
78     return 0;
79 }

 

posted @ 2015-07-22 08:46  Deribs4  阅读(179)  评论(0)  编辑  收藏  举报