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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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。

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <stack>
 #include <iostream>
 #include <cmath>
 using namespace std;
 #define exp 1e-9
 struct node{
     int e;
     double c;
     node *next;
 };
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     node *a,*b,*head;
     head=new node();
     head->next=NULL;
     int na;
     scanf("%d",&na);
     int i;
     a=new node[na];
     for(i=;i<na;i++){
         scanf("%d %lf",&a[i].e,&a[i].c);
     }
     int nb;
     scanf("%d",&nb);
     b=new node[nb];
     for(i=;i<nb;i++){
         scanf("%d %lf",&b[i].e,&b[i].c);
     }
     int j,num=;
     for(i=;i<na;i++){
         for(j=;j<nb;j++){
             int e=a[i].e+b[j].e;
             double c=a[i].c*b[j].c;
             node *q=head,*t=head->next;
             while(t){
                 if(t->e<e){//q->e>=e
                     break;
                 }
                 q=t;
                 t=q->next;
             }
             if(q!=head&&q->e==e){
                 q->c+=c;
             }
             else{
                 t=new node();
                 t->c=c;
                 t->e=e;
                 t->next=q->next;
                 q->next=t;
             }
         }
     }
     node *p;
     p=head->next;
     while(p){//系数可能为0！！  这里注意!!
         if(abs(p->c)>exp)
         num++;
         p=p->next;
     }
     p=head->next;
     printf("%d",num);
     while(p){
         if(abs(p->c)>exp)//系数可能为0！！
         printf(" %d %.1lf",p->e,p->c);
         head->next=p->next;
         delete p;
         p=head->next;
     }
     printf("\n");
     delete []a;
     delete []b;
     delete head;
     return ;
 }