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=0;i<na;i++){
        scanf("%d %lf",&a[i].e,&a[i].c);
    }
    int nb;
    scanf("%d",&nb);
    b=new node[nb];
    for(i=0;i<nb;i++){
        scanf("%d %lf",&b[i].e,&b[i].c);
    }
    int j,num=0;
    for(i=0;i<na;i++){
        for(j=0;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("
");
    delete []a;
    delete []b;
    delete head;
    return 0;
}