pat 1009. Product of Polynomials

最新推荐文章于 2022-02-20 15:19:12 发布

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最新推荐文章于 2022-02-20 15:19:12 发布

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分类专栏： pat记录文章标签： pat 1009

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本文介绍了一种计算两个多项式相乘的算法，并通过C语言实现。输入为两个多项式的系数和指数，输出为相乘后的多项式形式。算法遍历每个多项式的项并计算乘积。

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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

#include <stdio.h>
#include <stdlib.h>

int main()
{
  //freopen("F:/CppProject/data/1009.txt","r",stdin);
  int k,exp;
  double a[1002],result[2002]={0.0},coe;
  scanf("%d",&k);
  int first_exps[1000];//存储第一行中所有的系统，用于乘法时循环。
  for(int i =0;i<k;i++)
  {
    scanf("%d%lf",&exp,&coe);
    a[exp]=coe;
    first_exps[i]=exp;
  }
  scanf("%d",&k);
  int tmp=0;
  double mul=0;
  for(int i=0;i<k;i++)
  {
    scanf("%d%lf",&exp,&coe);
    for(int j=0;j<k;j++)
    {
      tmp=exp+first_exps[j];
      mul=coe*a[first_exps[j]];
      result[tmp]=result[tmp]+mul;
    }
  }
  int count=0;
  for(int i=2001;i>=0;i--)
  {
    if(result[i]!=0.0)
      count++;
  }
  printf("%d",count);
  for(int i=2001;i>=0;i--)
  {
    if(result[i]!=0.0)
      printf(" %d %.1lf",i,result[i]);
  }
  
  return 0;
}