PAT A 1009. Product of Polynomials (25)

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 A map基本操作文章标签： PAT

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本文介绍了一种使用C++实现的高效算法来计算两个多项式的乘积。通过使用map结构来存储多项式的指数和系数，该算法能自动按指数排序并输出结果。此算法适用于处理多项式中稀疏的非零项。

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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

// 题目分析：

采用map处理多项式，用键值对存储指数和系数，由于map会按照键值的自然顺序自动排序（底层红黑树），最终逆序输出即可；

练习了map的判断、添加、删除、遍历等操作；

效率很高；

// 代码

#include <iostream>
#include <vector>
#include <map>
#include <cmath>
using namespace std;
struct Node{
  int exp;
  double coeff;
};

int main(){
  vector<Node> v1;
  vector<Node> v2;
  int K1,K2;
  Node node;
  cin>>K1;
  for(int i=0;i<K1;i++){
    scanf("%d%lf",&node.exp,&node.coeff);
    v1.push_back(node);
  }
  cin>>K2;
  for(int i=0;i<K2;i++){
    scanf("%d%lf",&node.exp,&node.coeff);
    v2.push_back(node);
  }
  map<int,double> m;
  for(int i=0;i<K1;i++){
    for(int j=0;j<K2;j++){
      int exp=v1[i].exp+v2[j].exp;
      double coeff=v1[i].coeff*v2[j].coeff;
      if(m.count(exp)){
        m[exp]=m[exp]+coeff;
      }else{
        m[exp]=coeff;
      }
      if(abs(m[exp]-0)<0.00000001){
        m.erase(exp);
      }
    }
  }
  cout<<m.size();
  map<int,double>::iterator it=m.end();
  it--;
  if(m.size()!=0){
    while(1){
      printf(" %d %.1f",it->first,it->second);
      if(it==m.begin())
        break;
      it--;
    }
  }
  system("pause");
  return 0;
}
// 结果