PAT A 1009. Product of Polynomials (25)

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.

2 1 2.4 0 3.22 2 1.5 1 0.5Sample 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