1002. A+B for Polynomials

1002. A+B for Polynomials (25)

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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

For each test case you should output the sum 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 to 1 decimal place.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

---

 #include <iostream>
 #include <cstdio>
 #include <vector>

 using namespace std;
 class node{
 public:
     int expon;//exponents
     double coe;//coefficients
     node(int _expon, double _coe) :expon(_expon), coe(_coe){}
     ~node(){}
 };

 int main(void)
 {
     int N1;
     cin >> N1;
     vector<node> List1;
     for (size_t i = ; i < N1; i++)
     {
         int expon; double coe;
         cin >> expon >> coe;
         List1.push_back(node(expon, coe));
     }
     int N2;
     cin >> N2;
     vector<node> List2;
     for (size_t i = ; i < N2; i++)
     {
         int expon; double coe;
         cin >> expon >> coe;
         List2.push_back(node(expon, coe));
     }

     vector<node>::iterator it1, it2;
     it1 = List1.begin(); it2 = List2.begin();

     vector<node> List3;
     while( (it1 != List1.end()) && (it2  !=List2.end()) )
     {

         if ((*it1).expon == (*it2).expon)
         {

             double coe_sum = (*it1).coe + (*it2).coe;
             if (coe_sum != )
                 List3.push_back(node((*it1).expon, coe_sum));
             it1++; it2++;
         }
         else if ((*it1).expon > (*it2).expon)
         {
             List3.push_back(node((*it1).expon, (*it1).coe));
             it1++;
         }
         else if ((*it1).expon < (*it2).expon)
         {
             List3.push_back(node((*it2).expon, (*it2).coe));
             it2++;

         }

     }
     while (it1 != List1.end()){ List3.push_back(node((*it1).expon, (*it1).coe)); it1++; }
     while (it2 != List2.end()){ List3.push_back(node((*it2).expon, (*it2).coe)); it2++; }

     cout << List3.size();
     for (size_t i = ; i < List3.size(); i++)
     {
         cout << " " << List3[i].expon;
         printf(" %0.1f", List3[i].coe);
     }
     cout << endl;
     return ;
 }

分析： 只需要注意一下输出的格式即好，这里用了C里面的输出函数