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.

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

题目要求：给出两个多项式的系数和指数 ，求两个多项式的乘积
系数为0的项不输出
这里用map不是很简洁，要遍历两次map才能得到正确答案，用数组似乎简洁一些，但是占用更多的内存

 1 #include<iostream>
 2 #include<map>
 3 using namespace std;
 4 struct node{
 5   int exp;
 6   double coe;
 7 };
 8 map<int, double> mmap;
 9 int main(){
10   node v1[11], v2[11];
11   int n, m, i;
12   cin>>n;
13   for(i=0; i<n; i++)cin>>v1[i].exp>>v1[i].coe;
14   cin>>m;
15   for(i=0; i<m; i++)cin>>v2[i].exp>>v2[i].coe;
16   for(i=0; i<n; i++){
17     for(int j=0; j<m; j++){
18       int exp=v1[i].exp+v2[j].exp;
19       double coe=v1[i].coe*v2[j].coe;
20       if(mmap.count(exp)==0) mmap[exp]=coe;
21       else mmap[exp]+=coe;
22     }
23   }
24   int cnt=0;
25   for(auto it=mmap.begin(); it!=mmap.end(); it++)
26     if(it->second!=0.0) cnt++;
27   cout<<cnt;
28   for(auto it=mmap.rbegin(); it!=mmap.rend(); it++)if(it->second!=0.0)printf(" %d %.1f", it->first, it->second);
29   return 0;
30 }