PAT (Advanced Level) 1009. Product of Polynomials (25) 多项式相乘

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 3.22 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

使用vector<pair<int,double> > 存储两个多项式a和b，使用map<int,double>，以指数为key，底数为value，遍历a和b进行迭代，注意去除底数为0的项。

/*2015.7.19cyq*/#include <iostream>#include <vector>#include <map>using namespace std;int main(){int ka,kb;vector<pair<int,double> > a,b;cin>>ka;int tmp=ka;int c;double d;while(tmp--){cin>>c>>d;a.push_back(make_pair(c,d));//多项式a}cin>>kb;tmp=kb;while(tmp--){cin>>c>>d;b.push_back(make_pair(c,d));//多项式b}map<int,double> map;for(int i=0;i<ka;i++){for(int j=0;j<kb;j++){int exp=a[i].first+b[j].first;//指数相加map[exp]+=a[i].second*b[j].second;//底数相乘if(map[exp]==0)map.erase(exp);}}cout<<map.size();for(auto it=map.rbegin();it!=map.rend();++it){printf(" %d %0.1f",(*it).first,(*it).second);}return 0;}