PAT_1009

1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include <vector>#include <map>#include <iostream>#include <iomanip>using namespace std;int main(){int n;cin>>n;vector<pair<int,double> > cot1;vector<pair<int,double> > cot2;for(int i=1;i<=n;++i){int exp;double coe;cin>>exp>>coe;cot1.push_back(make_pair(exp,coe));}cin>>n;for(int i=1;i<=n;++i){int exp;double coe;cin>>exp>>coe;cot2.push_back(make_pair(exp,coe));}map<int, double> ret;for(vector<pair<int,double> >::size_type i = 0; i < cot1.size(); ++i){double coe1 = cot1[i].second;int exp1 = cot1[i].first;for(vector<pair<int,double> >::size_type j = 0; j < cot2.size(); ++j){int exp2 = cot2[j].first;double coe2 = cot2[j].second;int exp = exp1 +exp2;double coe = coe1 * coe2;if(coe != 0){if(ret.find(exp) == ret.end()){ret[exp] = coe;}else{ret[exp] += coe;}}}}map<int, double>::iterator it = ret.begin();while(it != ret.end()){if((*it).second == 0)ret.erase(it++);else++it;}cout<<fixed<<setprecision(1);cout<<ret.size();map<int, double>::reverse_iterator o = ret.rbegin();while(o != ret.rend()){cout<<" "<<(*o).first<<" "<<(*o).second;++o;}}