PAT甲级1009

1009. Product of Polynomials (25)


时间限制
400 ms
内存限制
65536 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 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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

#include<iostream>#include<stdio.h>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;int main(){int K; pair<int, double> p;vector<pair<int, double> > v1,v2,v,vss;cin >> K;set<int> s; vector<int> vs;for (int i = 0; i < K; i++){cin >> p.first >> p.second;v1.push_back(p);}//把指数和系数当成一对加入到容器中，下同cin >> K;for (int i = 0; i < K; i++){cin >> p.first >> p.second;v2.push_back(p);}for (int k = 0; k < v1.size(); k++){for (int m = 0; m < v2.size(); m++){p.first = v1[k].first + v2[m].first;p.second = v1[k].second *v2[m].second;if (p.second){v.push_back(p);s.insert(p.first);}}}for (set<int>::iterator it = s.begin(); it != s.end(); it++){vs.push_back(*(it));}reverse(vs.begin(),vs.end());double sum = 0;for (int j = 0; j < vs.size(); j++){sum = 0;for (int i = 0; i < v.size(); i++){if (v[i].first == vs[j]){p.first = vs[j];sum += v[i].second;}}p.second = sum;if(sum)//我被坑在这里了，要注意到合并同类项时可能系数为0vss.push_back(p);}cout << vss.size();for (int k = 0; k < vss.size(); k++){//cout << v[k].first << " " << v[k].second;printf(" %d %.1f", vss[k].first, vss[k].second);//控制格式}v.clear(); v1.clear(); v2.clear(); vss.clear(); s.clear(); vs.clear();}