PAT Advanced 1009 Product of Polynomials (25分)
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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 N?1?? a?N?1???? N?2?? a?N?2???? ... N?K?? a?N?K????
where K is the number of nonzero terms in the polynomial, N?i?? and a?N?i???? (,) are the exponents and coefficients, respectively. It is given that 1, 0.
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
多项式相乘,我们需要注意去使用map进行存储,去除结果为0的,然后输出即可
#include <iostream> #include <vector> #include <map> using namespace std; struct poly{ double coeff; int expo; }; int main() { int M;poly tmp; vector<poly> poly1,poly2; /**input*/ cin>>M; while(M--){ cin>>tmp.expo>>tmp.coeff; poly1.push_back(tmp); } cin>>M; while(M--){ cin>>tmp.expo>>tmp.coeff; poly2.push_back(tmp); } map<int,double,greater<int>> m; /**计算*/ for(int i=0;i<poly1.size();i++) for(int j=0;j<poly2.size();j++) m[poly1[i].expo+poly2[j].expo]+=poly1[i].coeff*poly2[j].coeff; /**去0*/ for(auto it=m.begin();it!=m.end();it++) if(it->second==0) m.erase(it); /**输出*/ cout<<m.size(); for(auto it=m.begin();it!=m.end();it++) printf(" %d %.1f",it->first,it->second); system("pause"); return 0; }
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