PAT Advanced 1009 Product of Polynomials (25 分)(vector删除元素用的是erase)
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 (,) 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
#include <iostream> #include <vector> #include <algorithm> using namespace std; struct poly{ int expo; double coef; }; void add(vector<poly>& vec,poly po){ for(int i=0;i<vec.size();i++){ if(vec[i].expo==po.expo){ vec[i].coef+=po.coef; if(vec[i].coef==0){ vec.erase(vec.begin()+i); /**这边需要判断0多项式,进行erase掉*/ } return; } } if(po.coef!=0) vec.push_back(po); } bool cmp(poly p,poly p2){ return p.expo>p2.expo; } int main(){ /** * 注意点 * 1.保留一位小数 */ int M,N;vector<poly> res; cin>>M; poly p[M]; for(int i=0;i<M;i++){ cin>>p[i].expo>>p[i].coef; } cin>>N; poly p2[N]; for(int i=0;i<N;i++){ cin>>p2[i].expo>>p2[i].coef; } for(int i=0;i<M;i++){ for(int j=0;j<N;j++){ poly temp; temp.expo=p[i].expo+p2[j].expo; temp.coef=p[i].coef*p2[j].coef; add(res,temp); } } /**这边需要sort一下*/ sort(res.begin(),res.end(),cmp); cout<<res.size(); for(int i=0;i<res.size();i++){ printf(" %d %.1f",res[i].expo,res[i].coef); } system("pause"); return 0; }
需要记住vector进行删除元素,用的是erase(iter*)