PAT 1009 Product of Polynomials (25)

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<iostream>#include<iomanip>#include<math.h> using namespace std;struct Polynomials {int exponents;double coefficients;};int main(){int i,j,n,m,exponents,cnt=0;double coefficients;Polynomials a[11],b[11],c[2001];cin>>n;for(i=0;i<n;i++){cin>>a[i].exponents>>a[i].coefficients;}cin>>m;for(i=0;i<m;i++){cin>>b[i].exponents>>b[i].coefficients;}for(i=0;i<=2000;i++){c[i].coefficients=0;c[i].exponents=i;}for(i=0;i<n;i++){for(j=0;j<m;j++){//cout<<a[i].exponents<<" "<<a[i].coefficients<<endl;//cout<<b[j].exponents<<" "<<b[j].coefficients<<endl;exponents=a[i].exponents+b[j].exponents;//指数相加coefficients=a[i].coefficients*b[j].coefficients;//系数相乘c[exponents].coefficients+=coefficients;//cout<<exponents<<" "<<coefficients<<endl;}}for(i=2000;i>=0;i--){if(fabs(c[i].coefficients)!=0){cnt++;}}cout<<cnt;for(i=2000;i>=0;i--){if(fabs(c[i].coefficients)!=0){cout<<" "<<c[i].exponents<<" "<<fixed<<setprecision(1)<<c[i].coefficients;}}    return 0;}