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 思路：与多项式加法类似，乘法是遍历多项式每一非空项，与另外的多项式每一项相加，数值相乘。使用数值可以实现。
#include <iostream>#include <iomanip>using namespace std;const int size = 1004; //数组元素量int main(){int na, nb;float a[size]= {0}, b[size] = {0}, c[size*2+1]={0};int count = 0, n;cin >> na;         //输入多项式A,Bfor (int i=0; i!=na; ++i) {cin >> n;cin >> a[n];}cin >> nb;for (int i=0; i!=nb; ++i) {cin >> n; cin >> b[n];}for (int i=0; i!=size; ++i) for (int j=0; j!=size; ++j) if (a[i] != 0 || b[i] != 0)  //多项式乘法，项相加，数值相乘        c[i+j] += a[i] * b[j];for (int i=2*size; i>=0; --i) //统计多少项非0if (c[i] != 0)            ++count;cout << count;for (int i=2*size; i>=0; --i) //输出格式控制{if (c[i] != 0)cout << " " << i << " " << setiosflags(ios::fixed) << setprecision(1) << c[i]; }cout << endl;//system("pause");return 0;}