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

题目的意思就是给出多项式A,多项式B,求A*B。样例:

A =2.4*x+3.2

B = 1.5*x^2 + 0.5*x

A*B = 3.6*x^3 + 6.0*x^2 + 1.6*x

Solution

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <iomanip>
using namespace std;
double c[5000] = { 0 };
double a[5000] = { 0 };
double b[5000] = { 0 };
int main() {
	//inputA
	int na;
	cin >> na;
	for (int i = 0; i < na; i++) {
		int index;
		cin >> index;
		cin >> a[index];
	}
	//inputB
	int nb;
	cin >> nb;
	for (int i = 0; i < nb; i++) {
		int index;
		cin >> index;
		cin>> b[index];
	}
	//A*B
	int cnt = 0;
	for (int i = 4999; i >=0; i--) {
		for (int k = 4999; k >=0; k--) {
			if (a[i] != 0 && b[k] != 0) {
				c[i + k] += a[i] * b[k];
			}
		}
	}
	for (int i = 0; i < 5000; i++) {
		if (c[i] != 0) {
			cnt++;
		}
	}
	cout << cnt;
	for (int i = 4999; i>=0; i--) {
		if (c[i] != 0) {
			cout << " " << i;
			cout << fixed << setprecision(1) << " " << c[i];
		}
	}
	//system("pause");
	return 0;
}