1009 Product of Polynomials（25 分）(PAT甲级）

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​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N​K​​<⋯<N​2​​<N​1​​≤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

多项式的乘法。

#include <iostream>
using namespace std;
int main() {
	int n1, n2, a, cnt = 0;
	scanf("%d", &n1);
	double b, arr[1001] = {0.0}, ans[2001] = {0.0};
	for(int i = 0; i < n1; i++) {
		scanf("%d %lf", &a, &b);
		arr[a] = b;
	}
	scanf("%d", &n2);
	for(int i = 0; i < n2; i++) {
		scanf("%d %lf", &a, &b);
		for(int j = 0; j < 1001; j++)
			ans[j + a] += arr[j] * b;
	}
	for(int i = 2000; i >= 0; i--) {
		if(ans[i] != 0.0) {
			cnt++;
		}
	}
	printf("%d", cnt);
	for(int i = 2000; i >= 0; i--) {
		if(ans[i] != 0.0) {
			printf(" %d %.1f", i, ans[i]);
		}
	}
	return 0;
}