PAT-A-1009-Product of Polynomials

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

1 2	2 1 2.4 0 3.2 2 2 1.5 1 0.5

Sample Output:

1	3 3 3.6 2 6.0 1 1.6

Code

//思路：f2的每一项与f1的每一项相乘；利用数组的索引值当作项的指数
#include <iostream>

using namespace std;
double f1[1001]={0.0};
double res[2001]={0.0};
int main(){
	
	int n, m;
	//多项式中的项数
	scanf("%d", &n);
	int i, j;
	int exp;
	double coe;
	for (i = 0;i < n;i++) {
		scanf("%d%lf", &exp, &coe);
		f1[exp] = coe;
	}
	//输入多项式二
	scanf("%d", &m);
	for (i = 0;i < m;i++) {
		scanf("%d %lf", &exp, &coe);
		//输入项与f1的每一项相乘
		for (j = 0;j < 1001;j++) {
			//先判断f1中这一项存不存在
			if (f1[j] != 0.0)
				res[exp + j] += f1[j] * coe;
		}
	}

	//统计结果的项数
	int count = 0;
	for (i = 0;i < 2001;i++) {
		if (res[i]!=0.0)
			count++;
	}

	printf("%d", count);

	for (i = 2000;i >= 0;i--) {
		if (res[i]!=0.0) 
			printf(" %d %.1f", i, res[i]);
	}

	return 0;

}