PAT：A1009 Product of Polynomials （25 分）

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本文介绍了一个PAT题目的解决方案——计算两个多项式的乘积。输入包含两个多项式的系数和指数，输出它们相乘后的结果。采用数组存储中间结果，并筛选出非零项进行输出。

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PAT：A1009 Product of Polynomials （25 分）

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

代码：

#include<stdio.h>

struct Poly{
	int exp; 
	double cof;  
}ploy[1001];

double con[2001];
int main() {
	int n,m;	
	int num = 0;	
	scanf("%d", &m);
	for(int i = 0; i < m; i++) {
		scanf("%d%lf", &ploy[i].exp, &ploy[i].cof);	 
	} 
	scanf("%d", &n);
	for(int j = 0; j < n; j++) {
		int exp;
		double cof;
		scanf("%d%lf", &exp, &cof);
		for(int k = 0; k < m; k++) {
			con[ploy[k].exp + exp] += (cof * ploy[k].cof);
		}
	}
	// 查询con中不为0的数
	for(int i = 0; i < 2001; i++) {
		if(con[i] != 0.0) num++;
	} 
	printf("%d", num);
	for(int i = 2000; i >= 0; i--) {
		if(con[i] != 0.0) printf(" %d %.1f", i, con[i]);
	}
	return 0;
}