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 … N_K _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:

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

Meaning of the questions:

Calculating two polynomials a, b multiplied result;

Input:

a non-zero entries polynomial coefficient indices ... ...
B polynomial coefficients nonzero entry indices ... ...

Output:

polynomial c (a * b) non-zero entries coefficient indices ... ...

Ideas:

(1) takes three floating-point arrays were stored polynomials a, b, c, subscript index store, a corresponding coefficient value;
(2) evaluation polynomial c attention accumulated during the non-zero coefficient.

Code:

#include <cstdio>
const int max_=1010;
const int maxn=2010;//两个多项式相乘的指数最大为2000
int main(){
	int k,x,count=0;
	double y,a[max_]={0},b[max_]={0},c[maxn]={0}; 
	scanf("%d",&k);
	for(int i=0;i<k;i++){
		scanf("%d %lf",&x,&y);
		a[x]=y;//x存放指数，a[x]存放对应的系数 
	}
	scanf("%d",&k);
	for(int i=0;i<k;i++){
		scanf("%d %lf",&x,&y);
		b[x]=y;
		for(int j=0;j<max_;j++){
			if(a[j]!=0){
				//if(c[x+j]==0)count++;//计数非零项(用此方法计数非零项，oj的第一个测试点通不过，因为系数相加为0的情况下也进行计数会出错)
				c[x+j]+=a[j]*b[x];//x+j存放相乘后多项式的指数，c[x+j]存放对应的系数	
			}
		}
	} 
	for(int i=0;i<maxn;i++){
		if(c[i]!=0)
			count++;
	}
	printf("%d",count); 
	for(int i=maxn-1;i>=0;i--){
		if(c[i]!=0){
			printf(" %d %.1f",i,c[i]);
		} 
	}
	return 0;
}