1009 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 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

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

一，注意點

1，開的陣列長度是max_n，應當從max_n-1開始遍歷到下標為0的，不能從max_n開始。

二，程式碼

#include<cstdio>
using namespace std;
const int max_n =2020;
double arr[max_n]={0};
struct Polynomial{
	int ex;
	double coe;
}pol[13];
int main(){
	int num_1=0;
	int num_2=0;
	int K=0;
	double Nk=0;
	scanf("%d",&num_1);
	for(int i=0;i<num_1;i++){
		scanf("%d %lf",&pol[i].ex,&pol[i].coe);
	}
	scanf("%d",&num_2);
	for(int i=0;i<num_2;i++){
		int a=0;
		double b=0;
		scanf("%d %lf",&K,&Nk);
		for(int j=0;j<num_1;j++){
			a=K+pol[j].ex;
			b=Nk*pol[j].coe;
			arr[a]+=b;
		}
	}
	int ans=0;
        /************************************
         ************************************
        從max_n-1開始遍歷*********************/
	for(int i=max_n-1;i>=0;i--){
		if(arr[i]!=0){
			ans++;
		}
	}
	printf("%d",ans);
	for(int i=max_n-1;i>=0;i--){
		if(arr[i]!=0){
			printf(" %d %.1f",i,arr[i]);
		}
	}
	return 0;
}