多项式乘法细节+数组越界+输出细节 1009 Product of Polynomials (25分)

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
​​

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
​​ <⋯ ​2
​​ ​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

解题
给定两个多项式，求其相乘结果；
逐项相乘——指数相加，系数相乘；
注意点
结果数组要开两倍的空间，因为相乘后最大指数会变成两倍；（1000+1000=2000）

#include
#include
using namespace std;

struct Num{
	int exp;       //指数相加 
	double coef;   //系数相乘 
}; 

Num Poly1[11];
Num Poly2[11];
int K1,K2;

void input()
{
	cin>>K1;
	for(int i=0;i<K1;i++)
		cin>>Poly1[i].exp>>Poly1[i].coef;
	cin>>K2;
	for(int i=0;i<K2;i++)
		cin>>Poly2[i].exp>>Poly2[i].coef;
}

double result[2002];      //其中数组下标代表指数，数组元素存储系数
//注意结果数组C的大小要大于2000，以为A,B中指数的最大值为1000，相乘后的值不大于2000
void Mult()
{	
	fill(result,result+2002,0);
	for(int i=0;i<K1;i++)
		for(int j=0;j<K2;j++)
			result[Poly1[i].exp+Poly2[j].exp]+=Poly1[i].coef*Poly2[j].coef;
}

void output()
{
	stack <int>res;
	for(int i=0;i<2002;i++)
		if(result[i]!=0)        //可能有小于0的系数 
			res.push(i);

	printf("%d",res.size());
	if(res.empty()) printf("\n");
	while(!res.empty())
		{
			printf(" %d %.1f",res.top(),result[res.top()]);
			res.pop();
		}
}

int main()
{
	input();
	Mult();
	output();
}

总结
结构函数可用两个数组替代，多项式储存方式不唯一；