A1009 Product of Polynomials (25 point(s))

多项式相乘

1. 原文

原题

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 $N_1$ $a_{N1}$ $N_2$ $a_{N2}$ … $N_K$ $a_{NK}$

where K is the number of nonzero terms in the polynomial, $N_i$ and $a_{Ni}$ (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 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

2. 解析

两个多项式相乘，多项式的每一项都需要乘以另一个多项式的每一项，指数相加，系数相乘。

指数相加，系数相乘

ans[a+i]+=b*num[i];

3. AC代码

#include<cstdio>
const int maxn = 1010;
double num[maxn]={};
double ans[2*maxn]={};
int main(){
	int k;
	scanf("%d",&k);
	int a;
	double b;
	for (int i = 0; i < k; ++i)
	{
		scanf("%d%lf",&a,&b);
		num[a]+=b;
	}
	scanf("%d",&k);
	while(k--)
	{
		scanf("%d%lf",&a,&b);
		for (int i = 0; i < maxn; ++i)
		{
			if (num[i]!=0)
			{
				ans[a+i]+=num[i]*b;
			}
		}
	}
	int cnt = 0;
	for (int i = 0; i < 2*maxn; ++i)
	{
		if (ans[i]!=0)
		{
			cnt++;
		}
	}
	printf("%d",cnt);
	for (int i = 2*maxn-1; i >= 0; i--)
	{
		if (ans[i]!=0)
		{
			printf(" %d %.1f",i,ans[i]);
		}
	}
	printf("\n");

	return 0;
}