1009 Product of Polynomials -PAT甲级

最新推荐文章于 2024-12-17 09:10:08 发布

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最新推荐文章于 2024-12-17 09:10:08 发布

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本文介绍了一种计算两个多项式乘积的方法，通过系数相乘和指数相加的方式，实现多项式乘法的高效计算。文章提供了一个示例输入和输出，并给出了一个完整的C++代码实现，展示了如何处理多项式的乘法运算。

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This time, you are supposed to find A×BA\times BA×B where AAA and BBB 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:

KKK N1N_1N1 aN1a_{N_1}aN1 N2N_2N2 aN2a_{N_2}aN2 ... NKN_KNK aNKa_{N_K}aNK

where KKK is the number of nonzero terms in the polynomial, NiN_iNi and aNia_{N_i}aNi (i=1,2,⋯,Ki=1, 2, \cdots , Ki=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤101\le K \le 101≤K≤10, 0≤NK<⋯<N2<N1≤10000 \le N_K < \cdots < N_2 < N_1 \le 10000≤NK<⋯<N2<N1≤1000.

Output Specification:

For each test case you should output the product of AAA and BBB 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

按照题中的意思进行多项式的计算即可，系数相乘，指数相加，指数相加后的结果大于2000，开大于2000的数组即可

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=2005;
int p[2];
struct Node{
	int e=0;
	double c=0;
	Node(){}
	Node(int ee,double cc){
		e=ee;
		c=cc;
	}
}a[N],b[N],res[N];
int main(){
	int e;
	double c;
	scanf("%d",&p[0]);
	for(int i=0;i<p[0];i++){
		cin>>e>>c;
		a[i]={e,c};
	}
	scanf("%d",&p[1]);
	for(int i=0;i<p[1];i++){
		cin>>e>>c;
		b[i]={e,c};
	}
	for(int i=0;i<p[0];i++){
		for(int j=0;j<p[1];j++){
			int ee=a[i].e+b[j].e;
			double cc=a[i].c*b[j].c;
			res[ee].e=ee;
			res[ee].c+=cc;
		}
	}
	int sum=0;
	for(int i=0;i<N;i++){
		if(res[i].c!=0){
			sum++;
		}
	}
	printf("%d",sum);
	for(int i=N;i>=0;i--){
		if(res[i].c!=0)
			printf(" %d %.1f",i,res[i].c);
	}
	printf("\n");
	return 0;
}