1009 Product of Polynomials

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本文介绍了一种使用链表存储多项式并计算两个多项式乘积的方法，避免了数组存储导致的空间浪费，适用于多项式稀疏的情况。通过实例演示了如何构建链表和进行乘法运算。

This time, you are supposed to find $A \times B$ 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:

$K, N_1, a_{N_1}, N_2a_{N_2}, ... , N_Ka_{N_K}$

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 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

题解：

对于多项式乘积问题，如果采用数组存储则可能导致过于稀疏，对空间和时间都是极大的浪费，所以此处采用链表方式来存储 $A$ ，对于乘积结果 $m u t i l$ 可直接用数组存储，方便同级项相加；
需要注意的是，对于数组 $m u t i l$ ，其下标上限应为 $A, B$ 两多项式乘积中的最大级数，此题中为 2001；

注意：

请熟练掌握链表的初始化与构建；

#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;

typedef struct PolyNode *Polynomial;
struct PolyNode {
	int exp;
	double coef;
	Polynomial next;
};

int main()
{
    Polynomial A, Next, Temp;
	A = (Polynomial)malloc(sizeof(struct PolyNode));
	A->next = NULL;
	Next = A;
	double mutil[2001] = {0};

	int n;
	int exp; double coef;
	// 读入 A
	cin >> n;
	while (n--) {
		Temp = (Polynomial)malloc(sizeof(struct PolyNode));
		cin >> exp >> coef;
		Temp->exp = exp;
		Temp->coef = coef;
		Temp->next = NULL;
		Next->next = Temp;
		Next = Temp;		//带有头节点
	}
	// 释放空节点
	Temp = A;
	A = A->next;
	free(Temp);

	// 读入 B，同时进行相乘相加
    cin >> n;
	while (n--) {
		cin >> exp >> coef;
		Temp = A;
		while(Temp){
            mutil[Temp->exp + exp] += Temp->coef * coef;
            Temp = Temp->next;
		}
	}
	int cont = 0;
	for(int i=0;i<2001;i++){
        if(mutil[i]!=0){
            cont++;
        }
	}
	cout<<cont;
	for(int i=2000;i>=0;i--){
        if(mutil[i]!=0){
            cout<<" "<<i<<" "<<fixed<<setprecision(1)<<mutil[i];
        }
	}
}