[C/C++] 1009 Product of Polynomials (25 分)

最新推荐文章于 2024-03-14 17:55:08 发布

原创最新推荐文章于 2024-03-14 17:55:08 发布 · 209 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C #C++

PAT-Advanced 专栏收录该内容

21 篇文章

订阅专栏

本文详细介绍了一种解决多项式乘法问题的高效算法。通过使用数组存储多项式的系数和指数，文章展示了如何通过两层循环实现两个多项式的相乘，并提供了两种不同的实现方式：一种是先存储两个多项式再进行处理，另一种是在读取第二个多项式的同时直接进行处理。代码实例清晰，便于理解和实践。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 N1 a_N1 N2 a_N2… NK a_NK
where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N_K<⋯<N₂ <N₁ ≤1000.

Output Specification:
For each test case you should output the sum 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 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

1.建立两个数组，分别存储A和B；建立一个数组存结果(注意大小是A+B的大小）；系数相乘，次数相加

#include<stdio.h> 

int main()
{   
    const int NUM = 1010;
	double a[NUM] = {0};
	double b[NUM] = {0};
	double pdt[2*NUM] = {0}; 
	int k;
	int ept;
	double cft;
	int cnt = 0;
	scanf("%d",&k);
	for(int i=0; i<k; i++){
		scanf("%d%lf",&ept,&cft);
		a[ept] = cft;
	}
	scanf("%d",&k);
	for(int i=0; i<k; i++){
		scanf("%d%lf",&ept,&cft);
		b[ept] = cft;
	}
	
	for(int i=0; i<NUM; i++){
		for(int j=0; j<NUM; j++){
			if( a[i]!=0.0 && b[j]!=0.0 ){
				pdt[i+j] += a[i]*b[j];
			}
		}
	}
	
	for( int i=0; i<2*NUM; i++){
		if( pdt[i]!= 0) cnt++;
	}
	printf("%d",cnt);
	for(int i=2*NUM-1; i>=0; i--){
		if( pdt[i]!=0 ){
			printf(" %d %.1f",i,pdt[i]);
		}
	}

    return 0;
 }

2.没有必要开两个数组存放两个多项式再处理，只需要第二个多项式的系数边读边处理
替换代码如下:

scanf("%d",&k);
	for(int i=0; i<k; i++){
		scanf("%d%lf",&ept,&cft);
		for(int j=0; j<NUM; j++){
			if( a[j]!=0.0 && cft!=0.0 ){
				pdt[j+ept] += a[j]*cft;
			}
		}
	}