1009 Product of Polynomials

最新推荐文章于 2021-02-21 15:20:58 发布

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本文介绍了一种计算两个多项式相乘的算法实现，输入包含两个多项式的系数和指数，输出为乘法结果的多项式形式。算法首先读取两个多项式的参数，然后通过双重循环计算乘积并存储结果。

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 aN1 N2 aN2 ... NK aNK

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

#include<iostream>
#include<cmath>

using namespace std;

float a[1001], b[1001], c[2001];

int main(){
	int n, exponent, cnt = 0;
	float coefficient;
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%d %f", &exponent, &coefficient);
		a[exponent] = coefficient;
	}
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%d %f", &exponent, &coefficient);
		b[exponent] = coefficient;
	}
	for(int i = 0; i < 1001; i++){
		if(abs(a[i]) > 1e-7){
			for(int j = 0; j < 1001; j++){
				if(abs(b[j]) > 1e-7){
					if(abs(c[i + j]) < 1e-7) cnt++;
					c[i + j] += a[i] * b[j];
					if(abs(c[i + j]) < 1e-7) cnt--;
				}
			}
		}
	}
	printf("%d", cnt);
	for(int i = 2000; i >= 0; i--){
		if(abs(c[i]) > 1e-7){
			printf(" %d %.1f", i, c[i]);
		}
	}
	return 0;
}