PTA 甲 1009 Product of Polynomials

1009 Product of Polynomials

题目原题

目录

• 1009 Product of Polynomials
• 题目原题
  • Input Specification:
  • Output Specification:
  • Sample Input:
  • Sample Output:
  • 注意点:
  • 一开始的代码
  • 改进后如下

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 ... N**K aNK

where K is the number of nonzero terms in the polynomial, N**i and aNi (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 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

生词一览:

polynomials 多项式

exponents 指数

coefficients 系数

题目大意,就是两个没有零项式的多项式相乘,输出他们的结果,

注意点:

① 这道题目不需要用两个数组

② 在计算他们的乘积的时候,要先把他们的乘积的数组算出来.

这个非常的重要,Sumc一定要在C[e]都算好了在统计个数
因为C[e]在计算中,可能会出现一正一负的,可以消去的同类项.
会多计算两次C[e] 一般这就是测试点0通不过的原因

③ 最好在输出的时候,是不可以输出非零项的.

一开始的代码

代码如下

#include<cstdio>
#include<iostream>
using namespace std;
int main(void) {
	int K1, K2, Maxa = 0, Maxb = 0, e, Sumc = 0;	//Sumc就是计算乘起来之后一共有几个非零的多项式
	double  c;
	double A[1005] = { 0 }, B[1005] = { 0 }, C[2005] = { 0 };
	scanf("%d", &K1);
	for (int i = 0; i < K1; i++) {
		scanf("%d%lf", &e, &c);
		if (e > Maxa) Maxa = e;			//这是计算乘积的时候的循环最大值
		A[e] = c;
	}
	scanf("%d", &K2);
	for (int i = 0; i < K2; i++) {
		scanf("%d%lf", &e, &c);
		if (e > Maxb) Maxb = e;			//这是计算乘积的时候的循环最大值
		B[e] = c;
	}
	for (int i = 0; i <= Maxa; i++) {
		if (A[i] != 0.0) {
			for (int j = 0; j <= Maxb; j++) {
				if (B[j] != 0.0) {
					e = i + j;
					c = A[i] * B[j];
					if (C[e] == 0.0) {
						C[e] = c;
					}
					else
						C[e] = C[e] + c;
				}
			}
		}
	}
	//这个非常的重要,Sumc一定要在C[e]都算好了在统计个数
	//因为C[e]在计算中,可能会出现一正一负的,可以消去的同类项.
	//会多计算两次C[e]
	for (int i = 0; i <= Maxa + Maxb; i++) {
		if (C[i] != 0) {
			Sumc++;
		}
	}
	//第一个测试点有0 项,要删除,不能输出.
	printf("%d", Sumc);
	for (int i = Maxa + Maxb; i >= 0; i--) {
		if (C[i] != 0.0) {
			printf(" %d %.1lf", i, C[i]);
		}
	}
	return 0;
}

改进后如下

#include<cstdio>
#include<iostream>
using namespace std;
int main(void) {
	int K1, K2,e, Sumc = 0;	//Sumc就是计算乘起来之后一共有几个非零的多项式
	double  c;
	double A[1005] = { 0 }, B[1005] = { 0 }, C[2005] = { 0 };
	scanf("%d", &K1);
	for (int i = 0; i < K1; i++) {
		scanf("%d%lf", &e, &c);
		A[e] = c;
	}
	scanf("%d", &K2);

	for (int i = 0; i < K2; i++) {
		scanf("%d%lf", &e, &c);
		int Te = e;
		double Tc = c;					//每一次写好了,要重新赋值
		for (int j = 0; j <= 1002; j++) {

			e = j + e;
			c = A[j] * c;
			if (c != 0) {			//不计算非零项
				if (C[e] == 0.0) {
					C[e] = c;
				}
				else
					C[e] = C[e] + c;
			}
			e = Te; c = Tc;					//重新赋值
		}

	}
	//这个非常的重要,Sumc一定要在C[e]都算好了在统计个数
	//因为C[e]在计算中,可能会出现一正一负的,可以消去的同类项.
	//会多计算两次C[e]
	for (int i = 0; i <= 2002; i++) {
		if (C[i] != 0) {
			Sumc++;
		}
	}
	//第一个测试点有0 项,要删除,不能输出.
	printf("%d", Sumc);
	for (int i = 2002; i >= 0; i--) {
		if (C[i] != 0.0) {
			printf(" %d %.1lf", i, C[i]);
		}
	}
	return 0;
}