【PAT (Advanced Level) Practice】刷题1002 A+B for Polynomials

本文链接：https://blog.youkuaiyun.com/qq_43533623/article/details/107777467

本文介绍了一种解决多项式加法问题的算法实现，输入两个多项式的非零项系数和指数，输出它们的和。算法使用哈希表存储每个多项式的项，并进行逐项相加，最后按降序输出结果。

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A+B for Polynomials

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 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.

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	float hash[1001];
	int num = 0, num1, num2;
	int i, e;
	float c;
	for (i = 0;i < 1001;i++) {
		hash[i] = 0;
	}
	cin >> num1;
	for (i = 0;i < num1;i++)
	{
		cin >> e >> c;
		hash[e] = hash[e] + c;
	}
	cin >> num2;
	for (i = 0;i < num2;i++)
	{
		cin >> e >> c;
		hash[e] = hash[e] + c;
	}
	for (i = 0;i < 1001;i++)
	{
		if (hash[i] != 0)
		{
			num++;
		}
	}
	if (num == 0)
	{
		cout << num<<endl;
		return 0;
	}
	cout << num << " ";
	for (i = 1000;i >= 0;i--)
	{
		if (hash[i] != 0)
		{
			if (num == 1)
			{
				cout << i << " ";
				printf("%.1f", hash[i]);
				break;
			}
			else
			{
				cout << i << " " ;
				printf("%.1f ", hash[i]);
				num--;
			}
		}
	}
	cout << endl;
	return 0;
}