PAT甲级1002 A+B for Polynomials（简单数组循环）-优快云博客

本文链接：https://blog.youkuaiyun.com/threeh20/article/details/87453605

本文详细介绍了一种解决多项式加法问题的算法实现。通过输入两个多项式的非零项，包括指数和系数，该算法能够准确计算并输出这两个多项式的和。采用数组循环运算，确保结果精确到小数点后一位，适用于计算机科学与数学计算领域。

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1002 A+B for 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 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.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

简单的数组循环运算。

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a,b;
	cin>>a;
	double dat[1111];
	memset(dat,0,sizeof(dat));
	double x;
	int s;
	for(int i=0;i<a;i++)
	{
		cin>>s>>x;
		dat[s]+=x;
	} 
	cin>>b;
	for(int i=0;i<b;i++)
	{
		cin>>s>>x;
		dat[s]+=x;
	}
	int js=0; 
	for(int i=0;i<=1000;i++)
	if(dat[i]!=0)js++;
	cout<<js;
	for(int i=1000;i>=0;i--)
	{
		if(dat[i]!=0)
		{
			cout<<" "<<i<<" ";
			printf("%.1lf",dat[i]);
		}
	}
	cout<<endl;
	return 0;
}