PAT (Advanced Level) Practice 1002 A+B for Polynomials_pta pat (advanced level) practice 1002-优快云博客

本文链接：https://blog.youkuaiyun.com/honesty1024/article/details/88572684

本文详细介绍了一种解决多项式加法问题的算法实现，包括输入多项式的读取、多项式相加的处理以及结果输出的过程。算法使用了C++语言，并通过两个不同版本的代码示例展示了如何有效地进行多项式运算。

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1002 A+B for Polynomials （25 分)

时间限制: 400 ms 内存限制: 64 MB 代码长度限制: 16 KB

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

AC算法一：

#include<iostream>
#include<cstring>
using namespace std;
#define MAX 1001
double a[MAX];
void nInit();
void nInput(int k);
void nAdd(int k);
void nOutput();
int main()
{
	int nA,nB;
	while(cin>>nA)
	{
		nInit();
		nInput(nA);
		cin>>nB;
		nInput(nB);
		nOutput();
	}
	return 0;
}
void nInit()
{
	memset(a,0,sizeof(a));
}
void nInput(int k)
{
	int i,n;
	double temp;
	for(i=1;i<=k;i++)
	{
		cin>>n;
		cin>>temp;
		a[n]+=temp;
	}	
}
void nOutput()
{
	int i;
	int count;
	count=0;
	for(i=1000;i>=0;i--)
	{
		if(a[i]==0)
		{
			continue;
		}
		else
		{
			count++;
		}
	}
	cout<<count;
	for(i=1000;i>=0;i--)
	{
		if(a[i]==0)
		{
			continue;
		}
		else
		{
			printf(" %d %.1lf",i,a[i]);
		}
	}
	cout<<endl;
}

AC算法二：

#include<bits/stdc++.h>
using namespace std;
#define MAX 1001
double Na[MAX],Nb[MAX],N[MAX];
int num;
int ka,kb;

void vInit();
void vInput();
void vAdd();
void vOutput();

int main()
{
	while(cin>>ka)
	{
		vInit();
		vInput();
		vAdd();
		vOutput();
	}
	return 0;
}
void vInit()
{
	memset(Na,0,sizeof(Na));
	memset(Nb,0,sizeof(Nb));
	memset(N,0,sizeof(N));
	num=0;
}
void vInput()
{
	int na,nb;
	double ana,anb;
	for(int i=1;i<=ka;i++)
	{
		cin>>na>>ana;
		Na[na]=ana;
	}	
	cin>>kb;
	for(int i=1;i<=kb;i++)
	{
		cin>>nb>>anb;
		Nb[nb]=anb;
	}	 
}
void vAdd()
{
	for(int i=0;i<MAX;i++)
	{
		N[i]=Na[i]+Nb[i];
	}
	for(int i=0;i<MAX;i++)
	{
		if(N[i]!=0)
		{
			num++;
		}
	}
}
void vOutput()
{
	cout<<num;
	for(int i=MAX-1;i>=0;i--)
	{
		if(N[i]!=0)
		{
			printf(" %d %.1lf",i,N[i]);
		}
	}
	cout<<endl;
}