A+B for Polynomials

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.

Output

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

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3 2 1.5 1 2.9 0 3.2

题目大意：给出两个多项式求二者的和

思路：用一个结构体数组来保存多项式的系数和指数，其中指数充当结构体数组的下标(由题意知，多项式每一项的指数不超过1000)，即若指数为i，则{指数、系数}保存在下标为i的地方，初始所有指数成员均置为-1,(与多项式中真正包含的项数区分开)，输入第二个多项式时之间将指数相同的项相加，其中注意若相加为0，则应将该项的指数置为-1(即最终结果暂时删除这一项)，最后按指数从大到小方式输出结果多项式(只输出指数不为-1的)

code:

#include<iostream>
#include<stdio.h>
using namespace std;
struct Polynomials{
	int exponents;//指数 
	double coefficients;//底数 
};
struct Polynomials p1[1005];//保存第一个多项式，并与第二个多项式相加后保存结果 
void Initial(){//对结构体数组进行初始化，将指数均标记为-1有助于之后的加法计算 
	int i;
	for(i=0;i<1005;i++){
		p1[i].exponents=-1;
		p1[i].coefficients=0.0;
	}
}
int main(){
	int i,j,p1num,p2num,e;
	double c;
	int count=0;
	Initial();
	//读入两组多项式 
	cin>>p1num;
	for(i=0;i<p1num;i++){
		cin>>e>>c;
		if(p1[e].exponents==-1){
			count++;//统计个数 
			p1[e].exponents=e;
			p1[e].coefficients=c;  
		}
		else{ 
			p1[e].coefficients+=c;
			if(p1[e].coefficients==0){
				count--;
				p1[e].exponents=-1; 
			}
		}
	}
	cin>>p2num;
	for(i=0;i<p2num;i++){
		cin>>e>>c;
		if(p1[e].exponents==-1){
			count++;//统计个数 
			p1[e].exponents=e;
			p1[e].coefficients=c;  
		}
		else{ 
			p1[e].coefficients+=c;
			if(p1[e].coefficients==0){
				count--;
				p1[e].exponents=-1; 
			} 
		}
	}
	cout<<count;
	for(i=1004;i>=0;i--){
		if(p1[i].exponents>=0){
			printf(" %d %.1f",p1[i].exponents,p1[i].coefficients);  
		}
	}
	return 0;
}

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