【PAT甲级】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≤N​K<⋯

【翻译】：计算多项式的和，每个样例有两行，每一行代表一个多项式，K是非零整数，Ni是指数，aNi是系数，K范围是1到10，NK范围是0到1000，输出一行，行尾不可有空格

举栗说明：

2 1 2.4 0 3.2表示 多项式有两个项，分别是指数为1，系数为2.4和指数为0系数为3.2的项。

这题肥肠简单了……但是有几个坑

1.系数为负数时要输出，不然测试点2过不去

2.输出要精确到小数点后一位

#include 
#include 
using namespace std;
double a[1005];
int main(){
	int n,k,x,m,count = 0;
	double y;
	cin>>m;
	for(int i = 0; i < m; i++){
		cin>>x>>y;
		a[x] += y;
	}
	cin>>m;
	for(int i = 0; i < m; i++){
		cin>>x>>y;
		a[x] += y;
	}
	for(int i = 0; i < 1001; i++){
		if(a[i] != 0) count++;
	} 
	cout<= 0; i--){
		if(a[i] != 0){
			cout<<" "<