1002 A+B for Polynomials (25)

1002 A+B for Polynomials (25)（25 分）

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

3 2 1.5 1 2.9 0 3.2

//input: 多项式A：非零项个数 指数 系数 
//       多项式B：非零项个数 指数 系数
//output:A+B : 非零项个数 指数 系数
//多项式：例如 2.4x
//polynomial:多项式  exponent:指数  coefficient:系数    本题关键是设立辅助数组 double p[1001] = {0};

//注意点：最后不能有空格，可以采用 “ ”a“ ”b 形式输出 // B与A中重复相加后，仍有可能为0，
//      精确度的问题

#include<stdio.h>
#include<iostream>
#include<iomanip> //设置精度

using namespace std;

int main()
{
    double p[1001] = {0}; //存放与指数对应的系数,置初值为0

    int n;//指数
    double a;//系数
    int count = 0;//多项式A+B中的非零项个数

    //输入多项式A
    int k;
    cin >> k;
    count += k;
    for(int i = 0;i < k;i++)
    {
        cin >> n;
        cin >> a;
        p[n] += a;
    }

    //输入多项式B
    cin >> k;
    count += k;
    for(int i = 0;i < k;i++)
    {
        cin >> n;
        cin >> a;

        if(p[n] != 0) //与A中重复
        {
            count--;
        }
        p[n] += a;
        if(p[n] == 0) //与A中相加后还是等于0
        {
            count--;
        }
    }

    //输出结果
    cout << count;
    for(int i= 1000;i >= 0;i--)
    {
        if(p[i] != 0)
        {
            cout  << " " << i <<" " << setiosflags(ios::fixed)<<setprecision(1)<< p[i];
        }
    }

    return 0;
}