1002. A+B for Polynomials (25)

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 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


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


 polynomials 多项式
 exponents 指数
 coefficients 系数

题目大意：计算A+B的和，A和B都是多项式

       思路： 整一个数组a，存系数，而a的下标表示指数，对于每一个输入，把指数下标的那个空间放系数之和，
       然后遍历一遍，找系数不是0的，把下标存到一个数组b中，遍历这个数组，输出b，并且输出相应b的a。
       
       注意，如果一开始b中的个数是0，说明系数都是0，那么就只输出0。



#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;


double a[1001];


int main()
{
    //初始化a，b
    memset(a,0.0,sizeof(a));


    //输入
    int ka, kb;
    cin >> ka;
    for (int i = 0; i < ka; i++)
    {
        int zhi;
        double xi;
        cin >> zhi >> xi;
        a[zhi] += xi;
    }
    cin >> kb;
    for (int i = 0; i < kb; i++)
    {
        int zhi;
        double xi;
        cin >> zhi >> xi;
        a[zhi] += xi;
    }




    //遍历一遍a数组，找不是0的
    int res[1001];
    int cou = 0;
    for (int i = 1001; i >= 0; --i)
    {
        if (a[i] != 0.0)
            res[cou++] = i;
    }


    //输出结果
    cout << cou;
    if (cou != 0)
        cout << " ";                    //如果是0的话后面的空格就不用输出了
    for (int i = 0; i < cou; i++)
    {
        if (i == 0)
            printf("%d %.1f",res[i],a[res[i]]);
        else
            printf(" %d %.1f",res[i],a[res[i]]);
    }


    return 0;
}