A1002 A+B for Polynomials 多项式相加

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 N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (,) are the exponents and coefficients, respectively. It is given that 1，0.

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

思路：

•用数组a存储最后结果，注意数组初始化；

•分别输入A和B，加到一个全空a上；遍历整个a，计算出非空项的个数；

•循环输出，注意空格（Notice that there must be NO extra space at the end of each line）和精确度（Please be accurate to 1 decimal place）要求。

#include <iostream>
using namespace std;
int main() {
    int k1, k2, e1, e2,count = 0;
    double c1, c2;
    double a[1005] = {0.0};//a保存最后结果 注意初始化
    //输入A
    cin >> k1;
    for (int i = 0; i < k1; i++) {
        cin >> e1 >> c1;
        a[e1] = c1;
    }
    //输入B
    cin >> k2;
    for (int i = 0; i < k2; i++) {
        cin >> e2 >> c2;
        a[e2] += c2;
    }
    //循环遍历计算非空项的个数
    for (int i = 1000; i >= 0; i--) {
        if (a[i] != 0) count++;
    }
    //输出 注意空格和精确度要求
    printf("%d", count);
    for (int i = 1000; i >= 0; i--) {
        if (a[i] != 0)
            printf(" %d %.1lf", i, a[i]);
    }
    return 0;
}