PAT1002 A+B for Polynomials

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

可耻地没读懂题目。。这道题多项式加法也太不按（我的）常理出牌了吧（嘤嘤嘤
每个样例包含两行，每行第一个数表示几组（指数，系数）
那么输出也是这样。第一个数表示几组（指数，系数）
这边开始我不明白起来了。。为啥是3呢= =
原来这个统计的是指数的个数（跪了好吧）
相同指数的系数相加
这个输出还有点顺序→指数从大到小排列
分析完了我们就可以写了（呜呜呜）

#include<iostream>
#include<iomanip>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
int main() {
    int  n, ex, cnt = 0;
    double co;
    double a[10000] = { 0 };
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> ex >> co;
        a[ex] += co;
    }
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> ex >> co;
        a[ex] += co;
    }
    for (int i = 0; i < 1001; i++) {
        if (a[i] != 0) {
            cnt++;
        }
    }
    cout << cnt;

    for (int i = 1000; i >= 0; i--) {
        if (a[i] != 0) {
            printf(" %d %.1lf", i, a[i]);
        }
    }
    system("pause");
    return 0;
}

呜呜呜，说几个卡着的点

第一次错是因为系数是浮点型（嘤嘤嘤！数组也记得开浮点型嗷

第二个就是系数记得保留一位小数！！

上面是AC代码嘤嘤嘤溜了溜了