pat 1002 A+B for Polynomials （25 分）

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 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​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

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

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define wzf ((1 + sqrt(5.0)) / 2.0)
#define INF 0x3f3f3f3f
#define eps 0.0000001
#define LL long long
using namespace std;

const int MAXN = 1e3 + 10;
int cnt = 0, book[MAXN] = {0}, n, a;
double A[MAXN] = {0.0}, B[MAXN] = {0.0}, C[MAXN] = {0.0}, b;

int main()
{
    freopen("Date1.txt", "r", stdin);
    scanf("%d", &n);
    while (n --)
    {
        scanf("%d%lf", &a, &b);
        A[a] = b;
    }
    scanf("%d", &n);
    while (n --)
    {
        scanf("%d%lf", &a, &b);
        B[a] += b;
    }

    for (int i = 1005; i >= 0; -- i)
    {
        if (A[i] != 0 || B[i] != 0)
            C[i] = A[i] + B[i];
        if (C[i] != 0)
            ++ cnt;
    }
    printf("%d", cnt);
    for (int i = 1005; i >= 0; -- i)
    {
        if (C[i] != 0)
            printf(" %d %.1f", i, C[i]);
    }
    printf("\n");
    return 0;
}