PTA-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

題意：輸入兩個係數爲浮點型的多項式，輸出兩個多項式的和

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 1005
#define eps 1e-6
double sum[maxn];
int main()
{
    int k;
    scanf("%d",&k);
    int x;
    double y;
    for(int i = 0; i < k; i ++)//輸入第一個多項式
    {
        scanf("%d %lf",&x,&y);
        sum[x] += 1.0*y;
    }
    scanf("%d",&k);
    for(int i = 0; i < k; i ++)//輸入第二個多項式
    {
        scanf("%d %lf",&x,&y);
        sum[x] += 1.0*y;
    }
    int counts = 0;
    for(int i = 1000; i >= 0; i --)
    {
        if(sum[i]>eps||sum[i]<-eps)//判斷的時候不能忘了係數爲負的情況
            counts++;
    }
    printf("%d",counts);
    for(int i = 1000; i >= 0; i --)
    {
        if(sum[i]>eps||sum[i]<-eps)//輸出也是
            printf(" %d %.1f",i,sum[i]);
    }
    return 0;
}