PAT 1002 A+B for Polynomials (25分)

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本文介绍了一个简单的算法，用于计算两个多项式的和，并给出了完整的C++实现代码。输入为两个多项式，每个多项式由一系列非零项组成，输出为这两个多项式的和，格式与输入相同。

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 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 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 <cstdio>
using namespace std;
int main(){
    const int LEN = 1000+1;
    float a[LEN]={0};

    int k;
    int exponent;
    float coefficient;
    for(int i = 0; i < 2; i++){
        scanf("%d", &k);
        for(int j = 0; j < k; j++){
            scanf("%d %f", &exponent, &coefficient);
            a[exponent] = a[exponent] + coefficient;
        }
    }
    int ctr = 0;
    for(int i = LEN - 1; i >= 0; i--){
        if(a[i] != 0)
            ++ctr;
    }
    printf("%d", ctr);
    for(int i = LEN - 1; i >= 0; i--){
        if(a[i] != 0){
            printf(" %d %.1f", i, a[i]);
        }
    }
     return 0;
}