Pat甲级 1002 A+B for Polynomials

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这篇博客主要介绍了如何实现一个多项式求和的算法，包括输入规格、输出规格以及样例输入和输出。代码使用C++编写，通过读取两个多项式并进行相加，最后以特定格式输出结果。博客中提供了完整的源代码，并欢迎读者提出建议和指正。

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1002 A+B for Polynomials (25 point(s))

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 <stdio.h>
#define MAXN 1010
using namespace std;

float res[MAXN];

int main() {
    int k;
    int a;
    float b;
    for (int i = 0; i < 2; i++) {
        scanf("%d", &k);
        while (k--) {
            scanf("%d%f", &a, &b);
            res[a] += b;
        }
    }
    int count = 0;
    for (int i = 0; i < MAXN; i++) {
        if (res[i] != 0)
            count++;
    }
    printf("%d",count);
    for (int i = MAXN-1; i >=0;i--) {
        if (res[i] != 0)
            printf(" %d %.1f", i, res[i]);
    }
    return 0;
}