算法笔记：PTA A1002 多项式相加

题目

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 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

代码实现

关于最后的空格输出，有很多小技巧，我是选了一件看上去最笨的方法。（也是最简单的）
PS. 《算法笔记》上给的代码有点问题，所以只能作为参考用。


/* A1002
思路很简单，就是相同系数相加
*/
#include <cstdio>
const int MAXN = 1001;
int main()
{
    int k, counts = 0;
    double exp[MAXN] = {0};
    for (int i = 0; i < 2; i++) {
        scanf("%d", &k);
        for (int i = 0; i < k; i++) {
            int x;
            double a;
            scanf("%d%lf", &x, &a);
            exp[x] += a;
        }
    }
    for (int i = 0; i < MAXN; i++) {
        if (exp[i] != 0) counts++;
    }
    printf("%d", counts);
    for (int i = MAXN - 1; i >= 0; i--) {
        if (exp[i] != 0) {
            printf(" %d %.1f", i, exp[i]);
            //if (i != 0) printf(" ");
        }

    }
    return 0;
}