PAT A1002

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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

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"
//#include "algorithm"
//using namespace std;
int main(){
    double a[1010] = {0};
    int k;
    double n;//n为系数
    int e, count = 0;//e为指数,count计数不为零的导数项个数
    scanf("%d", &k);
    while (k--) {
        scanf("%d%lf", &e, &n);
        a[e] += n;
    }
    scanf("%d", &k);
    while (k--) {
        scanf("%d%lf", &e, &n);
        a[e] += n;
    }
    for (int i = 0; i <= 1000; i++) {
        if (a[i] != 0) {
            count ++;
        }
    }
    printf("%d", count);
    for (int i = 1000; i >= 0; i--) {
        if(a[i] != 0) printf(" %d %.1lf", i, a[i]);
    }
    return 0;
}

收获:
此题与PAT B1010一元多项式求导对比:
B1010题需要考虑零次项的求导,且得从低次枚举到高次;
此题需要考虑用count计数,需要在相加之后(考虑正负相加)计数;