A1002. A+B for Polynomials

Title Description

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

Input Format

　　Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a_N1 N2 a_N2 ... NK a_NK , where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1,2,...,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N_k<...<N₂<N₁≤1000

Output Format

　　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

The meaning of problems

Given two rows, each row represents a polynomial, the first number indicates the number of entries polynomial coefficients of non-zero entries, each represented by two digits after one, two numbers represent the coefficients and the power of the. Determine and two polynomials, and outputting the same results to the same format as previously

Sample interpretation

2 is a first means two polynomial coefficients have non-zero entries, 1 item 2.4 represents a coefficient represented by the coefficient 3.2 to 2.4,0 (i.e. constant term) zero term is 3.2, that is F1 (x) = 2.4 x + 3.2
Likewise the second row can be written as a polynomial representation F2 (X) = 1.5x ² + 0.5x
Then Fl (X) + F2 of (X) = 1.5x ² + 2.9x + 3.2

#include <bits/stdc++.h>
const int max_n = 1111;
double p[max_n] = {};
int main(int argc, char *argv[]) {
    int k, n, count = 0;
    double a;
    scanf("%d", &k);
    for(int i = 0; i < k; i++){
        scanf("%d %1f", &n, &a);
        p[n] += a;
    }
    scanf("%d", &k);
    for(int i = 0; i < k; i++){
        scanf("%d %1f", &n, &a);
        p[n] += a;
    }
    for(int i = 0; i < max_n; i++){
        if(p[i] != 0){
            count++;
        }
    }
    printf("%d", count);
    for(int i = max_n - 1; i >= 0; i--){
        if(p[i] != 0){
            printf( "%d %.1f", i, p[i]);
        }
    }
    return 0;    
}