A1002. A+B for Polynomials

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 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 <= 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<cstdio>
#include<iostream>
using namespace std;
int main(){
    int K, n, count = 0;
    double poly1[1001] = {0}, poly2[1001] = {0}, a;
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d%lf", &n, &a);
        poly1[n] = a;
    }
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d%lf", &n, &a);
        poly2[n] = a;
    }
    for(int i = 1000; i >= 0; i--){
        poly1[i] = poly1[i] + poly2[i];
        if(poly1[i] != 0) 
            count++;
    }
    printf("%d", count);
    for(int i = 1000; i >= 0; i--){
        if(poly1[i] != 0)
            printf(" %d %.1lf", i, poly1[i]);
    }
    cin >> K;
    return 0;

}

总结：

1、多项式加法，在项数不多的情况下直接开数组。

2、注意不用输出系数为0的项