1002. A+B for Polynomials (25)

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

主要存在几个点：
　　可以输入重复的项
　　输入的项相加后可能为0，此时项数减1
　　当项目为0时，只输出项数，不要加空格
　　输出最后一项不要加空格
　　最终的结果多项式项数最多为20

#include <stdio.h>
#include<string.h>
int main()
{
    float a[1001];
    int i,k;
    float temp;
    // 初始化数组
    for(i = 0; i <= 1000; i++){
        a[i] = 0.0f;
    }
    // 分别输入两个多项式
    scanf("%d", &k);
    while(k--){
        scanf("%d%f", &i, &temp);
        a[i] += temp;
    }
    scanf("%d", &k);
    while(k--){
        scanf("%d%f", &i, &temp);
        a[i] += temp;
    }
    // 判断当前多项式的项数
    k = 0;
    for(i = 0; i <= 1000; i++){
        if(a[i]!=0.0){
                k++;
        }
    }
    printf("%d", k);
    // 项数为0则只输出k，且不带空格
    if(k != 0)
        printf(" ");
    for(i=1000; i >= 0; i--){
        if(a[i]!=0.0){
            printf("%d ", i);
            printf("%0.1f", a[i]);
            k--;
            // 输出最后一项后不带空格
            if(k != 0)
                printf(" ");
        }
    }
    return 0;
}