PAT (Advanced Level) Practise：1002. A+B for Polynomials

【题目链接】

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

题意分析：

两个多项式加法，然后按指数项从高到低分别输出“指数 系数”。

提交代码：

#include <stdio.h>
#include <string.h>

int main(void)
{
    float a[10000+1];
    float b[10000+1];
    int i, cnt;
    int K, Nk;
    float aNk;

    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));

    scanf("%d", &K);
    for(i = 0; i < K; i++)
    {
        scanf("%d %f", &Nk, &aNk);
        a[Nk] = aNk;
    }
    scanf("%d", &K);
    for(i = 0; i < K; i++)
    {
        scanf("%d %f", &Nk, &aNk);
        b[Nk] = aNk;
    }

    cnt = 0;
    for(i = 0; i < sizeof(a)/sizeof(a[0]); i++)
    {
        a[i] = a[i] + b[i];
        if(a[i] != 0.0)
            cnt++;
    }

    printf("%d", cnt);
    for(i = sizeof(a)/sizeof(a[0]) - 1; i >= 0; i--)
    {
        if(a[i] != 0)
            printf(" %d %.1f", i, a[i]);
    }

    return 0;
}