PAT Advanced 1009. Product of Polynomials (25) (C语言实现)

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题目

This time, you are supposed to find $A\times B$ where $A$ and $B$ are two
polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each
line contains the information of a polynomial:

$K$ $N_1$ $a_{N_1}$ $N_2$ $a_{N_2}$ … $N_K$ $a_{N_K}$

where $K$ is the number of nonzero terms in the polynomial, $N_i$ and
$a_{N_i}$ ( $\cdots , K$ ) are the exponents and coefficients,
respectively. It is given that $1\le K \le 10$ , $\le N_K < \cdots < N_2 < N_1 \le 1000$ .

Output Specification:

For each test case you should output the product 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 up to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

思路

为了思路简单，就直接开大数组遍历了，没有什么难点

代码

最新代码@github，欢迎交流

#include <stdio.h>
int main()
{
    int N, exp, count = 0;
    float coef, A[1001] = {0}, B[1001] = {0}, MUL[2001] = {0};

    scanf("%d", &N);
    while(N--){ scanf("%d %f", &exp, &coef); A[exp] = coef; }
    scanf("%d", &N);
    while(N--){ scanf("%d %f", &exp, &coef); B[exp] = coef; }

    for(int i = 0; i < 1001; i++)
        for(int j = 0; j < 1001; j++)
            MUL[i + j] += A[i] * B[j];

    for(int i = 0; i < 2001; i++)
        if(MUL[i]) count++;

    printf("%d", count);
    for(int i = 2000; i >= 0; i--) if(MUL[i])
        printf(" %d %.1f", i, MUL[i]);

    return 0;
}