1009. Product of Polynomials

Product of Polynomials

this time, you are supposed to find A*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 N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=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

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题意

多项式乘法。

思路

用3个double数组去记录多项式。需要注意的点有：

1. 结果多项式的最大指数为2000，相应的要提高结果数组的容量；
2. 系数可能为负；
3. 更加规范的double类型比较，应当要设置一个精度再进行比较。

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代码实现

#include <cstdio>

double a[1001];
double b[1001];
double ans[2001];

int main()
{
    int k, exp;
    double coe;
    int num = 0;                // 统计系数非零项个数

    scanf("%d", &k);
    for (int i = 0; i < k; i++)
    {
        scanf("%d %lf", &exp, &coe);
        a[exp] = coe;
    }
    scanf("%d", &k);
    for (int i = 0; i < k; i++)
    {
        scanf("%d %lf", &exp, &coe);
        b[exp] = coe;
    }

    for (int i = 0; i < 1001; i++)
        for (int j = 0; j < 1001; j++)
            ans[i + j] += a[i] * b[j];      // 注意这里是+=而不是=

    for (int i = 0; i < 2001; i++)
        if (ans[i] != 0)                    // 注意系数可能为负(图简便这里的比较并不规范)
            num++;

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

    return 0;
}