PTA A1009&A1010

第五天

A1009 Product of Polynomials (25 分)

题目内容

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 N a N a... Na
where K is the number of nonzero terms in the polynomial, Nand a(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N<⋯<N<N≤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

单词

product

英 /'prɒdʌkt/ 美 /'prɑdʌkt/

n. 产品；结果；[数] 乘积；作品

题目分析

多项式相乘，类似于A1002的多项式相加，之前也写过用的开结构体数组的方法，很笨，所以用之前在A1002学到的开一个数组，用下标代表指数，对应元素代表系数的方式来存，不过因为是乘法所以要多开两个数组把数据临时保存一下。代码如下。

具体代码

#include<stdio.h>
#include<stdlib.h>

double p1[1001];
double p2[1001];
double res[2002];
int N, M;
int main(void)
{
    scanf("%d", &N);
    for (int i = 0; i < N; i++)
    {
        int e;
        double c;
        scanf("%d %lf", &e, &c);
        p1[e] = c;
    }
    scanf("%d", &M);
    for (int i = 0; i < M; i++)
    {
        int e;
        double c;
        scanf("%d %lf", &e, &c);
        p2[e] = c;
    }
    for (int i = 0; i < 1001; i++)
    {
        if (p1[i] != 0)
            for (int j = 0; j < 1001; j++)
                if (p2[j] != 0)
                    res[i + j] += p1[i] * p2[j];
    }
    int num = 0;
    for (int i = 0; i < 2002; i++)
    {
        if (res[i] != 0)
            num++;
    }
    printf("%d", num);
    for (int i = 2001; i >= 0; i--)
    {
        if (res[i] != 0)
            printf(" %d %0.1f", i, res[i]);
    }
    system("pause");
}