PAT甲级1009. Product of Polynomials (25)

计算多项式的乘积并输出

看题目

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

要求很简单，输入两个多项式，输出乘积，结果的系数保留一位小数。

//为了输出方便，本次使用c
#include <cstdio>
#define MAX 2002 //定义会用到的最大的数组元素个数

int main(void)
{
int N = 0, exp = 0, count = 0;//分别代表输入总数，幂次，最后输出总数目
double cof = 0.0f;//使用double表示系数

//三个数组，初始化为0，虽然浮点数不能精确表示数但是0还是能的
double pA[MAX / 2] = { 0, };
double pB[MAX / 2] = { 0, };
double result[MAX] = { 0, };

//输入总数
scanf("%d", &N);
//输入pA的数据
for (int i = 0; i < N; ++i)
{
scanf("%d %lf", &exp, &cof);
pA[exp] = cof;
}

//输入pB
scanf("%d", &N);
for (int i = 0; i < N; ++i)
{
scanf("%d %lf", &exp, &cof);
pB[exp] = cof;
}

//开始计算乘积
for (int i = 0; i < MAX / 2; ++i)
{
//pA[i] != 0，能节省点时间就行
for (int j = 0; j < MAX / 2 && pA != 0; ++j)
{
//参考多项式乘法法则
result[i + j] += pA[i] * pB[j];
}
}

//记录不为0的result个数
for (int i = 0; i < MAX; ++i)
{
//这个精确度
a9b4
够了，
if (result[i] >= 0.1f || result[i] <= -0.1f)
{
++count;
}
}

//输出count
printf("%d", count);

//输出多项式
for (int i = MAX - 1; i >= 0; --i)
{
if (result[i] != 0.0f)
{
//精确到小数点后1位
printf(" %d %.1f", i, result[i]);
--count;
}
}

//换行，虽然没必要
printf("\n");

//结束
return 0;
}

好好学习，天天向上