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。
#include <cstdio>
#define MAX 2002
int main(
void)
{
int N =
0,
exp =
0, count =
0;
double cof =
0.0f;
double pA[MAX /
2] = {
0, };
double pB[MAX /
2] = {
0, };
double result[MAX] = {
0, };
scanf(
"%d", &N);
for (
int i =
0; i < N; ++i)
{
scanf(
"%d %lf", &
exp, &cof);
pA[
exp] = cof;
}
scanf(
"%d", &N);
for (
int i =
0; i < N; ++i)
{
scanf(
"%d %lf", &
exp, &cof);
pB[
exp] = cof;
}
for (
int i =
0; i < MAX /
2; ++i)
{
for (
int j =
0; j < MAX /
2 && pA !=
0; ++j)
{
result[i + j] += pA[i] * pB[j];
}
}
for (
int i =
0; i < MAX; ++i)
{
if (result[i] >=
0.1f || result[i] <= -
0.1f)
{
++count;
}
}
printf(
"%d", count);
for (
int i = MAX -
1; i >=
0; --i)
{
if (result[i] !=
0.0f)
{
printf(
" %d %.1f", i, result[i]);
--count;
}
}
printf(
"\n");
return 0;
}
好好学习,天天向上
转载请注明原文地址: https://ju.6miu.com/read-16497.html