一、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
二、题目大意
多项式乘法
三、考点
模拟、数组
类似题目:多项式加法【笨方法学PAT】1002 A+B for Polynomials(25 分)
四、解题思路
1、使用 float 数组保存系数,指数作为数组下标;
2、读取数据,二次遍历,指数相加,系统相乘;
3、计算非0项的个数,从后向前输出非0项。
4、读取、计数和输出操作可以写在一个循环里,也可以分开。写在一起,节省运行时间;分开写的话,思路更加清晰,减少出错概率。
【注意】数组最好使用 fill 函数显式初始化,避免有未初始化的情况。
五、代码
#include<iostream>
#define N 2010
using namespace std;
int main() {
float f1[N], f2[N], f3[N];
fill(f1, f1 + N, 0.0);
fill(f2, f2 + N, 0.0);
fill(f3, f3 + N, 0.0);
//读入数据
int n,m;
cin >> n;
for (int i = 0; i < n;++i) {
int a;
float b;;
cin >> a >> b;
f1[a] = b;
}
cin >> m;
for(int i=0;i<m;++i) {
int a;
float b;;
cin >> a >> b;
f2[a] = b;
}
//循环
for (int i = 0; i < N/2; ++i) {
for (int j = 0; j < N/2; ++j) {
f3[i + j] += f1[i] * f2[j];
}
}
//计算非0个数
int ans_num = 0;
for (int i = 0; i < N; ++i) {
if (f3[i] != 0) {
ans_num++;
}
}
//输出
cout << ans_num;
for (int i = N-1; i >= 0; --i) {
if (f3[i] != 0) {
printf(" %d %.1f", i, f3[i]);
}
}
system("pause");
return 0;
}