PAT甲级 1009 Product of Polynomials

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本文介绍了如何通过两种不同的方法实现多项式的相乘，并提供了具体的C++代码示例。第一种方法使用两个输入数组和一个结果数组，而第二种方法则是在输入第二个数组的同时直接将结果存入结果数组。

1009 Product of Polynomials (25)（25 分）

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 a~N1~ N2 a~N2~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (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

方法一:

通过两个输入数组和一个结果数组进行。

代码如下:

#include <iostream>
#include <stdio.h>
using namespace std;

int main(){
   float f_1[1001] = {0.0};//都是浮点数直接用浮点数组，下标要是1001 
   
   int k,a;
   float b;
   cin>>k;
   for(int i=0;i<k;i++){
		cin>>a;
		cin>>b;
		f_1[a] += b;
   }
   
   float f_2[1001] = {0.0};
   cin>>k;
   for(int i=0;i<k;i++){
		cin>>a;
		cin>>b;
		f_2[a] += b;
   }
   
   float res[2001] = {0.0};
   for(int i=0;i<1001;i++){//需要注意下标是1001 
		for(int j=0;j<1001;j++){
			if(f_1[i] != 0.0)
			{
				if(f_2[j] != 0.0)//当两个系数都不为0时才计算 
				{
					res[i+j] += (f_1[i] * f_2[j]);
				}
			}		
		}
   }
   
    int cnt = 0;
	for(int i = 0;i<2001;i++)
		if(res[i] != 0.0)
				cnt++;
		
	cout<<cnt;
	for(int i = 2000;i>=0;i--)
		if(res[i] != 0.0)
				printf(" %d %.1f",i,res[i]);

   return 0;
}

方法二：

通过一个输入数组，在输入第二个数组的数时直接将结果存入一个结果数组。

代码如下:

#include <iostream>
#include <stdio.h>
using namespace std;

int main(){
   float f_1[1001] = {0.0};//都是浮点数直接用浮点数组，下标要是1001 
   
   int k,a;
   float b;
   cin>>k;
   for(int i=0;i<k;i++){
		cin>>a;
		cin>>b;
		f_1[a] += b;
   }
   
   float f_2[1001] = {0.0};
   cin>>k;
   for(int i=0;i<k;i++){
		cin>>a;
		cin>>b;
		f_2[a] += b;
   }
   
   float res[2001] = {0.0};
   for(int i=0;i<1001;i++){//需要注意下标是1001 
		for(int j=0;j<1001;j++){
			if(f_1[i] != 0.0)
			{
				if(f_2[j] != 0.0)//当两个系数都不为0时才计算 
				{
					res[i+j] += (f_1[i] * f_2[j]);
				}
			}		
		}
   }
   
    int cnt = 0;
	for(int i = 0;i<2001;i++)
		if(res[i] != 0.0)
				cnt++;
		
	cout<<cnt;
	for(int i = 2000;i>=0;i--)
		if(res[i] != 0.0)
				printf(" %d %.1f",i,res[i]);

   return 0;
}