1009 Product of Polynomials (25分) C++-优快云博客

本文链接：https://blog.youkuaiyun.com/Jay_fearless/article/details/106172482

本文详细介绍了一种解决多项式乘法问题的高效算法。通过使用三个数组分别存储两个输入多项式的非零项及其系数，以及计算结果的非零项，文章提供了一个清晰的C++代码实现。该算法能准确地计算出两个多项式的乘积，并输出非零项的个数及具体值。

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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 N1aN1N2aN2… NKaNKwhere 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

题意：多项式乘法最终输出非零项的个数和非零项。

思路：用2个数组用来存储各个项，最后用第3个数组存储输出项即可。

#include<bits/stdc++.h>
using namespace std;
#define MAX 2500
int main(){
	int k1,mp1=-1,mp2=-1,n,i;
	double p,po1[MAX],po2[MAX],po3[MAX];
	cin>>k1;
	for(i=0;i<k1;i++)
	{
		scanf("%d",&n);
		scanf("%lf",&p);
		po1[n]=p;
		if(n>mp1) mp1=n;
	}
	cin>>k1;
	for(i=0;i<k1;i++)
	{
		scanf("%d",&n);
		scanf("%lf",&p);
		po2[n]=p;
		if(n>mp2) mp2=n;
	}
	k1=-1;
	for(i=mp1;i>=0;i--)
	{	
		n=0;
		for(int j=mp2;j>=0;j--)
		{
			n=i+j;
			if(n>k1) k1=n;
			po3[n]+=po1[i]*po2[j];
		}
	}
	n=0;
	for(i=k1;i>=0;i--)
	{
		if(po3[i]!=0)
		{
			n++;
		}
	}
	cout<<n;
	for(i=k1;i>=0;i--)
	{
		if(po3[i]!=0)
		{
			printf(" %d %.1lf",i,po3[i]);
		}
	}
	return 0;
}