PTA(Advanced Level)1009.Product of Polynomials

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 ... N**K aNK

where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N**K<⋯<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

思路

• 多项式相乘，就是逐项展开乘即可，所以我们需要分别记录两个多项式的系数，指数，然后手动模拟展开即可。

代码

#include<bits/stdc++.h>
using namespace std;
struct node
{
	double a;
	int n;
	node(){}
	node(double _a, int _n): a(_a), n(_n){}
};//a*x^n
vector<node> v1;
vector<node> v2;   //分别用来记录里2个多项式
double a[2010] = {0};

int main()
{
	int k;

	cin >> k;
	int id;
	double tmp;
	while(k--)
	{
		cin >> id >> tmp;
		v1.push_back(node(tmp, id));
	}
	cin >> k;
	while(k--)
	{
		cin >> id >> tmp;
		v2.push_back(node(tmp, id));
	}
	double ans = 0.0;
	int index = 0;
	int base = 0;
	for(int i=0;i<v1.size();i++)
	{
		for(int j=0;j<v2.size();j++)
		{
			ans = v1[i].a * v2[j].a;
			index = v1[i].n + v2[j].n;  //系数相乘，指数相加
			//printf("ans:%d index:%d
", ans, index);
			a[index] += ans;
		}
	}  //逐项展开的过程
	int cnt = 0;
	for(int i=0;i<2010;i++)
	{
		if(a[i] != 0)
			cnt++;
	}
	cout << cnt;
	for(int i=2009;i>=0;i--)
	{
		if(a[i] != 0)
            printf(" %d %.1f", i, a[i]);
	}
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805509540921344