1009 Product of Polynomials

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最新推荐文章于 2021-08-13 11:12:42 发布

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本文介绍了一种使用STL模板中的map数据结构来计算两个多项式乘积的方法。通过将指数作为键，系数作为值，可以有效地进行指数相加和系数相乘的操作，同时避免了重复项并确保了输出格式的正确性。

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1009 Product of Polynomials （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 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

题型分类：模拟、数学运算

题目大意：计算两个多项式的乘积

解题思路：使用STL模板里面的map，指数相加，系数相乘。需要注意的是，最后输出polyResult时，要判断系数是否为0，系数为0的项不计入输出项内。

#include <cstdio>
#include <map>

using namespace std;

const int maxn = 1010;

map<int, double> poly, polyResult;

int main(int argc, char** argv) {
	int K1, K2, expo;
	double coef;
	scanf("%d", &K1);
	while(K1--){
		scanf("%d %lf", &expo, &coef);
		poly[expo] = coef;
	}
	scanf("%d", &K2);
	while(K2--){
		scanf("%d %lf", &expo, &coef);
		for(map<int, double>::iterator it = poly.begin(); it != poly.end(); it++){
			polyResult[expo + it->first] += coef * it->second; 
		}
	}
	int cnt = 0;
	for(map<int, double>::iterator it = polyResult.begin(); it != polyResult.end(); it++){
		if(it->second != 0.0) cnt++;
	}
	printf("%d", cnt);
	for(map<int, double>::reverse_iterator it = polyResult.rbegin(); it != polyResult.rend(); it++){
		if(it->second != 0.0) printf(" %d %.1f", it->first, it->second);
	}
	return 0;
}