PAT --甲级1009（1009 Product of Polynomials ）

1009 Product of Polynomials （25 分)

This time, you are supposed to find $A×BA\times B$

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:

$...$

where $is the number of nonzero terms in the polynomial, and () are the exponents and coefficients, respectively. It is given that, .$

Output Specification:

For each test case you should output the product of $and in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.$

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

给两个多项式，格式为：k（该多项式的项数）指数系数
求这两个多项式相乘的结果，输出项数以及各个指数从大到小的非零项。

思路：模拟，数据规模不大，用数组即可（如果数据规模大的话可能要用键值对树存储）。注意有个坑点：结果要输出非零项的项数以及非零项，项数的问题很容易忽略！然后指数它的测试数据为整数（如果为浮点数的话注意考虑精度问题）

Code：

#include <bits/stdc++.h>

using namespace std;
const double esp = 1e-6;

struct Node {
	double zhishu;
	double xishu;
	
	Node operator * (const Node& A) const {
		Node tmp;
		tmp.zhishu = zhishu + A.zhishu;
		tmp.xishu = xishu * A.xishu;
		return tmp;
	}
	
	bool operator == (const Node& A) const { //指数相同
		if (fabs(zhishu - A.zhishu) <= esp) {
			return true;
		}
		return false;
	}
	
	bool operator < (const Node& A) const {
		return zhishu > A.zhishu;
	}
};

int k1,k2;
vector<Node> A, B, C;

int main() {
	Node tmp;
	A.clear(); B.clear(); C.clear();
	cin >> k1;
	for (int i = 0; i < k1; i++) {
		cin >> tmp.zhishu >> tmp.xishu;
		A.push_back(tmp);
	}
	cin >> k2;
	for (int i = 0; i < k2; i++) {
		cin >> tmp.zhishu >> tmp.xishu;
		B.push_back(tmp);
	}
	for (int i = 0; i < k1; i++) {
		for (int j = 0; j < k2; j++) {
			tmp = A[i] * B[j];
			int z = 0;
			for (; z < C.size(); z++) {
				if (tmp == C[z]) {
					break;
				}
			}
			if (z < C.size()) {
				C[z].xishu += tmp.xishu;
			} else {
				C.push_back(tmp);
			}
		}
	}
	for (vector<Node>::iterator it = C.begin(); it != C.end(); it++) {
		if (fabs(it->xishu) <= esp) {
			C.erase(it);
		}
	}
	sort(C.begin(), C.end());
	printf("%d", C.size());
	for (int i = 0; i < C.size(); i++) {
			printf(" %.0lf %.1lf", C[i].zhishu, C[i].xishu);
	}
	printf("\n");
	return 0;
}