PAT (Advanced Level) 1002 A+B for Polynomials （25 分）

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1002 A+B for 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 N​₁​​ a​_N​1​​​​ N_​2​​ a_​N​2​​​​ … N_​K​​ a_​N​K​​​​
where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N_​K​​<⋯<N​₂​​<N_​1​​≤1000.

Output Specification:

For each test case you should output the sum 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 to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

Code

#include <iostream>
#include <string>
#include <map>
#include <iomanip>

using namespace std;

int main()
{
	const int numOfPoly = 2; // 相加的多项式个数
	int numOfTerms[numOfPoly];
	map<int, float> polynomials[numOfPoly]; // 用来保存加数多项式的map， <exponent， coefficient>
	//-------------------输入加数多项式-----------------//
	for (int i = 0; i < numOfPoly; i++)
	{
		cin >> numOfTerms[i];
		for (int j = 0; j < numOfTerms[i]; j++)
		{
			int exp;
			float coeff;
			cin >> exp >> coeff;
			polynomials[i][exp] = coeff;
		}
	}
	//--------------------输入完成----------------------//
	map<int, float> sumPolynomial; // 用来保存和多项式的map
	for (int i = 0; i < numOfPoly; i++) // 遍历每个加数多项式
	{
		for (auto it : polynomials[i]) // 遍历每个加数多项式的每一项
		{
			auto itOfSum = sumPolynomial.find(it.first); // 查找和多项式中是否有exponent相同的项
			if (itOfSum != sumPolynomial.end()) // 如果有，coefficient相加
			{
				itOfSum->second += it.second;
				if (itOfSum->second == 0) // 注意，如果相加等于0， 删去这一项
					sumPolynomial.erase(itOfSum);
			}
			else                                // 如果没有， 在和多项式中加入这一项
			{
				sumPolynomial[it.first] = it.second;
			}
		}
	}
	cout << sumPolynomial.size(); // 输出项数
	map<int, float>::reverse_iterator rIter;
	for (rIter = sumPolynomial.rbegin(); rIter != sumPolynomial.rend(); rIter++) //反向遍历输出，注意精确到小数点后一位
	{
		cout << " " << rIter->first << " " << setiosflags(ios::fixed) << setprecision(1) <<rIter->second;
	}
	cout << endl;
	return 0;
}

思路

本题目要求模拟多项式相加，如果用map容器，以指数（exponent）为键（key），以它的系数（coefficient）为值（value），那么，A、B两式相加时，A可以直接通过每一项的指数，找到B是否有指数相同的项，如果有就把值相加。具体实现方式请见代码注释。

warning

1. Please be accurate to 1 decimal place. 注意精确到小数点后一位，如果没有注意这个，测试点1会错。
2. 如果系数相加等于0那么要把这一组<exponent, coeffcient>删掉， 如果没有注意这个， 测试点3~6都会错。

关于C++ STL map的使用方法可以去
https://www.cnblogs.com/fnlingnzb-learner/p/5833051.html
学习
关于C++11 auto遍历可以去
https://blog.csdn.net/qq_40213457/article/details/80720170
学习

以上