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

• 编程题

1002 A+B for Polynomials (25)（25 分）

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.

Output

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

题解1：

该题首先需要注意数据类型。

使用数组标签的方式记录数据，result[exponent]=coefficient。

考虑到最后需要先输出多项式的项数，每次读取result的时候记录项数。这里使用总数a+b减去多余项数的方式，即，如果第二组中有和第一组中相同的项，总数减1，如果两项系数相加和为0，总数再减1。

也可以先把所有项的系数相加，最后统计项数。

最后输出项数和所有系数不为0的项。输出保留1位小数，使用setprecision(n)<<std::fixed。

源代码1：

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
	int a, b;
	int expo;
	double coef;
	int j;
    double result[1001] = { 0 };
	int count = 0;
	cin >> a;
	for (j = 0; j < a; j++)
	{
		cin >> expo;
		cin >> result[expo];
	}


	cin >> b;
	for (j = 0; j < b; j++)
	{
		cin >> expo >> coef;
		if ((result[expo]) != 0)
		{
			count++;
			result[expo] += coef;
			if (result[expo] == 0)
				count++;
		}
		else
			result[expo] = coef;
	}
	cout << a + b - count;
	for (j = 1000; j >=0; j--)
	{
		if (result[j] != 0)
			cout << " " << j << " " << setprecision(1)<< std::fixed << result[j];
	}

	return 0;
}