PAT 1002. A+B for Polynomials (25)

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最新推荐文章于 2024-01-28 22:42:58 发布

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分类专栏： PAT 文章标签： pat Polynomials a+b

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本文介绍了一种计算两个多项式相加的方法，并提供了一个具体的C++实现案例。输入包括两个多项式的系数和指数，输出为两个多项式相加后的结果。程序通过读取输入数据，进行多项式的加法运算，并按照指定格式输出结果。

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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.
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    int na, nb;
	float a[1005]= {0}, b[1005] = {0}, c[1005]={0};
	int count = 0, n;

	cin >> na;
	for (int i=0; i!=na; ++i) { //输入两组数据
		cin >> n;
	    cin >> a[n];
	}
	cin >> nb;
	for (int i=0; i!=nb; ++i) {
		cin >> n; 
	    cin >> b[n];
	}

	for (int i=0; i!=1005; ++i) {  //多项式相加并统计非零项
		c[i] = a[i] + b[i];
		if (c[i] != 0)
		   ++count;
	}
	cout << count;
	for (int i=1004; i>=0; --i)  {
        if (c[i] != 0)
	   cout << " " << i << " " 
		       << setiosflags(ios::fixed) << setprecision(1) << c[i];  //保留一位小数
	}
	cout << endl;
return 0;
}