1002A+B for Polynomials (25)

最新推荐文章于 2024-01-28 22:42:58 发布

原创最新推荐文章于 2024-01-28 22:42:58 发布 · 171 阅读

CC 4.0 BY-SA版权

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。该程序能够接收两个多项式的系数和指数作为输入，然后计算并输出它们的和。程序使用数组来存储每个多项式的项，并通过遍历数组的方式找到所有非零项，最终输出结果保持与输入相同的格式。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述

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

输入描述:

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.

输出描述:

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.

输入例子:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

输出例子:

3 2 1.5 1 2.9 0 3.2

#include <iostream>
#include <math.h>
using namespace std;

int main() {
int K1, K2, n, count = 0;
float a;
float x[1001] = {0}, y[1001] = {0}, z[1001] = {0};
cin >> K1;
for (int i = 0; i < K1; i++) {
cin >> n >> a;
x[n] = a;
}
cin >> K2;
for (int i = 0; i < K2; i++) {
cin >> n >> a;
y[n] = a;
}
for (int i = 0; i < 1001; i++) {
z[i] = x[i] + y[i];
if (z[i] != 0)
count++;
}
cout << count;
for (int i = 1001; i >= 0; i--) {
if (z[i] != 0) {
if (z[i] == round(z[i]))
cout << " " << i << " " << z[i] << ".0";
else
cout << " " << i << " " << z[i];
}
}
return 0;
}