PAT-1002 A+B for Polynomials

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 N1 aN1 *N2 *aN2 … NK aNK

where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N**K<⋯<N2<N1≤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:

1 2	2 1 2.4 0 3.2 2 2 1.5 1 0.5

Sample Output:

1	3 2 1.5 1 2.9 0 3.2

Code

#pragma once
#include <iostream>
#include <stack>

using namespace std;
//注意点
//coefficients可以是负数
//系数为0不进行输出
//如果size为0，直接输出0
//最后没有空格
float poly[1001];

void _1002() {
	stack<float> coefficients;
	stack<int> exponents;
	//每行几个多项式
	int n, exp;
	float cof;
	//共两行
	for (int i = 0; i < 2; ++i) {
		(void)scanf("%d", &n);
		for (int i = 0; i < n; ++i) {
			(void)scanf("%d%f", &exp, &cof);
			poly[exp] += cof;
		}
	}

	//次数从高到底
	for (int i = 0; i <= 1000; i++) {
		if (poly[i] != 0) {
			coefficients.push(poly[i]);
			exponents.push(i);
		}
	}
	
	int count = coefficients.size();
	printf("%d", count);
	for (int i = 0; i < count; i++) {
		float coe = coefficients.top();
		int exp = exponents.top();
		coefficients.pop();
		exponents.pop();
		printf(" %d %.1f", exp, coe);
	}
}