1002 A+B for Polynomials

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本文介绍了一种解决多项式加法问题的算法，通过使用数组实现类似基数排序的方法，在O(1)时间内高效求解两个多项式的和。该算法首先读取两个多项式的系数和指数，然后将它们相加并输出结果。

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, 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.

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

Analyse:

I wrote an algorithm whose time complexity is O(N), but it only passed three samples. Therefore, I had to change my code into better one which can satisfy the problem's requests in O(1) time. We can use an array to do this and that is so called "radix sort".

#include<iostream>
#include<cmath>

using namespace std;

int main(){
	float a[1001] = {0};
	int K, cnt = 0, exponent;
	float coefficient;
	for(int i = 0; i < 2; i++){
		scanf("%d", &K);
		for(int j = 0; j < K; j++){
			scanf("%d %f", &exponent, &coefficient);
			if(abs(a[exponent]) <= 1e-7) cnt++;
			a[exponent] += coefficient;
			if(abs(a[exponent]) <= 1e-7) cnt--;
		}
	}
	printf("%d", cnt);
	for(int i = 1000; i >= 0; i--){
		if(abs(a[i]) >= 1e-7) printf(" %d %.1f", i, a[i]);
	}
	return 0;
}