pat1002 甲级 A+B for Polynomials （25 分）

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本文详细解析了一种多项式求和的算法实现，通过读取输入文件中两个多项式的系数和指数，进行求和运算，并输出结果。重点介绍了如何处理非零项，确保输出保留一位小数的细节。

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

就是多项式求和，k为多项式系数（一开始以为是非零项什么鬼），n为指数，An为系数。

一开始因为，输出时数组没有开到1001错了几次。

后面又因为没有注意到输出是要保留一位小数，错了。

#include<iostream>
#include<string.h>
using namespace std;
int main() {
	int k,n;
	double cof[1005],An;
	while (cin >> k) {
		memset(cof, 0, sizeof(cof)); 
		for (int i = 0 ; i < k; i++) {
			cin >> n >> An;
			cof[n] += An;
		}
		cin >> k;
		for (int i = 0; i < k; i++) {
			cin >> n >> An;
			cof[n] += An;
		}
		 n = 0;//这里的n初始化放在循环里就错了...不太懂
		for (int i = 0 ; i <=1000; i++) {
			if (cof[i] !=0) n++;//记录n个多项式
		}
		cout << n;
		for (int i =1000; i >=0; i--) {
			if (cof[i] != 0)
				printf(" %d %.1lf",i, cof[i]);
		}
		cout << endl;
	}
	return 0;
}