1002 A+B for Polynomials -PAT甲级

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本文介绍了一种解决多项式加法问题的算法实现，输入两个多项式的非零项系数和指数，输出它们的和。算法首先初始化一个足够大的数组来存储结果，然后逐项读取并累加两个多项式的每一项，最后输出结果。该算法适用于处理最大指数不超过1000的多项式。

This time, you are supposed to find A+BA+BA+B where AAA and BBB 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:

KKK N1N_1N1 aN1a_{N_1}aN1 N2N_2N2 aN2a_{N_2}aN2 ... NKN_KNK aNKa_{N_K}aNK

where KKK is the number of nonzero terms in the polynomial, NiN_iNi and aNia_{N_i}aNi (i=1,2,⋯,Ki=1, 2, \cdots , Ki=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤101 \le K \le 101≤K≤10，0≤NK<⋯<N2<N1≤10000 \le N_K < \cdots < N_2 < N_1 \le 10000≤NK<⋯<N2<N1≤1000.

Output Specification:

For each test case you should output the sum of AAA and BBB 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

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
double exe[N];
int main(){
	for(int i=0;i<N;i++)
		exe[i]=0;
	int k,e;
	double c;
	for(int i=0;i<2;i++){
		scanf("%d",&k);
		for(int j=1;j<=k;j++){
			cin>>e>>c;
			exe[e]+=c;
		}
	}
	int sum=0;
	for(int i=0;i<N;i++){
		if(exe[i]!=0){
			sum++;
		}
	}
	printf("%d",sum);
	for(int j=N-1;j>=0;j--)
		if(exe[j]!=0)
			printf(" %d %.1f",j,exe[j]);
	return 0;
}