PAT-A1002 A+B for Polynomials

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本文介绍了一种简单的方法来计算两个多项式的和。通过输入两个多项式的非零项系数及指数，程序能够准确地计算并输出结果。示例展示了如何进行输入输出。

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1002 A+B for Polynomials（25 分）

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

也没什么思路，就是简单的相加

#include <stdio.h>

int main()
{
	int k1,k2,exp;
	double eft; 
	double A[1001]={0}; //初始化为零
	scanf("%d",&k1);
	for(int i=0;i<k1;i++){
		scanf("%d %lf",&exp,&eft);
		 A[exp]+=eft;
	}
	scanf("%d",&k2);
	for(int i=0;i<k2;i++){
		scanf("%d %lf",&exp,&eft);
		A[exp]+=eft; 
	}
	int count = 0;
	for(int i=0;i<1001	;i++){
		if(A[i]!=0){
			count++;
		}
	}
	printf("%d",count);
	for(int i=1000;i>=0;i--){
		if(A[i]!=0){
			printf("% d %.1f",i,A[i]);
		}
		
	}
	return 0;
}