PAT 甲级 1002 A+B for Polynomials (25)（25 分）

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本文介绍了一种解决多项式相加问题的算法实现方法，输入包含两个多项式的系数及指数，通过处理输出两个多项式相加后的结果。具体实现了输入解析、多项式相加逻辑及格式化输出。

1002 A+B for Polynomials (25)（25 分）

This time, you are supposed to find A+B where A and B are two polynomials.

Input

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a~N1~ N2 a~N2~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.

Output

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

# include <cstdio>
# include <cstring>

void input(double sum[]){
	int k, a;
	double b;
	scanf("%d", &k);
	while(k--){
		scanf("%d%lf", &a, &b);
		sum[a] += b;
	}
}

int main(){
//	freopen("C:\\1.txt", "r", stdin);
	int count = 0;
	double sum[1001] = {0};
	memset(sum, 0, sizeof(sum));
	
	input(sum);
	input(sum);
	
	for(int i = 1000; i >= 0; i--){
		if(sum[i] != 0){
			count++;
		}
	} 
	printf("%d", count);
	
	for(int i = 1000; i >= 0; i--){
		if(sum[i] != 0){
			printf(" %d %.1lf", i, sum[i]);
		}
	}
	return 0;
}