题目

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:

$K$ $N_1$ $a_{N_1}$ $N_2$ $a_{N_2}$ … $N_K$ $a_{N_K}$

where $K$ is the number of nonzero terms in the polynomial, $N_i$ and $a_{N_i}$ ( $i=1, 2, \cdots , K$ ) are the exponents and coefficients, respectively. It is given that $1 \le K \le 10$ ， $0 \le N_K < \cdots < N_2 < N_1 \le 1000$ .

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:

1
2

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

1

3 2 1.5 1 2.9 0 3.2

思路

最初由于数学概念的理解错误，纠结了半天。0是零多项式，在题中将之归于“0项”多项式。当然了，实际是没有0项多项式的，只有零多项式，但是非要输出个结果，0还是合理的。我一开始把0当成只有常数项的一项多项式，结果最后一测试点老过不去。。。。

很多人都用1000长度的数组存储多项式，确实思路简单，且并不是很浪费空间。我用了另一种思路，用数组模仿链表的实现。将A、B两多项式由次数从高到低依次计算，存入新数组。

疑问：最初最后一个测试点没过的时候，我以为又是小负数近似格式的问题（见PAT Basic 1051. 复数乘法 (15)（C语言实现）），便将系数绝对值小于0.05的项全忽略了。当然这么做也通过了，可能输入保证了精度也只有1位小数，这样就没有这个必要了。

代码

Github最新代码，欢迎交流

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45

#include <stdio.h>
#include <math.h>

typedef struct Poly {
	double coef;
	int exp;
} Poly[20];

int main()
{
	int KA, KB, Ksum = 0;
	Poly A, B, sum;

	scanf("%d", &KA);
	for (int i = 0; i < KA; i++)
		scanf("%d %lf", &A[i].exp, &A[i].coef);

	scanf("%d", &KB);
	for (int i = 0; i < KB; i++)
		scanf("%d %lf", &B[i].exp, &B[i].coef);

	int i = 0, j = 0;
	while (i < KA || j < KB) {
		if (i == KA || (j < KB && A[i].exp < B[j].exp)) {
			sum[Ksum].exp = B[j].exp;
			sum[Ksum].coef = B[j++].coef;
		} else if (j == KB || (i < KA && A[i].exp > B[j].exp)) {
			sum[Ksum].exp = A[i].exp;
			sum[Ksum].coef = A[i++].coef;
		} else {
			sum[Ksum].exp = A[i].exp;
			sum[Ksum].coef = A[i++].coef + B[j++].coef;
		}

		if (fabs(sum[Ksum].coef) >= 0.05)
			Ksum++;
	}

	printf("%d", Ksum);

	for (int i = 0; i < Ksum; i++)
		printf(" %d %.1lf", sum[i].exp, sum[i].coef);

	return 0;
}