PAT Advanced 1002. A+B for Polynomials (25) (C语言实现)

题目

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 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

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

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

最新代码@github，欢迎交流 ^_^

#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;
}