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
Code:
using namespace std;
int main(int argc, char *argv[]) {
int K1;
scanf("%d", &K1);
K1 = 2 * K1;
double *N1 = new double[K1 + 1];
for (int i = 1; i <= K1; i++)
scanf("%lf", &N1[i]);
int K2;
scanf("%d", &K2);
K2 = 2 * K2;
double *N2 = new double[K2 + 1];
for (int i = 1; i <= K2; i++)
scanf("%lf", &N2[i]);
double *N3 = new double[K1 + K2];
int i = 1, j = 1, k = 0;//三个数组的下标
int zhishu;
double xishu;
bool flag = false;//多余的量
for (; i <= K1; i = i + 2) {
zhishu = N1[i];
while (zhishu <= int(N2[j]) && j < K2) {
if (int(N1[i]) == int(N2[j])) {//N1+N2->N3
xishu = N1[i + 1] + N2[j + 1];
if (fabs(xishu) > 1e-6) {
N3[k] = zhishu;
N3[k + 1] = xishu;
k = k + 2;
}
flag = true;
}
else {//N2->N3
if (fabs(N2[j + 1]) > 1e-6) {
N3[k] = int(N2[j]);
N3[k + 1] = N2[j + 1];
k = k + 2;
}
flag = false;
}
j = j + 2;
}
if (!flag) {//N1->N3
if (fabs(N1[i + 1]) > 1e-6) {
N3[k] = zhishu;
N3[k + 1] = N1[i + 1];
k = k + 2;
}
}
if (i > K1 - 2 && j < K2) {//N2->N3
while (j < K2) {
if (fabs(N2[j + 1]) > 1e-6) {
N3[k] = int(N2[j]);
N3[k + 1] = N2[j + 1];
j = j + 2;
k = k + 2;
}
}
}
if (j > K2) {
i = i + 2;
break;
}
}
while (i < K1) {//N1->N3
if (fabs(N1[i + 1]) > 1e-6) {
N3[k] = int(N1[i]);
N3[k + 1] = N1[i + 1];
i = i + 2;
k = k + 2;
}
}
printf("%d", k / 2);
for (i = 0; i < k; i = i + 2) {
printf(" %d %.1lf", int(N3[i]), N3[i + 1]);
}
system("pause");
return 0;
}
//我一开始是想用指数作为数组下标，那样会简单，参考答案的思路也都是如此，但我想让这个代码支持计算double指数，尝试增加了点难度
//测试点提示：没记错的话4/5是指数为0，6是结果只有一个0，0和2较简单，1和3卡了我最久，是逻辑上的错误，我用的测试数据是
/*
1 1000 0.7
5 1075 0.3 1000 -0.6 4 0.7 3 0.2 2 0.1
*/
//我一开始以为是没考虑到大指数，其实是当数组1算完了，数组2还有值的情况没考虑全，完全调通看到测试点全红的一刻简直爽爆
//好像各大OJ只有CF和TC会给测试数据，“不给测试数据更能锻炼思维的严谨性”，刚开始做的时候不太理解为什么从来看不到测试点信息，
//“对于一个oj来说，最值钱的不是这个系统，而是每道题的测试数据。”现在的我深以为然，另外，我这里也不提供简单版本的题解了
//因为下一道题，多项式乘法（这道题是多项式加法）我会用那种做法。