PAT1002. A+B for Polynomials (25)

题目如下:

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

题意就是要实现多项式的相加。那么我首先想到用链表来做，每个结点储存指数，系数以及指向下一个结点的指针，先将第一行数据构造成一条链表，然后再读第二行数据，遇到指数比当前结点大的，则在该结点之前插入新结点；遇到指数和当前结点一样的，那么直接相加；遇到指数比当前结点小的，则指针向后指；如果遇到空，则在后面创建新结点。大致代码如下:

#include <iostream>
#include <iomanip>
using namespace std;

struct Polynomial {
int exponent;
double coefficient;
Polynomial* next;
};

int main()
{
Polynomial* head = new Polynomial;
int exponent;
double coefficient;
head->next = nullptr;
Polynomial *p = head;
int k, k1, k2; //k记录相加后的多项式项数
cin >> k1;
k = k1;
for (int i = 0; i < k1; i++)
{
cin >> exponent >> coefficient;
p->next = new Polynomial;
p = p->next;
p->exponent = exponent;
p->coefficient = coefficient;
p->next = nullptr;
}

cin >> k2;
p = head;
for (int i = 0;i < k2;i++)
{
cin >> exponent >> coefficient;
if (p->next == nullptr)
{
p->next = new Polynomial;
p = p->next;
p->exponent = exponent;
p->coefficient = coefficient;
p->next = nullptr;
k++;
}
else
{
while (p->next != nullptr && exponent < p->next->exponent ) //当前读的指数小于当前结点且非空，则一直向后。
{
p = p->next;
}

if (p->next == nullptr)
{
p->next = new Polynomial;
p = p->next;
p->exponent = exponent;
p->coefficient = coefficient;
p->next = nullptr;
k++;
}
else if (exponent > p->next->exponent) //当前读的指数比链表上当前结点的大，那么需要插在该结点前面。
{
Polynomial* temp = p->next;
p->next = new Polynomial;
p->next->exponent = exponent;
p->next->coefficient = coefficient;
p = p->next;
p->next = temp;
k++;
}
else if (exponent == p->next->exponent) //指数相等则指数不变系数相加
{
p->next->coefficient += coefficient;
p = p->next;
}

}
}

cout << k;
p = head;
while (p->next != nullptr)
{
p = p->next;
cout << " " << p->exponent << " ";
cout << setiosflags(ios::fixed) << setprecision(1) << p->coefficient; //保留一位小数
}

return 0;
}

但是这样运行用只有基本的测试点通过了，后面4个测试点答案错误：



那么我再仔细想了想，发现忽略了当指数相同时，若系数相加为0，则该项需要删掉的情况，经过修改得到了现在的代码：

#include <iostream>
#include <iomanip>
using namespace std;

struct Polynomial {
int exponent;
double coefficient;
Polynomial* next;
};

int main()
{
Polynomial* head = new Polynomial;
int exponent;
double coefficient;
head->next = nullptr;
Polynomial *p = head;
int k, k1, k2; //k记录相加后的多项式项数
cin >> k1;
k = k1;
for (int i = 0; i < k1; i++)
{
cin >> exponent >> coefficient;
p->next = new Polynomial;
p = p->next;
p->exponent = exponent;
p->coefficient = coefficient;
p->next = nullptr;
}

cin >> k2;
p = head;
for (int i = 0;i < k2;i++)
{
cin >> exponent >> coefficient;
if (p->next == nullptr)
{
p->next = new Polynomial;
p = p->next;
p->exponent = exponent;
p->coefficient = coefficient;
p->next = nullptr;
k++;
}
else
{
while (p->next != nullptr && exponent < p->next->exponent ) //当前读的指数小于当前结点且非空，则一直向后。
{
p = p->next;
}

if (p->next == nullptr)
{
p->next = new Polynomial;
p = p->next;
p->exponent = exponent;
p->coefficient = coefficient;
p->next = nullptr;
k++;
}
else if (exponent > p->next->exponent) //当前读的指数比链表上当前结点的大，那么需要插在该结点前面。
{
Polynomial* temp = p->next;
p->next = new Polynomial;
p->next->exponent = exponent;
p->next->coefficient = coefficient;
p = p->next;
p->next = temp;
k++;
}
else if (exponent == p->next->exponent) //指数相等则指数不变系数相加
{
p->next->coefficient += coefficient;
if (p->next->coefficient == 0) //如果相加后为0 则要删除该项
{
Polynomial* temp = p->next;
p->next = temp->next;
delete temp;
k--;
}
else
p = p->next;
}

}
}

cout << k;
p = head;
while (p->next != nullptr)
{
p = p->next;
cout << " " << p->exponent << " ";
cout << setiosflags(ios::fixed) << setprecision(1) << p->coefficient; //保留一位小数
}

return 0;
}

经过这样的修改，就通过了所有的测试点。