一元多項式加法計算問題

一元多項式加法計算問題

總體上來說一元多項式加法的計算問題難度一般並沒有什麼新的算法只是細節上需要注意，在加法上擴展的減法和乘除法都不算太難只需要在原代碼上進一步擴展就可以了
測試數據：
5,3 7,8 9,15 0,0
2,0 6,3 -7,8 0,0

3,1 5,3 7,8 9,15 0,0
2,0 6,3 -7,8 0,0

#include<iostream>
#include<stdlib.h>
using namespace std;
typedef struct Node
{
    int factor;
    int power;
    Node *next;
}Linklist;

void Combine(Linklist *&A, Linklist *&B)
{
    Linklist *p = A, *t = p->next, *q = B->next, *s;
    while (q->next != NULL)
    {
        if (q->power == t->power)
        {
            int sum = t->factor + q->factor;
            if (sum == 0)
            {
                p->next = t->next;
                free(t);
                t = p->next;
            }
            else
            {
                t->factor = sum;
                p = p->next;
                t = t->next;
            }
            q = q->next;
        }
        else if (q->power < t->power)
        {
            s = (Linklist *)malloc(sizeof(Linklist));
            s->next = NULL;
            s->factor = q->factor;
            s->power = q->power;
            s->next = p->next;
            p->next = s;
            p = s;
            q = q->next;
            continue;
        }
        else if (q->power > t->power)
        {
            if (q->power > t->power&&t->next->next == NULL)
            {
                s = (Linklist *)malloc(sizeof(Linklist));
                s->next = NULL;
                s->factor = q->factor;
                s->power = q->power;
                s->next = t->next;
                t->next = s;
                q = q->next;
            }
            t = t->next;
        }
    }

}
void Displist(Linklist *L)
{
    Linklist *p = L->next;
    while (p->next->next != NULL)
    {
        cout << p->factor << 'x' << '^' << p->power;
        if (p->next->factor > 0)
            cout << '+';
        p = p->next;
    }
    cout << p->factor << 'x' << '^' << p->power;
}
int main()
{
    int i;
    char c;
    Linklist *A, *B, *t, *p, *q, *s;

    A = (Linklist *)malloc(sizeof(Linklist));
    A->next = NULL;
    p = A;
    while (1)
    {
        t = (Linklist *)malloc(sizeof(Linklist));
        cin >> t->factor >> c >> t->power;
        p->next = t;
        p = t;
        if (p->factor == 0 && p->power == 0)
            break;
    }
    p->next = NULL;

    B = (Linklist *)malloc(sizeof(Linklist));
    B->next = NULL;
    q = B;
    while (1)
    {
        s = (Linklist *)malloc(sizeof(Linklist));
        cin >> s->factor >> c >> s->power;
        q->next = s;
        q = s;
        if (q->factor == 0 && q->power == 0)
            break;
    }
    q->next = NULL;

    Combine(A, B);
    Displist(A);
    return 0;
}