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浙大PAT甲級-1002

這道題是大一的時候就做過的,當時剛學運算子過載。現在我用的是連結串列來做的,也算是複習一下快忘光的資料結構吧。
#include <iostream>
#include<iomanip>
using namespace std;

//節點(多項式的項)
struct listNode{
	int exp;
	double coe;
	listNode *link;
	listNode(){}
	listNode(int e, double c){
		exp = e;
		coe = c;
		link = NULL;
	}
};

class list{
public:
	listNode *first;
	listNode *last;   //方便後面直接用就沒設為private啦
	list(){

		first = NULL;
		last = NULL;
	}
	//輸入,初始化連結串列
	void init(int n){
		int e;
		double c;
		for (int i = 0; i<n; i++){
			cin >> e >> c;
			insert(e, c);
		}
	}
    //插入節點
	void insert(int e, double c){
		listNode *newNode = new listNode(e,c);
		if (first == NULL){
			newNode->link = first;
			first =  newNode;
			last = newNode;
		}
		else{
			last->link = newNode;
			last = newNode;
		}
		last->link = NULL;
	}
	//將係數為0的節點都剪掉
	void merge(){
		while (first!=NULL && first->coe == 0){
			first = first->link;
		}
		if (first != NULL){
			listNode *m = first->link;
			listNode *n = first;
			while (m != NULL){
				if (m->coe == 0){
					n->link = m->link;
				}
				else{
					n = n->link;
				}
				m = m->link;
			}
		}
	}
	//連結串列長度
	int Length(){
		listNode *p = first;
		int cnt = 0;
		while (p != NULL){
			p = p->link;
			cnt++;
		}
		return cnt;
	}
};
int main()
{
	int p, q;
	while (cin >> p){
		list a, b, c;
		a.init(p);
		cin >> q;
		b.init(q);

		//多項式加法
		listNode *ca = a.first;
		listNode *cb = b.first;
		while (ca != NULL&&cb != NULL){
			if (ca->exp > cb->exp){
				c.insert(ca->exp, ca->coe);
				ca = ca->link;
			}
			else if (ca->exp == cb->exp){
				c.insert(ca->exp, ca->coe + cb->coe);
				ca = ca->link;
				cb = cb->link;
			}
			else{
				c.insert(cb->exp, cb->coe);
				cb = cb->link;
			}
		}

		if (ca != NULL){
			while (ca != NULL){
				c.insert(ca->exp, ca->coe);
				ca = ca->link;
			}
		}
		else if (cb != NULL){
			while (cb != NULL){
				c.insert(cb->exp, cb->coe);
				cb = cb->link;
			}
		}

		c.merge();
		cout << c.Length();
		listNode *m = c.first;
		while (m != NULL){
			cout << " " << m->exp << " " << setiosflags(ios::fixed)
				<< setprecision(1) << m->coe;//輸出是有精度要求的哦
			m = m->link;
		}
		cout << endl;
	}
	return 0;
}