PAT 1002. A+B for Polynomials

PAT 1002. A+B for Polynomials

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

代码如下(复杂的链表）

#include<iostream>
using namespace std;
struct node{
	int exp;
	double coe;
	node* next;
};
int main(){
	int N,cnt=0;
	cin>>N;
	node* l1=new node();
	node* ptr1=l1;
	for(int i=0;i<N;i++){
		node *lnode=new node();
		lnode->next=NULL;
		cin>>lnode->exp>>lnode->coe;
		ptr1->next=lnode;
		ptr1=ptr1->next;
	}
    cin>>N;
    node* l2=new node();
    node* ptr2=l2;
    for(int i=0;i<N;i++){
		node *lnode=new node();
		cin>>lnode->exp>>lnode->coe;
		lnode->next=NULL;
		ptr2->next=lnode;
		ptr2=ptr2->next;
	}
    ptr1=l1->next; ptr2=l2->next;
    node* l=new node(); node* ptr=l;
    while(ptr1!=NULL&&ptr2!=NULL){
    	if(ptr1->exp>ptr2->exp){
    		if(ptr1->coe!=0) cnt++;
    		ptr->next=ptr1;
    		ptr=ptr->next;
    		ptr1=ptr1->next;
		}
		else if(ptr1->exp<ptr2->exp){
    		if(ptr2->coe!=0) cnt++;
    		ptr->next=ptr2;
    		ptr=ptr->next;
    		ptr2=ptr2->next;
		}else{
			ptr1->coe+=ptr2->coe;
			if(ptr1->coe!=0) cnt++;
			ptr->next=ptr1;
    		ptr=ptr->next;
    		ptr1=ptr1->next;
    		ptr2=ptr2->next;	
		}	
	}
	while(ptr1!=NULL){
		if(ptr1->coe!=0) cnt++;
		ptr->next=ptr1;
		ptr=ptr->next;
		ptr1=ptr1->next;
	}	
		while(ptr2!=NULL){
		if(ptr2->coe!=0) cnt++;
		ptr->next=ptr2;
		ptr=ptr->next;
		ptr2=ptr2->next;
	    }
	ptr=l->next;
	cout<<cnt;
	while(ptr!=NULL){
		if(ptr->coe!=0)
		printf(" %d %.1f",ptr->exp,ptr->coe);
		ptr=ptr->next; 
	}
}

分析

其实用其他容器都可以很好的解决这个问题，vector,map等等；下面附上我的另一种解法

代码如下

#include<iostream>
#include<map>
using namespace std;
int main(){
	map<int,double> map1,map2;
	int N,exp,cnt=0;
	double coe;
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>exp>>coe;
		map1[exp]=coe;
	}
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>exp>>coe;
		map2[exp]=coe;
	}
    auto p1=map1.begin();
    for(;p1!=map1.end();p1++){
    	map2[p1->first]+=p1->second;
	}
	for(auto p2=map2.rbegin();p2!=map2.rend();p2++){
		if(p2->second!=0)
		cnt++;
	}
	cout<<cnt;
	for(auto p2=map2.rbegin();p2!=map2.rend();p2++){
		if(p2->second!=0)
		printf(" %d %.1f",p2->first,p2->second);
	}
}