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 3.22 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
    
     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
     
      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
      
       >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
       
        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<
        
         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
    
     #include