题目：

1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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 题目大意就是给定两个多项式 F1=aN1*x^N1+aN2*x^N2+aN3*x^N3+.....和F2=bN1*x^N1+bN2*x^N2+bN3*x^N3+..... 求两个多项式的和

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;template 
           
            class Link{public: T data1,data2; Link
            
              *next; Link(const T d1,const T d2,Link
             
               *nex=NULL) { data1=d1; data2=d2; next=nex; } Link() { data1=0; data2=0; next=NULL; }};template 
              
               class InkLink{private: Link
               
                 *head; int lon;public: InkLink() { head=new Link
                
                 ; lon=0; } void display2() { Link
                 
                   *p; p=head->next; cout << lon ; while(p!=NULL) { printf(" "); printf("%d %.1lf",(int)p->data1,p->data2); p=p->next; } cout <
                  
                   next==NULL) { head->next=new Link
                   
                    (value1,value2); lon++; return true; } Link
                    
                      *p,*q; p=head->next; q=head; while(p!=NULL&&((p->data1)>value1)) { q=p; p=p->next; } if(p==NULL||((p->data1)
                     
                      next=new Link
                      
                       (value1,value2,p); lon++; } else if(p->data1==value1) { p->data2+=value2; if(p->data2==0) { if(p->next!=NULL)q->next=p->next; else q->next=NULL; delete p; lon--; } } return true; } void clearLink() { Link
                       
                         *p,*q; p=head->next; q=p; while(p!=NULL) { p=p->next; delete q; q=p; } lon=0; head->next=NULL; }};int main(){ InkLink
                        
                          l; int n; double a,b; while(scanf("%d",&n)!=EOF) { for(int i=0; i
                        
                       
                      
                     
                    
                   
                  
                 
                
               
              
             
            
           
          
         
        
       
      

 这道题交了无数回，踩过了所有的坑。。。首先多项式的系数可以为负数，当系数减到0的时候要删掉这个节点。还有就是精确到小数点后一位。最后要控制格式，答案末尾直接输出回车，不要有空格。最后一组样例的结果应该是0，同样不能输出空格，没过的可能需要注意一下这点。