1002. A+B for Polynomials (25)

题目链接：https://www.patest.cn/contests/pat-a-practise/1002

原题如下：

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 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2
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　　这道题我感觉就是一元多项式的加法，因此用了之前文章中的方法，集用两个链表分别存储两行元素，每次比较然后相加，可是有三个测试点超时了……代码如下：

 #include<stdio.h>
 #include<stdlib.h>

 typedef struct Node{
     struct Node *Next;
     int expon;
     float coef;
 }PNode;

 void Insert(PNode *P,PNode **PtrRear)
 {
     //printf("er");
     PNode *tmp=(PNode *)malloc(sizeof(struct Node));
     tmp->expon=P->expon;tmp->coef=P->coef;tmp->Next=NULL;
     (*PtrRear)->Next=tmp;
     *PtrRear=tmp;
     //printf("sd");
     return ;
 }

 PNode * ReadP(int K)
 {
     int i,e;
     float c;
     PNode *P1=(PNode *)malloc(sizeof (struct Node)); P1->Next=NULL;
     PNode *tmp=(PNode *)malloc(sizeof (struct Node));tmp->Next=NULL;
     tmp=P1;
     for (i=;i<K;i++)
     {
         scanf("%d %f",&e, &c);
         PNode *P=(PNode*)malloc(sizeof (struct Node));
         P->expon =e;P->coef=c;P->Next=NULL;
         P1->Next=P;
         P1=P;
     }
     tmp=tmp->Next;

     return tmp;
 }

 int main()
 {
     int K1,K2,i;
     PNode *rear,*front,*tmp;
     rear=(PNode *)malloc(sizeof (struct Node));front=rear;
     PNode *P1=(PNode *)malloc(sizeof (struct Node));
     PNode *P2=(PNode *)malloc(sizeof (struct Node));

     scanf("%d",&K1);P1=ReadP(K1);
     scanf("%d",&K2);P2=ReadP(K2); //printf("\n%d %.1lf %d %.1lf",P1->expon,P1->coef,P1->Next->expon,P1->Next->coef); printf("\n%d %.1lf %d %.1lf\n",P2->expon,P2->coef,P2->Next->expon,P2->Next->coef);

     int cnt=;
     while (P1 && P2)
     {
         if (P1->expon>P2->expon)
         {
         Insert(P1,&rear);
         P1=P1->Next;
         cnt++;
         }
         else if (P1->expon<P2->expon)
         {
             Insert(P2,&rear);
             P2=P2->Next;
             cnt++;
         }
         else if (P1->expon==P2->expon)
         {
             if (P1->coef+P2->coef)
             {
                 PNode *tmp=(PNode *)malloc(sizeof (struct Node));
                 tmp->expon=P1->expon;tmp->coef=P1->coef+P2->coef;
                 tmp->Next=NULL;
                 Insert(tmp,&rear);
                 P1=P1->Next;P2=P2->Next;
                 cnt++;
             }
         }
     }
     while (P1){Insert(P1,&rear);P1=P1->Next;cnt++;}
     while (P2){Insert(P2,&rear);P2=P2->Next;cnt++;}

     int flag=;
     tmp=front;
     front=front->Next;
     while (front)
     {
         if (!flag)
         {printf("%d %d %.1f",cnt,front->expon,front->coef);front=front->Next;flag=;}
         else
         {
             printf(" %d %.1f",front->expon,front->coef);front=front->Next;
         }
     }
     free(tmp);
     return ;
 }

希望哪位朋友能帮忙看下问题在哪……

在网上看了其他人的代码，真是简单精妙啊，直接开一个数组，指数即为数值下标，每个数组的值即为相应的每个指数对应系数的值，代码如下：

#include<stdio.h>
#define MaxN 1001

int main()
{
    int K,i,e;
    int line=;
    int cnt=;
    int Maxe=;
    double Input[MaxN]={};
    double c;
    while (line)
    {scanf("%d",&K);
    for (i=;i<K;i++)
    {
        scanf("%d %lf",&e,&c);
        Input[e]+=c;
        if (e>Maxe)Maxe=e;
    }
    line--;
    }

    for (i=Maxe;i>=;i--)
    {
        if (Input[i]!=)cnt++;
    }

    printf("%d",cnt);
    for (i=Maxe;i>=;i--)
    {
        if (Input[i]!=)printf(" %d %.1lf",i,Input[i]);
    }
    return ;
}