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=0;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("
%d %.1lf %d %.1lf",P1->expon,P1->coef,P1->Next->expon,P1->Next->coef); printf("
%d %.1lf %d %.1lf
",P2->expon,P2->coef,P2->Next->expon,P2->Next->coef);

    int cnt=0;
    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=0;
    tmp=front;
    front=front->Next;
    while (front)
    {
        if (!flag)
        {printf("%d %d %.1f",cnt,front->expon,front->coef);front=front->Next;flag=1;}
        else
        {
            printf(" %d %.1f",front->expon,front->coef);front=front->Next;
        }
    }
    free(tmp);
    return 0;
}

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=2;
    int cnt=0;
    int Maxe=0;
    double Input[MaxN]={0};
    double c;
    while (line)
    {scanf("%d",&K);
    for (i=0;i<K;i++)
    {
        scanf("%d %lf",&e,&c);
        Input[e]+=c;
        if (e>Maxe)Maxe=e;
    }
    line--;
    }

    for (i=Maxe;i>=0;i--)
    {
        if (Input[i]!=0)cnt++;
    }

    printf("%d",cnt);
    for (i=Maxe;i>=0;i--)
    {
        if (Input[i]!=0)printf(" %d %.1lf",i,Input[i]);
    }
    return 0;
}