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 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
__________________________________________________________________________________________________________________________________________________
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 ;
- }
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