1002. A+B for Polynomials (25)

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分类专栏： PAT甲级练习文章标签： PAT C语言

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本文介绍了一种使用链表实现多项式加法的方法。通过定义链表节点结构，初始化链表并输入两个多项式的系数和指数，然后将两个多项式相加，并输出结果。该方法适用于计算机科学和编程教育。

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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 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

链表版：

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

struct Node;
typedef struct Node* PtrNode;
struct Node {
float coefficient;
int exponent;
PtrNode next;
int Num;
};

PtrNode Init(int N);
PtrNode Sum(PtrNode A, PtrNode B);

int main()
{
PtrNode A, B, C;
int N1,N2,N;
scanf("%d", &N1);
A=Init(N1);
scanf("%d", &N2);
B = Init(N2);
C=Sum(A, B);
printf("%d", C->Num);
N = C->Num;
if (C->next != NULL) {
C = C->next;
for (int i = 0; i < N; i++) {
printf(" %d", C->exponent);
printf(" %.1f", C->coefficient);
C = C->next;
}
}
return 0;
}

PtrNode Init(int N)
{
PtrNode Head = (PtrNode)malloc(sizeof(struct Node));
PtrNode Tmp = Head;
Head->coefficient = Head->exponent = -1;
Head->Num = N;
Head->next = NULL;
for (int i = 0; i < N; i++) {
PtrNode p = (PtrNode)malloc(sizeof(struct Node));
scanf("%d", &p->exponent);
scanf("%f", &p->coefficient);
p->next = NULL;
Tmp->next = p;
Tmp = p;
}
return Head;
}

PtrNode Sum(PtrNode A, PtrNode B)
{
A = A->next; B = B->next;
PtrNode Head = (PtrNode)malloc(sizeof(struct Node));
Head->coefficient = -1;
Head->exponent = -1;
Head->next = NULL;
Head->Num = 0;
PtrNode tmp = Head;
while (A != NULL && B != NULL) {
if (A->exponent > B->exponent) {
PtrNode p = (PtrNode)malloc(sizeof(struct Node));
p->coefficient = A->coefficient;
p->exponent = A->exponent;
p->next = NULL;
tmp->next = p;
tmp = p;
A = A->next;
Head->Num++;
}
else if (A->exponent < B->exponent) {
PtrNode p = (PtrNode)malloc(sizeof(struct Node));
p->coefficient = B->coefficient;
p->exponent = B->exponent;
p->next = NULL;
tmp->next = p;
tmp = p;
B = B->next;
Head->Num++;
}
else {
if (A->coefficient + B->coefficient != 0) {
PtrNode p = (PtrNode)malloc(sizeof(struct Node));
p->coefficient = A->coefficient + B->coefficient;
p->exponent = A->exponent;
p->next = NULL;
tmp->next = p;
tmp = p;
Head->Num++;
}
A = A->next;
B = B->next;
}
}
if (A == NULL && B != NULL) {
while (B != NULL) {
PtrNode p = (PtrNode)malloc(sizeof(struct Node));
p->coefficient = B->coefficient;
p->exponent = B->exponent;
p->next = NULL;
tmp->next = p;
tmp = p;
B = B->next;
Head->Num++;
}
}
if (A != NULL &&B == NULL) {
while (A != NULL) {
PtrNode p = (PtrNode)malloc(sizeof(struct Node));
p->coefficient = A->coefficient;
p->exponent = A->exponent;
p->next = NULL;
tmp->next = p;
tmp = p;
A = A->next;
Head->Num++;
}
}
return Head;
}