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题目:设计函数分别求两个一元多项式的乘积与和。
输入格式:
输入分2行,每行分别先给出多项式非零项的个数,再以指数递降方式输入一个多项式非零项系数和指数(绝对值均为不超过1000的整数)。数字间以空格分隔。
输出格式:
输出分2行,分别以指数递降方式输出乘积多项式以及和多项式非零项的系数和指数。数字间以空格分隔,但结尾不能有多余空格。零多项式应输出0 0
。
输入样例:
4 3 4 -5 2 6 1 -2 0
3 5 20 -7 4 3 1
输出样例:
15 24 -25 22 30 21 -10 20 -21 8 35 6 -33 5 14 4 -15 3 18 2 -6 1
5 20 -4 4 -5 2 9 1 -2 0
解答:
首先我们先构建出程序框架:
#include<iostream>
using namespace std;
int main(){
Polynomial P1,P2,PP,PS;
P1 = ReadPoly();
P1 = ReadPoly();
PP = Mult(P1,P2)
PrintPoly(PP);
PS = Add(P1,P2);
PrintPoly(PS);
return 0;
}
各个函数名字一看就很明确,这里就不说了。
因为视频里对乘法介绍很详细,没有视频可以去看看课件,其实有点基础的看看课件就都懂了。读入和打印也非常详细,所以这里只说说加法:
即通过循环遍历,因为是指数从大到小排列的,所以当其中一个多项式遍历到NULL时,说明另一个多项式末尾都可以直接挂到目标链表上。
若都不为空,当其中一个指数大于另一个,我们就把大的这个添加给目标链表,然后推进。如果指数相同,还得判断是不是同样大小。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
using namespace std;
typedef struct PolyNode *Polynomial;
struct PolyNode {
int coef;
int expon;
Polynomial link;
};
void Attach(int c, int e, Polynomial *pRear);
Polynomial ReadPoly();
void PrintPoly(Polynomial P);
Polynomial Mult(Polynomial P1, Polynomial P2);
Polynomial Add(Polynomial P1, Polynomial P2);
int main() {
Polynomial P1, P2, PP, PS;
P1 = ReadPoly();
P2 = ReadPoly();
PP = Mult(P1, P2);
PrintPoly(PP);
PS = Add(P1, P2);
PrintPoly(PS);
system("pause");
return 0;
}
void Attach(int c, int e, Polynomial *pRear)
{
Polynomial P;
P = (Polynomial)malloc(sizeof(struct PolyNode));
P->coef = c; // 对新结点赋值
P->expon = e;
P->link = NULL;
(*pRear)->link = P;
*pRear = P; // 修改pRear值
}
Polynomial ReadPoly()
{
Polynomial P, Rear, t;
int c, e, N;
scanf("%d", &N);
//临时创建一个头结点,以后还要删除
P = (Polynomial)malloc(sizeof(struct PolyNode)); // 链表头空结点
P->link = NULL;
Rear = P;
while (N--) {
scanf("%d %d", &c, &e);
Attach(c, e, &Rear); // 将当前项插入多项式尾部
}
t = P; P = P->link; free(t); // 删除临时生成的头结点
return P;
}
void PrintPoly(Polynomial P)
{ // 输出多项式
int flag = 0; // 辅助调整输出格式用
if (!P) { printf("0 0\n"); return; }
while (P) {
if (!flag)
flag = 1;
else
printf(" ");
printf("%d %d", P->coef, P->expon);
P = P->link;
}
printf("\n");
}
Polynomial Mult(Polynomial P1, Polynomial P2)
{
Polynomial P, Rear, t1, t2, t;
int c, e;
if (!P1 || !P2) return NULL;//如果有一个为NULL,就得返回了。因为没法做乘法。
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNode)); P->link = NULL;
Rear = P;
while (t2) { // 先用P1的第1项乘以P2,得到P
Attach(t1->coef*t2->coef, t1->expon + t2->expon, &Rear);
t2 = t2->link;
}
t1 = t1->link;
while (t1) {
t2 = P2; Rear = P;
while (t2) {
e = t1->expon + t2->expon;
c = t1->coef * t2->coef;
//先判断是不是指向NULL
while (Rear->link && Rear->link->expon > e) //从合成项中从前往后遍历,一直到不小于它的值
Rear = Rear->link;
if (Rear->link && Rear->link->expon == e) { //
if (Rear->link->coef + c)
Rear->link->coef += c;
else {
t = Rear->link;
Rear->link = t->link;
free(t);
}
}
else {
t = (Polynomial)malloc(sizeof(struct PolyNode));
t->coef = c; t->expon = e;
t->link = Rear->link;
Rear->link = t; Rear = Rear->link;
}
t2 = t2->link;
}
t1 = t1->link;
}
t2 = P; P = P->link; free(t2);
return P;
}
Polynomial Add(Polynomial P1, Polynomial P2) {
Polynomial P, Rear, t1, t2, t;
int c, e;
if (!P1 && !P2) return NULL;//如果都为NULL,就得返回了。因为没法做加法。
t1 = P1;
t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNode)); P->link = NULL;
Rear = P;
//注意该链表没有头结点
while (t1 !=NULL && t2 != NULL) {
if (t1->expon < t2->expon)
{
Rear->link = t2;
t2 = t2->link;
Rear = Rear->link;
}
else if (t2->expon < t1->expon) {
Rear->link = t1;
t1 = t1->link;
Rear = Rear->link;
}
else //相等
{
t2->coef += t1->coef;
if (t2->coef != 0) //为0就抵消了
{
Rear->link = t2;
t2 = t2->link;
Rear = Rear->link;
}
else {
t = t2;
t2 = t2->link;
free(t);
}
t = t1;
t1 = t1->link;
free(t);//之前不小心写成了 free(t1)。结果一直段错误!真是郁闷。
}
}
if (t1 == NULL) {
Rear->link = t2;
}
else if(t2 == NULL){
Rear->link = t1;
}
t2 = P; P = P->link; free(t2);
return P;
}
测试全部通过。