一、题意理解
设计函数分别求两个一元多项式的乘积与和，例：
[ ext{已知以下两个多项式：} \
egin{align}
& 3x^4-5x^2+6x-2 \
& 5x^{20}-7x^4+3x
end{align}
]
[ ext{多项式和为：} \
egin{align}
5x^{20}-4x^4-5x^2+9x-2
end{align}
]
假设多项式的乘积为((a+b)(c+d)=ac+ad+bc+bd)，则多项式的乘积如下：
[egin{align}
15x^{24}-25x^{22}+30x^{21}-10x^{20}-21x^8+35x^6-33x^5+14x^4-15x^3+18x^2-6x
end{align}
]
通过上述题意理解，我们可以设计函数分别求两个一元多项式的乘积与和。
输入样例：
[egin{align}
& 3x^4-5x^2+6x-2 quad --> quad ext{4个}\,3\,4\,-5\,2\,6\,1\,-2\,0 \
& 5x^{20}-7x^4+3x quad --> quad ext{3个}\,5\,20\,-7\,4\,3\,1 \
end{align} \
]
输出样例：
[egin{align}
& 15x^{24}-25x^{22}+30x^{21}-10x^{20}-21x^8+35x^6-33x^5+14x^4-15x^3+18x^2-6x \
& 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
end{align}
]
二、求解思路
- 多项式表示
- 程序框架
- 读多项式
- 加法实现
- 乘法实现
- 多项式输出
三、多项式的表示
仅表示非零项
3.1 数组
优点：编程简单、调试简单
缺点：需要事先确定数组大小
一种比较好的实现方法是：动态数组（动态更改数组的大小）
3.2 链表
优点：动态性强
缺点：编程略为复杂、调试比较困难
数据结构设计：
/* c语言实现 */
typedef struct PolyNode *Polynomial;
struct PolyNode{
int coef;
int expon;
Polynomial link;
}
四、程序框架搭建
/* c语言实现 */
int main()
{
读入多项式1;
读入多项式2;
乘法运算并输出;
加法运算并输出;
return 0;
}
int main()
{
Polynomial P1, P2, PP, PS;
P1 = ReadPoly();
P2 = ReadPoly();
PP = Mult(P1, P2);
PrintPoly(PP);
PS = Add(P1, P2);
PrintPoly(PS);
return 0;
}
需要设计的函数：
- 读一个多项式
- 两多项式相乘
- 两多项式相加
- 多项式输出
五、如何读入多项式
/* c语言实现 */
Polynomial ReadPoly()
{
...;
scanf("%d", &N);
...;
while (N--) {
scanf("%d %d", &c, &e);
Attach(c, e, &Rear);
}
...;
return P;
}
Rear初值是多少？
两种处理方法：
- Rear初值为NULL：在Attach函数中根据Rear是否为NULL做不同处理
- Rear指向一个空结点
/* c语言实现 */
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 */
/* c语言实现 */
Polynomial ReadPoly()
{
Polynomial P, Rear, t;
int c, e, N;
scanf("%d", &N);
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;
}
六、如何将两个多项式相加
/* c语言实现 */
Polynomial Add(Polynomial P1, Polynomial P2)
{
...;
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNode));
P->link = NULL;
Rear = P;
while (t1 && t2){
if (t1->expon == t2->expon){
...;
}
else if (t1->expon > t2->expon){
...;
}
else{
...;
}
}
while (t1){
...;
}
while (t2){
...;
}
...;
return P;
}
七、如何将两个多项式相乘
方法：
- 将乘法运算转换为加法运算
将P1当前项(ci, ei)乘P2多项式，再加到结果多项式里
/* c语言实现 */
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNode)); P->link = NULL;
Rear = P;
while (t2){
Attach(t1->coef * t2->coef, t1->expon + t2->expon, &Rear);
t2 = t2->link;
}
- 逐项插入
将P1当前项(c1_i, e1_i)乘P2当前项(c2_i, e2_i)，并插入到结果多项式中。关键是要找到插入位置
初始结果多项式可由P1第一项乘P2获得（如上）
/* c语言实现 */
Polynomial Mult(Polynomial P1, Polynomial P2)
{
...;
t1 = P1; t2 = P2;
...;
while (t2){ // 先用P1的第一项乘以P2，得到P
...;
}
t1 = t1->link;
while (t1){
t2 = P2; Rear = P;
while (t2){
e = t1->expon + t2->expon;
c = t1->coef * t2->coef;
...;
t2 = t2->link;
}
t1 = t1->link;
}
...;
}
/* c语言实现 */
Polynomial Mult(Polynomial P1, Polynomial P2)
{
Polynomial P, Rear, t1, t2, t;
int c, e;
if (!P1 || !P2) return NULL;
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNOde)); P->link = NULL;
Rear = P;
while (t2){ // 先用P1的第一项乘以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;
...;
t2 = t2->link;
}
t1 = t1->link;
}
...;
}
/* c语言实现 */
Polynomial Mult(Polynomial P1, Polynomial P2)
{
Polynomial P, Rear, t1, t2, t;
int c, e;
if (!P1 || !P2) return NULL;
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNOde)); P->link = NULL;
Rear = P;
while (t2){ // 先用P1的第一项乘以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 = t2->coef * t2->coef;
while (Rear->link && Rear->link->expon > e)
Rear = Rear->link;
if (Rear->link && Rear->link->expon == e){
...;
}
else{
...;
}
t2 = t2->link;
}
t1 = t1->link;
}
...;
}
/* c语言实现 */
Polynomial Mult(Polynomial P1, Polynomial P2)
{
Polynomial P, Rear, t1, t2, t;
int c, e;
if (!P1 || !P2) return NULL;
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNOde)); P->link = NULL;
Rear = P;
while (t2){ // 先用P1的第一项乘以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 = t2->coef * t2->coef;
while (Rear->link && Rear->link->expon > e)
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;
}
...;
}
/* c语言实现 */
Polynomial Mult(Polynomial P1, Polynomial P2)
{
Polynomial P, Rear, t1, t2, t;
int c, e;
if (!P1 || !P2) return NULL;
t1 = P1; t2 = P2;
P = (Polynomial)malloc(sizeof(struct PolyNOde)); P->link = NULL;
Rear = P;
while (t2){ // 先用P1的第一项乘以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 = t2->coef * t2->coef;
while (Rear->link && Rear->link->expon > e)
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;
}
八、如何将多项式输出
/* c语言实现 */
void PrintPoly(Polynomial P)
{
// 输出多项式
int flag = 0; // 辅助调整输出格式用，判断输出加法还是乘法
if (!P) {printf("0 0
"); return ;}
while (P) {
if (!flag)
flag = 1;
else
printf(" ");
printf("%d %d", P->coef, P->expon);
P = P->link;
}
printf("
");
}