小白專場-多項式乘法與加法運算-c語言實現
阿新 • • 發佈：2019-09-04
目錄
- 一、題意理解
- 二、求解思路
- 三、多項式的表示
- 3.1 陣列
- 3.2 連結串列
- 四、程式框架搭建
- 五、如何讀入多項式
- 六、如何將兩個多項式相加
- 七、如何將兩個多項式相乘
- 八、如何將多項式輸出
一、題意理解
設計函式分別求兩個一元多項式的乘積與和，例：
\[
\text{已知以下兩個多項式：} \\
\begin{align}
& 3x^4-5x^2+6x-2 \\
& 5x^{20}-7x^4+3x
\end{align}
\]
假設多項式的乘積為\((a+b)(c+d)=ac+ad+bc+bd\)，則多項式的乘積如下：
\[
\begin{align}
15x^{24}-25x^{22}+30x^{21}-10x^{20}-21x^8+35x^6-33x^5+14x^4-15x^3+18x^2-6x
\end{align}
\]
通過上述題意理解，我們可以設計函式分別求兩個一元多項式的乘積與和。
輸入樣例：
\[
\begin{align}
& 3x^4-5x^2+6x-2 \quad --> \quad \text{4個}\,3\,4\,-5\,2\,6\,1\,-2\,0 \\
& 5x^{20}-7x^4+3x \quad --> \quad \text{3個}\,5\,20\,-7\,4\,3\,1 \\
\end{align} \\
\]
\[ \begin{align} & 15x^{24}-25x^{22}+30x^{21}-10x^{20}-21x^8+35x^6-33x^5+14x^4-15x^3+18x^2-6x \\ & 15 \, 24 \, -25 \, 22 \, 30 \, 21 \, -10 \, 20 \, -21 \, 8 \, 35 \, 6 \, -33 \, 5 \, 14 \, 4 \, -15 \, 3 \, 18 \, 2 \, -6 \, 1 \, 5 \, 20 \, -4 \, 4 \, -5 \, 2 \, 9 \, 1 \, -2 \, 0 \end{align} \]
二、求解思路
- 多項式表示
- 程式框架
- 讀多項式
- 加法實現
- 乘法實現
- 多項式輸出
三、多項式的表示
僅表示非零項
3.1 陣列
優點：程式設計簡單、除錯簡單
缺點：需要事先確定陣列大小
一種比較好的實現方法是：動態陣列（動態更改陣列的大小）
3.2 連結串列
優點：動態性強
缺點：程式設計略為複雜、除錯比較困難
資料結構設計：
/* c語言實現 */
typedef struct PolyNode *Polynomial;
struct PolyNode{
int coef;
int expon;
Polynomial link;
}
四、程式框架搭建
/* c語言實現 */
int main()
{
讀入多項式1;
讀入多項式2;
乘法運算並輸出;
加法運算並輸出;
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;
}需要設計的函式：
- 讀一個多項式
- 兩多項式相乘
- 兩多項式相加
- 多項式輸出
五、如何讀入多項式
/* c語言實現 */
Polynomial ReadPoly()
{
...;
scanf("%d", &N);
...;
while (N--) {
scanf("%d %d", &c, &e);
Attach(c, e, &Rear);
}
...;
return P;
}
Rear初值是多少？
兩種處理方法：
- Rear初值為NULL：在Attach函式中根據Rear是否為NULL做不同處理
- Rear指向一個空結點
/* c語言實現 */
void Attach(int c, int e, Polynomial *pRear)
{
Polynomial P;
P = (Polynomial)malloc(sizeof(struct PolyNode));
p->coef = c; /* 對新結點賦值 */
p->expon = e;
p->link = NULL;
(*pRear)->link = P;
(*pRear) = P; /* 修改pRear值 */
/* c語言實現 */
Polynomial ReadPoly()
{
Polynomial P, Rear, t;
int c, e, N;
scanf("%d", &N);
P = (Polynomial)malloc(sizeof(struct PolyNode)); // 連結串列頭空結點
P->link = NULL;
Rear = P;
while (N--) {
scanf("%d %d", &c, &e);
Attach(c, e, &Rear); // 將當前項插入多項式尾部
}
t = P; P = P->link; free(t); // 刪除臨時生成的頭結點
return P;
}
六、如何將兩個多項式相加
/* c語言實現 */
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、如何將兩個多項式相乘
方法：
- 將乘法運算轉換為加法運算
將P1當前項(ci, ei)乘P2多項式，再加到結果多項式裡
/* c語言實現 */
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;
}- 逐項插入
將P1當前項(c1_i, e1_i)乘P2當前項(c2_i, e2_i)，並插入到結果多項式中。關鍵是要找到插入位置
初始結果多項式可由P1第一項乘P2獲得（如上）
/* c語言實現 */
Polynomial Mult(Polynomial P1, Polynomial P2)
{
...;
t1 = P1; t2 = P2;
...;
while (t2){ // 先用P1的第一項乘以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語言實現 */
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的第一項乘以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語言實現 */
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的第一項乘以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語言實現 */
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的第一項乘以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語言實現 */
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的第一項乘以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、如何將多項式輸出
/* c語言實現 */
void PrintPoly(Polynomial P)
{
// 輸出多項式
int flag = 0; // 輔助調整輸出格式用，判斷輸出加法還是乘法
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");
}