小白专场-多项式乘法与加法运算-c语言实现
作者:互联网
目录
- 一、题意理解
- 二、求解思路
- 三、多项式的表示
- 3.1 数组
- 3.2 链表
- 四、程序框架搭建
- 五、如何读入多项式
- 六、如何将两个多项式相加
- 七、如何将两个多项式相乘
- 八、如何将多项式输出
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:
\[\text{已知以下两个多项式:} \\ \begin{align} & 3x^4-5x^2+6x-2 \\ & 5x^{20}-7x^4+3x \end{align} \]
\[\text{多项式和为:} \\ \begin{align} 5x^{20}-4x^4-5x^2+9x-2 \end{align} \]
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:
\[\begin{align} 15x^{24}-25x^{22}+30x^{21}-10x^{20}-21x^8+35x^6-33x^5+14x^4-15x^3+18x^2-6x \end{align} \]
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:
\[\begin{align} & 3x^4-5x^2+6x-2 \quad --> \quad \text{4个}\,3\,4\,-5\,2\,6\,1\,-2\,0 \\ & 5x^{20}-7x^4+3x \quad --> \quad \text{3个}\,5\,20\,-7\,4\,3\,1 \\ \end{align} \\ \]
输出样例:
\[\begin{align} & 15x^{24}-25x^{22}+30x^{21}-10x^{20}-21x^8+35x^6-33x^5+14x^4-15x^3+18x^2-6x \\ & 15 \, 24 \, -25 \, 22 \, 30 \, 21 \, -10 \, 20 \, -21 \, 8 \, 35 \, 6 \, -33 \, 5 \, 14 \, 4 \, -15 \, 3 \, 18 \, 2 \, -6 \, 1 \, 5 \, 20 \, -4 \, 4 \, -5 \, 2 \, 9 \, 1 \, -2 \, 0 \end{align} \]
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\n"); return ;} while (P) { if (!flag) flag = 1; else printf(" "); printf("%d %d", P->coef, P->expon); P = P->link; } printf("\n"); }
标签:专场,多项式,t2,t1,link,加法,Polynomial,expon,Rear 来源: https://blog.51cto.com/u_13804357/2708705