输入一元多项式的参数，然后根据一元多项式加法减法乘法的计算法则，求解最终结果。用到了结构体，指针，操作的时候要注意。
不多说，上代码：
1 #include <stdio.h>
2 #include <malloc.h>
3
4 typedef struct node{
5 float coef;
6 int expn;
7 struct node *next;
8 }PLOY;
9
10 void start(){
11 printf("*******************************\n");
12 printf("两个一元多项式相加/相减,相乘：\n");
13 printf("*******************************\n");
14 printf("请选择操作：\n");
15 printf("0.退出\n");
16 printf("1.两个一元多项式相加\n");
17 printf("2.两个一元多项式相乘\n");
18 printf("3.两个一元多项式相减\n");
19 }
20
21 void insert(PLOY *head,PLOY *inpt){
22 PLOY *pre,*now;
23 int signal=0;
24 pre=head;//pre定义为现在的前一个链接
25 if(pre->next==NULL){
26 pre->next=inpt;
27 }
28 else{
29 now=pre->next;
30 while(signal==0){
31 if(inpt->expn<now->expn){
32 if(now->next==NULL){
33 now->next=inpt;
34 signal=1;
35 }
36 else{
37 pre=now;
38 now=pre->next;
39 }
40 }
41 else{
42 if(inpt->expn > now->expn){
43 inpt->next=now;
44 pre->next=inpt;
45 signal=1;
46 }
47 else{
48 now->coef=now->coef+inpt->coef;
49 signal=1;
50 free(inpt);
51 if(now->coef==0){
52 pre->next=now->next;
53 free(now);
54 }
55 }
56 }
57 }
58 }
59 }
60
61 PLOY *creat(char ch){
62 PLOY *head,*inpt;
63 float x;
64 int y;
65 head=(PLOY *)malloc(sizeof(PLOY));
66 head->next=NULL;
67 printf("请输入一元多项式%c:(格式是系数 指数；以0 0结束！)\n",ch);
68 scanf("%f %d",&x,&y);
69 while(x!=0){
70 inpt=(PLOY *)malloc(sizeof(PLOY));
71 inpt->coef=x;
72 inpt->expn=y;
73 inpt->next=NULL;
74 insert(head,inpt);
75 printf("请输入一元多项式%c的下一项:(以0 0结束！)\n",ch);
76 scanf("%f %d",&x,&y);
77 }
78 return head;
79 }
80
81 PLOY *addPLOY(PLOY *head,PLOY *pre){
82 PLOY *inpt;
83 int flag=0;
84 while(flag==0){
85 if(pre->next==NULL)
86 flag=1;
87 else{
88 pre=pre->next;
89 inpt=(PLOY *)malloc(sizeof(PLOY));
90 inpt->coef=pre->coef;
91 inpt->expn=pre->expn;
92 inpt->next=NULL;
93 insert(head,inpt);
94 }
95 }
96 return head;
97 }
98
99 PLOY *minusPLOY(PLOY *head,PLOY *pre){
100 PLOY *inpt;
101 int flag=0;
102 while(flag==0){
103 if(pre->next==NULL){
104 flag=1;
105 }
106 else{
107 pre=pre->next;
108 inpt=(PLOY *)malloc(sizeof(PLOY));
109 inpt->coef=0-pre->coef;
110 inpt->expn=pre->expn;
111 inpt->next=NULL;
112 insert(head,inpt);
113 }
114 }
115 return head;
116 }
117
118 PLOY *byPLOY(PLOY *headl,PLOY *head2){
119 PLOY *inpt,*res,*pre;
120 int flag=0;
121 res=(PLOY *)malloc(sizeof(PLOY));
122 res->next=NULL;
123 headl=headl->next;
124 pre=head2;
125 while(flag==0){
126 if(pre->next==NULL){
127 pre=head2;
128 headl=headl->next;
129 continue;
130 }
131 if(headl==NULL){
132 flag=1;
133 continue;
134 }
135 pre=pre->next;
136 inpt=(PLOY *)malloc(sizeof(PLOY));
137 inpt->coef=pre->coef * headl->coef;
138 inpt->expn=pre->expn + headl->expn;
139 inpt->next=NULL;
140 insert(res,inpt);
141 }
142 return res;
143 }
144
145 void print(PLOY *fun){
146 PLOY *printing;
147 int flag=0;
148 printing=fun->next;
149 if(fun->next==NULL){
150 printf("0\n");
151 return ;
152 }
153 while(flag==0){
154 if(printing->coef>0&&fun->next!=printing)
155 printf("+");
156 if(printing->coef==1);
157 else if(printing->coef==-1)
158 printf("-");
159 else
160 printf("%f",printing->coef);
161 if(printing->expn!=0)
162 printf("x^%d",printing->expn);
163 else if((printing->coef==1)||(printing->coef==-1))
164 printf("1");
165 if(printing->next==NULL)
166 flag=1;
167 else
168 printing=printing->next;
169 }
170 printf("\n");
171 }
172
173 int main(){
174 PLOY *f,*g;
175 int sign=-1;
176 start();
177 while(sign!=0){
178 scanf("%d",&sign);
179 switch(sign){
180 case 0:
181 break;
182 case 1:
183 {
184 printf("你选择的操作是多项式相加：\n");
185 f=creat(‘f‘); //输入多项式f(x)
186 printf("f(x)=");
187 print(f);
188 g=creat(‘g‘); //输入多项式g(x)
189 printf("g(x)=");
190 print(g);
191 printf("F(x)=f(x)+g(x)=");
192 f=addPLOY(f,g);
193 print(f);
194 sign=-1;
195 start();
196 break;
197 }
198 case 2:
199 {
200 printf("你选择的操作是多项式相乘：\n");
201 f=creat(‘f‘);
202 printf("f(x)=");
203 print(f);
204 g=creat(‘g‘);
205 printf("g(x)=");
206 print(g);
207 printf("F(x)=f(x)*g(x)=");
208 f=byPLOY(f,g);
209 print(f);
210 sign=-1;
211 start();
212 break;
213 }
214 case 3:
215 {
216 printf("你选择的操作是多项式相减：\n");
217 f=creat(‘f‘);
218 printf("f(x)=");
219 print(f);
220 g=creat(‘g‘);
221 printf("g(x)=");
222 print(g);
223 printf("F(x)=f(x)-g(x)=");
224 f=minusPLOY(f,g);
225 print(f);
226 sign=-1;
227 start();
228 break;
229 }
230 default:
231 {
232 printf("输入有误！请重新选择操作！\n");
233 start();
234 break;
235 }
236 }
237
238 }
239 }
240
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时间： 2024-10-06 00:07:20