多项式的加减乘
【摘要】 这是我学数据结构写的第一个比较"大型"的程序，唉，还是有不足之处啊。。。
# include <stdio.h># include <iostream.h># include <stdlib.h># include <malloc.h> # define OK 1# define ERROR 0# define OVERFLO...
这是我学数据结构写的第一个比较"大型"的程序，唉，还是有不足之处啊。。。
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# include <stdio.h>
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# include <iostream.h>
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# include <stdlib.h>
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# include <malloc.h>
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# define OK 1
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# define ERROR 0
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# define OVERFLOW -1
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typedef struct Polynomial
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{
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float coef;
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int expn;
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struct Polynomial *next;
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}Polynomial,*LinkPolyn;
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void Create_Polyn(LinkPolyn &L)
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{
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int i,n;
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LinkPolyn p,s;
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L = (LinkPolyn) malloc (sizeof (Polynomial));
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if (!L)
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exit(OVERFLOW);
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L->next = NULL;
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p = L;
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cout<<" 请输入您要创建多项式的项数:";
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cin>>n;
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for (i=0; i<n; i++)
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{
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s = (LinkPolyn) malloc (sizeof (Polynomial));
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p->next = s;
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cout<<" 请输入第"<<i+1<<"项的系数:";
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cin>>s->coef;
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cout<<" 请输入第"<<i+1<<"项的指数:";
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cin>>s->expn;
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p = s;
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}
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p->next = NULL;
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}
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void Display(LinkPolyn L)
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{
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LinkPolyn p;
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if (!L->next )
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cout<<" 多项式的项数为0!无法显示"<<endl;
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p = L->next ;
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while (p)
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{
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if (p->coef == 0.0)
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cout<<"";
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else if (p->expn == 0 )
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cout<<p->coef ;
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else if (p->coef == 1 && p->expn == 1)
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cout<<"X";
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else if (p->coef == 1 && p->expn != 1)
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cout<<"X^"<<p->expn;
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else if (p->coef == -1 && p->expn == 1)
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cout<<"-X";
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else if (p->coef != 1 && p->expn == 1)
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cout<<p->coef<<"X";
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else if (p->coef == -1 && p->expn != 1)
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cout<<"-X^"<<p->expn ;
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else
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cout<<p->coef <<"X^"<<p->expn ;
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if (p->next!=NULL && p->coef !=0 && p->next->coef >=0)
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cout<<"+";
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p = p->next ;
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}
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cout<<endl;
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}
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void SortPolyn(LinkPolyn &L)
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{
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LinkPolyn p,q;
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int temp;
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float tmp;
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if (!L->next )
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cout<<" 多项式的项数为0,无法排序!"<<endl;
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for(p=L->next;p; p=p->next )
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for(q=L->next ;q; q=q->next)
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{
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if(p->expn < q->expn )
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{
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temp = p->expn ;
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p->expn = q->expn ;
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q->expn = temp;
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tmp = p->coef ;
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p->coef = q->coef ;
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q->coef = tmp;
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}
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else if (p != q && p->expn == q->expn)
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{
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p->coef +=q->coef ;
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q->coef = 0;
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q->expn = 0;
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-
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}
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}
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}
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void Diff_Polyn(LinkPolyn &L)
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{
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LinkPolyn p;
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if (!L->next )
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cout<<" 多项式的项数为0,无法求导!"<<endl;
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p = L->next ;
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while (p)
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{
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p->coef *= p->expn ;
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p->expn -=1;
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p = p->next ;
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}
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}
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void Int_Polyn(LinkPolyn &L)
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{
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LinkPolyn p;
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if (!L->next )
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cout<<" 多项式的项数为0,无法求积分!"<<endl;
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p = L->next ;
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while (p)
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{
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p->coef = p->coef/(p->expn+1);
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p->expn +=1;
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p = p->next ;
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}
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}
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int cmp (int a, int b)
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{
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if (a>b)
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return 1;
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else if (a == b)
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return 0;
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return -1;
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}
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int Add_Polyn(LinkPolyn &PA, LinkPolyn &PB)
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{
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LinkPolyn qc,qa,qb,temp;
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float sum;
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qc = PA;
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qa = PA->next ;
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qb = PB->next ;
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while (qa && qb)
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{
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switch(cmp(qa->expn ,qb->expn ))
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{
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case -1:
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qc->next = qa;
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qc = qa;
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qa = qa->next;
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break;
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case 0:
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sum = qa->coef + qb->coef ;
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if(sum !=0.0)
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{
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qa->coef = sum;
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qc->next = qa;
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qc = qa;
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qa = qa->next;
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}
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else
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{
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temp = qa;
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qa = qa->next ;
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free(temp);
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}
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temp = qb;
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qb = qb->next ;
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free(temp);
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break;
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case 1:
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qc->next = qb;
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qc = qb;
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qb = qb->next;
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break;
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}
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}
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while(qa)
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{
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qc->next = qa;
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qc = qa;
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qa = qa->next;
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}
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while(qb)
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{
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qc->next = qb;
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qc = qb;
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qb = qb->next;
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}
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free(qb);
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return OK;
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}
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int Sub_Polyn(LinkPolyn &PA, LinkPolyn &PB)
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{
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LinkPolyn qc,qa,qb,temp;
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float sum;
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qc = PA;
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qa = PA->next ;
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qb = PB->next ;
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while (qa && qb)
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{
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switch(cmp(qa->expn ,qb->expn ))
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{
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case -1:
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qc->next = qa;
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qc = qa;
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qa = qa->next;
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break;
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case 0:
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sum = qa->coef - qb->coef ;
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if(sum !=0.0)
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{
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qa->coef = sum;
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qc->next = qa;
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qc = qa;
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qa = qa->next;
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}
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else
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{
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temp = qa;
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qa = qa->next ;
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free(temp);
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}
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temp = qb;
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qb = qb->next ;
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free(temp);
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break;
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case 1:
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qb->coef =-qb->coef ;
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qc->next = qb;
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qc = qb;
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qb = qb->next;
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break;
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}
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}
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while(qa)
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{
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qc->next = qa;
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qc = qa;
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qa = qa->next;
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}
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while(qb)
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{
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qb->coef =-qb->coef ;
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qc->next = qb;
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qc = qb;
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qb = qb->next;
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}
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free(qb);
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return OK;
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}
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int Mul_Polyn(LinkPolyn pa,LinkPolyn pb,LinkPolyn &pc)
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{
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LinkPolyn qa,qb,qc,s;
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pc = (LinkPolyn) malloc (sizeof (Polynomial));
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pc->next = NULL;
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qc = pc;
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for(qa=pa->next;qa;qa=qa->next)
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{
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for(qb=pb->next;qb;qb=qb->next)
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{
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s = (LinkPolyn) malloc (sizeof (Polynomial));
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qc->next = s;
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s->coef=qa->coef*qb->coef;
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s->expn=qa->expn+qb->expn;
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qc = s;
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}
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}
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qc->next = NULL;
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SortPolyn(pc);
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return OK;
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}
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int main (void)
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{
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char i;
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LinkPolyn L;
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LinkPolyn PA,PB,PC;
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cout<<endl;
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cout<<" ********************************************************"<<endl;
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cout<<" | 欢迎使用本系统! |"<<endl;
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cout<<" | 制作人:风尘三侠 |"<<endl;
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cout<<" | 1:创建一个多项式 2:对一个多项式求导 |"<<endl;
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cout<<" | 3:对一个多项式求积分 4:两个多项式相加 |"<<endl;
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cout<<" | 5:两个多项式相减 6:两个多项式相乘 |"<<endl;
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cout<<" | 7:帮助 8:退出 |"<<endl;
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cout<<" ********************************************************"<<endl;
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cout<<endl;
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do
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{
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cout<<" 请选择您要执行的操作:";
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cin>>i;
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switch (i)
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{
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case '1':
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Create_Polyn(L);
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SortPolyn(L);
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cout<<" 您创建的多项式是: F(X) = ";
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Display(L);
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break;
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case '2':
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cout<<" 对一个多项式求导前请先创建一个多项式!"<<endl;
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Create_Polyn(L);
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SortPolyn(L);
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cout<<" 您创建的多项式是: F(X) = ";
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Display(L);
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Diff_Polyn(L);
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cout<<" 求导后的多项式是: F'(X) = ";
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Display(L);
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break;
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case '3':
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cout<<" 对一个多项式求积分前请先创建一个多项式!"<<endl;
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Create_Polyn(L);
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SortPolyn(L);
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cout<<" 您创建的多项式是: F(X) = ";
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Display(L);
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Int_Polyn(L);
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cout<<" 积分后的多项式是: F'(X) = ";
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Display(L);
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break;
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case '4':
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cout<<" 请先创建两个多项式!"<<endl;
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cout<<" 创建第一个多项式!"<<endl;
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Create_Polyn(PA);
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SortPolyn(PA);
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cout<<" 您创建的第一个多项式是: F(X) = ";
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Display(PA);
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cout<<" 创建第二个多项式!"<<endl;
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Create_Polyn(PB);
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SortPolyn(PB);
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cout<<" 您创建的第二个多项式是: G(X) = ";
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Display(PB);
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cout<<" 两个多项式相加后：F(X) + G(X) = ";
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Add_Polyn(PA, PB);
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Display(PA);
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break;
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case '5':
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cout<<" 请先创建两个多项式!"<<endl;
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cout<<" 创建第一个多项式!"<<endl;
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Create_Polyn(PA);
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SortPolyn(PA);
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cout<<" 您创建的第一个多项式是: F(X) = ";
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Display(PA);
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cout<<" 创建第二个多项式!"<<endl;
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Create_Polyn(PB);
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SortPolyn(PB);
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cout<<" 您创建的第二个多项式是: G(X) = ";
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Display(PB);
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cout<<" 两个多项式相减后：F(X) - G(X) = ";
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Sub_Polyn(PA, PB);
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Display(PA);
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break;
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case '6':
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cout<<" 请先创建两个多项式!"<<endl;
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cout<<" 创建第一个多项式!"<<endl;
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Create_Polyn(PA);
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SortPolyn(PA);
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cout<<" 您创建的第一个多项式是: F(X) = ";
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Display(PA);
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cout<<" 创建第二个多项式!"<<endl;
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Create_Polyn(PB);
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SortPolyn(PB);
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cout<<" 您创建的第二个多项式是: G(X) = ";
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Display(PB);
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cout<<" 两个多项式相乘后：F(X) * G(X) = ";
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Mul_Polyn(PA, PB,PC);
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Display(PC);
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break;
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case '7':
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cout<<" 创建多项式! "<<endl;
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cout<<" 先输入多项式的项数，要输入自然数!"<<endl;
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cout<<" 可以输入重复的项，先输入系数，再输入指数!"<<endl;
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cout<<" 系数可以是任意的数字，指数要输入自然数!"<<endl;
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break;
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case '8':
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cout<<" 谢谢使用,再见!"<<endl;
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exit(OVERFLOW);
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break;
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default:
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{
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cout<<" 您的输入有误，请重新输入!"<<endl;
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break;
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}
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}
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}while (i != 8);
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return 0;
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}
文章来源: blog.csdn.net，作者：悦来客栈的老板，版权归原作者所有，如需转载，请联系作者。
原文链接：blog.csdn.net/qq523176585/article/details/6818975
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