有优先级的运算
八进制与十六进制
十六进制与八进制
一元多项式求和
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define STACK_INIT_SIZE 20
#define STACKINCREMENT 10
char OP[]={'+','-','*','/','(',')','='};
typedef char ElemType;
struct node
{
int coef; //系数
int exp; //指数
struct node * next;//指针域
}chainLink;//创建chainLink的node结点对象
typedef struct
{
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;
typedef struct
{
int *base;
int *top;
int stackSize;
}SqStack;
void InitStack(sqStack *s)
{
s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));
if( !s->base )
{
exit(0);
}
s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}
void initStack(SqStack *s)
{
s->base = (int *)malloc(STACK_INIT_SIZE * sizeof(int));
if( !s->base )
{
exit(0);
}
s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}
void Push(sqStack *s, ElemType e)
{
if( s->top - s->base >= s->stackSize )
{
s->base = (ElemType *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(ElemType));
if( !s->base )
{
exit(0);
}
}
*(s->top) = e;
s->top++;
}
void Pop(sqStack *s, ElemType *e)
{
if( s->top == s->base )
{
return;
}
*e = *--(s->top);
}
void push(SqStack *s, int e)
{
if( s->top - s->base >= s->stackSize )
{
s->base = (int *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(int));
if( !s->base )
{
exit(0);
}
}
*(s->top) = e;
s->top++;
}
void pop(SqStack *s, int *e)
{
if( s->top == s->base )
{
return;
}
*e = *--(s->top);
}
int StackLen(sqStack s)
{
return (s.top - s.base);
}
int stackLen(SqStack s)
{
return (s.top - s.base);
}
char GetTop(sqStack *s)
{
char e;
if(s->top == s->base)
return -1;
e = *(s->top-1);
return e;
}
int getTop(SqStack *s)
{
int e;
if(s->top == s->base)
return -1;
e = *(s->top-1);
return e;
}
int judge(char theta)
{
int i;
if(theta == '+')
i=0;
else if(theta == '-')
i=1;
else if(theta == '*')
i=2;
else if(theta == '/')
i=3;
else if(theta == '(')
i=4;
else if(theta == ')')
i=5;
else
i=6;
return i;
}
int precede(char theta1,char theta2)
{
int std_cherry[7][7]=
{
1, 1,-1,-1,-1, 1, 1,
1, 1,-1,-1,-1, 1, 1,
1, 1, 1, 1,-1, 1, 1,
1, 1, 1, 1,-1, 1, 1,
-1,-1,-1,-1,-1, 0,-2,
1, 1, 1, 1,-2, 1, 1,
-1,-1,-1,-1,-1,-2, 0,
};
int i,j;
i = judge(theta1);
j = judge(theta2);
return std_cherry[i][j];
}
int OPerate(int a,char theta,int b)
{
int result;
switch(theta)
{
case '+':
result = a+b;
break;
case '-':
result = a-b;
break;
case '*':
result = a*b;
break;
case '/':
result = a/b;
break;
}
return result;
}
int In(char *c,char *OP)
{
int i,comp = 0;
for(i = 0; i < 7; i++)
{
if( c[0] == OP[i])
comp = 1;
}
return comp;
}
int EvaluateEcpression()
{
sqStack OPTR;
SqStack OPND;
char c[10],theta,x;
int a,b,put;
InitStack(&OPTR);
Push(&OPTR,'=');
initStack(&OPND);
scanf("%s",c);
while(c[0] != '=' || GetTop(&OPTR) != '=')
{
if(!In(c,OP))
{
put = atoi(c);
// printf("%d\n",put);
push(&OPND,put);
scanf("%s",c);
}
else
switch(precede(GetTop(&OPTR),c[0]))
{
case -1:
Push(&OPTR,c[0]);
scanf("%s",c);
break;
case 0:
Pop(&OPTR,&x);
scanf("%s",c);
break;
case 1:
Pop(&OPTR,&theta);
pop(&OPND,&b);
pop(&OPND,&a);
push(&OPND,OPerate(a,theta,b));
// printf("%d\n",OPerate(a,theta,b));
break;
}
}
return getTop(&OPND);
}
void BtoD()
{
ElemType c;
sqStack s;
int len, i, sum = 0;
InitStack(&s);
printf("请输入二进制数，输入#符号表示结束！\n");
scanf("%c", &c);
while( c != '#' )
{
Push(&s, c);
scanf("%c", &c);
}
getchar(); // 把'\n'从缓冲区去掉
len = StackLen(s);
printf("栈的当前容量是: %d\n", len);
for( i=0; i < len-1; i++ )
{
Pop(&s, &c);
sum = sum + (c-48) * pow(2, i);
}
printf("转化为十进制数是: %d\n", sum);
return 0;
}
void DtoB()
{
int c,d;
SqStack s;
int len, i;
InitStack(&s);
printf("请输入十进制数: ");
scanf("%d", &c);
printf("转化为二进制数是: ");
while(c > 0)
{
// printf("%d",c%2);
push(&s, c%2);
c = c - c%2;
c = c/2;
}
len = stackLen(s);
for(i = 0; i < len; i++)
{
pop(&s,&d);
printf("%d",d);
}
printf("\n栈的当前容量是: %d\n", len);
return 0;
}
void OtoH()
{
int i,j,k,s=0;
int a;
char p[30];
int result=0;
printf("输入你想要转化的八进制数: \n");
scanf("%d",&a);
for(i=1;a!=0;i*=8)
{
if(a%10>7)
{
s=1;
break;
}
else
{
result+=(a%10)*i;
a=a/10;
}
}
for(j=0;result!=0;j++)
{
p[j]=result%16;
result=result/16;
if(p[j]<10)
p[j]+=48;
else
{
switch(p[j])
{
case 10: p[j]='A';
break;
case 11: p[j]='B';
break;
case 12: p[j]='C';
break;
case 13: p[j]='D';
break;
case 14: p[j]='E';
break;
case 15: p[j]='F';
break;
}
}
}
if(s==1)
printf("您的输入有误！请重新输入\n");
else
{
printf("\n转换后的数为：");
for(k=j-1;k>=0;k--)
{
printf("%c",p[k]);
}
printf("\n");
}
}
void HtoO()
{
int i,j,k,s=0;
char a[10];
int result=0;
int b[30];
int p[30];
scanf("%s",&a);
for(k=0;;k++)
{
if(a[k]=='\0')
break;
}
for(i=0;i<k;i++)
{
if(a[i]<='9'&&a[i]>='1')
b[i]=a[i]-48;
else
{
switch(a[i])
{
case 'A': b[i]=10;
break;
case 'B': b[i]=11;
break;
case 'C': b[i]=12;
break;
case 'D': b[i]=13;
break;
case 'E': b[i]=14;
break;
case 'F': b[i]=15;
break;
case 'a': b[i]=10;
break;
case 'b': b[i]=11;
break;
case 'c': b[i]=12;
break;
case 'd': b[i]=13;
break;
case 'e': b[i]=14;
break;
case 'f': b[i]=15;
break;
default: s=1;
}
}
}
for(j=k-1,i=1;j>=0;j--,i*=16)
{
result+=b[j]*i;
}
for(j=0;result!=0;j++)
{
p[j]=result%8;
result=result/8;
}
if(s==1)
printf("您的输入有误！请重新输入\n");
else
{
printf("\n转换后的数为：");
for(k=j-1;k>=0;k--)
{
printf("%d",p[k]);
}
printf("\n");
}
}
struct node *create() //定义建立多项式函数，并返回链表头指针
{
struct node *h = NULL, *q, *p;//定义结构体头指针h，标记指针p和q，p是创造新节点的标记指针，q是链接链表的指针；
int i = 1, N; //定义多项式的项数N及标记数i
printf("请输入多项式的项数: ");
scanf("%d",&N);
p = 0;//指针初始化；
q = 0;//本地指针初始化；
for (; i <= N; i++)
{
p = (struct node *)malloc(sizeof(chainLink)); //为一个新节点分配空间
printf("请输入第 %i 项的系数：\t",i);
scanf("%d",&(*p).coef);
printf("请输入第 %i 项的指数: \t",i);
scanf("%d",&(*p).exp);
if (i == 1) { h = p; q = p; } //建立头节点
else
{
q->next = p;
q = p;
}
}
q->next = NULL;//p,q都成为新链表的尾指针；
p->next = NULL;
return h;
}
void display(struct node *h)
{
struct node *p; //定义标记指针，输出链表
p = h;
while (p != NULL)
{
if (p->coef == 0)
{
struct node *d;
d = h;
while (d->next != p)
{
d = d->next;
}
d->next = p->next;
p = p->next;//删去系数为0的节点；
}
else
{
if (p->coef<0) //系数为负时应加上括号,让式子容易看清；
{
if (p->exp == 0) printf("%d+",(*p).coef);
else printf("(%d)*X^%d+",(*p).coef,(*p).exp);
}
else
{
if (p->exp == 0) printf("%d+",(*p).coef);
else if (p->exp!=0&&p->exp!=1)
{
printf("%d*X^%d+",(*p).coef,(*p).exp);
}
else printf("%d*X+",(*p).coef);
}
p = p->next;
}
}
printf("\b \b \n");
}
struct node *order(struct node *h) //定义排序函数，并返回整理后的链表的头指针
{
struct node *p, *q, *t, *h1, *h2; //定义三个结构体标记指针和两个头指针
h1 = h; //建立一个新的链表，头指针为h，指向原链表的第一个节点，之后将原链表中的节点一个一个的插入此链表，进行排序
h2 = h1->next; //将原来的链表建立成待排序链表
h1->next = NULL; //截断第一个原链表中的节点
while (h2 != NULL)
{
q = h1;
p = q->next;
t = h2; //从待排序链表中选出一个节点准备插入到新链表中
h2 = h2->next; //移动待排序链表的头指针，便于进行下一次挑选
t->next = NULL;
if (t->exp>q->exp) //应插入头指针之前的情况
{
t->next = q;
h1 = t;
continue;
}
if (p == NULL&&t->exp <= q->exp) q->next = t; //应插入表尾的情况
while (p != NULL)
{
if (t->exp>p->exp)
{
q->next = t;
t->next = p;
break;
}
else
{
q = q->next;
p = p->next;
}
}
if (p == NULL) q->next = t;
}
return h1; //新链表即为按降幂顺序排好的链表，返回其头指针
}
struct node *add(struct node *h1, struct node *h2) //定义加法函数，并返回结果的链表的头指针
{
struct node *p, *q, *head, *r; //定义结构体头指针head和标记指针p,q,r
p = h1;
q = h2;
head = (struct node *)malloc(sizeof(chainLink)); //为结果多项式分配头指针
if (p->exp >= q->exp) { head = h1; p = p->next; }
else
{
if (p->exp<q->exp) { head = h2; q = q->next; }
else { p->coef = p->coef + q->coef; head = p; p = p->next; q = q->next; }
}
r = head;
while (p != NULL&&q != NULL)
{
if (p->exp>q->exp) { r->next = p; p = p->next; }
else
{
if (p->exp<q->exp) { r->next = q; q = q->next; }
else { p->coef = p->coef + q->coef; r->next = p; p = p->next; q = q->next; }
}
r = r->next;
}
if (p == NULL&&q == NULL) r->next = NULL;
else
{
if (p == NULL) r->next = q;
if (q == NULL) r->next = p;
}
return head;
}
void polynomail()
{
struct node *h1, *h2, *h3;//定义3个头指针；
printf("请输入第一个多项式：\n");
h1 = create();//创造第一个线性链表；
printf("\n您输入的第一个多项式为：\n");
display(h1);//把多项式显示出来；
h1 = order(h1);
printf("\n降幂排列后的第一个多项式为：\n");
display(h1);
printf("\n");
printf("\n请输入第二个多项式：\n");
h2 = create();//创造第二个线性链表；
printf("\n您输入的第二个多项式为：\n");
display(h2);//把多项式显示出来；
h2 = order(h2);
printf("\n降幂排列后的第二个多项式为：\n");
display(h2);
printf("\n");
h3 = add(h1, h2);//利用加函数将多项式相加；
printf("\n第一个多项式和第二个多项式相加后：\n");
display(h3);
system("pause");
return 0;
}
void phead(void)
{
printf("*******************************************************\n");
printf("********************Tom's calclater********************\n");
printf("*******************************************************\n");
printf("\n*************** 请选择需要的计算操作:*****************\n");
}
void thead(void)
{
printf("*******************************************************\n");
printf("请输入相应的数字,并进行一步计算: ");
}
void main()
{
int i,j,q=0;
int tom;
phead();
do
{
printf("0、退出\n1、进制转化\n2、计算器\n3、一元多项式加法\n");
thead();
scanf("%d",&i);
switch (i)
{
case 1: thead();
printf("\n请要转化的的进制:\n0、退出\n1、二进制到十进制\n2、十进制到二进制\n3、八进制到十六进制\n4、十六进制到八进制\n");
scanf("%d",&j);
switch(j)
{
case 1: printf("\n请输入您想要转化的数：");
BtoD();
break;
case 2: printf("\n请输入您想要转化的数：");
DtoB();
break;
case 3: printf("\n请输入您想要转化的数：");
OtoH();
break;
case 4: printf("\n请输入您想要转化的数：");
HtoO();
break;
case 0:
printf("谢谢使用！！");
}
break;
case 2: printf("\n请输入你想要计算的表达式（在以空格隔开每个输入！）：\n");
tom = EvaluateEcpression();
printf("表达式的计算结界是 %d\n",tom);
break;
case 3:
polynomail();
break;
case 0: printf("\n谢谢使用！\n");
break;
}
}while(q==1);
}