📄 5.cpp
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#include<stdlib.h>
#include<stdio.h>
#include<iostream.h>
#include<string.h>
#include<math.h>
#define stack_size 100
int zuhe[20];//变量的取值组合数组定义;
int N;//变量个数;
int k=0;
//根据表达式建立的二叉树的结点定义;
typedef struct tree{
char data;
struct tree *lch; //结点的左孩子;
struct tree *rch; //结点的右孩子.
}*BNode;
//识别表达式使用的堆栈定义,它存放的都是树的结构;
typedef struct stack{
struct tree **base; //栈中的元素都是树的结点结构;
struct tree **top;
int stacksize;
}sqstack;
//用于产生变量的各种取值组合;
void creatzuhe(int n)
{
int i;
for(i=N-1;i>=0;i--)
{zuhe[i]=n%2;
n=n/2; }
}
//自底向上地根据运算符地优先级来建立分子树函数;当逻辑表达式读完后-子根zigen就是一棵完整的二叉树
void create(BNode &zigen,BNode l,BNode r)
{
zigen->lch=l;
zigen->rch=r;
if(l&&r)
{
if(int(l->data)>=65&&int(l->data)<=90)
{
l->lch=NULL;
l->rch=NULL;
}
if(int(r->data)>=65&&int(r->data)<=90)
{
r->lch=NULL;
r->rch=NULL;
}
}
}
//逻辑运算符的优先级判别;
char paixu(char row,char col)
{
int i,j;
char compare[7][7]={' ','|','&','~','(',')','#',
'|','>','<','<','<','>','>',
'&','>','>','<','<','>','>',
'~','>','>','>','<','>','>',
'(','<','<','<','<','=',' ',
')','>','>','>',' ','>','>',
'#','<','<','<','<',' ','='};
for(i=0;i<7;i++)
if(compare[0][i]==row)
break;
for(j=0;j<7;j++)
if(compare[j][0]==col)
break;
return compare[j][i];
}
void creatstack(sqstack &S)
{
S.base=(BNode*)malloc(stack_size*sizeof(tree));
if(!S.base)
exit(0);
S.top=S.base;
S.stacksize=stack_size;
}
void push(sqstack &S,BNode e)
{
if(S.top-S.base<S.stacksize)
*S.top++=e;
else exit(0);
}
void pop(sqstack &S,BNode &e)
{
if(S.top==S.base)
exit(0);
e=*--S.top;
}
void gettop(sqstack &S,BNode &e)
{
if(S.top==S.base)
exit(0);
e=*(S.top-1);
}
//重言式识别
void creattree(char s[],BNode &tree)
{
sqstack variable; //变量栈;
sqstack logic; //逻辑运算符栈;
creatstack(variable);
creatstack(logic);
BNode logic_di,variables,logics,e,a,b,theta,parentheses; //定义栈中的元素;
logic_di=(BNode)malloc(sizeof(tree));
if(!logic_di) exit(0);
logic_di->data='#';
push(logic,logic_di);
while(*s!=NULL)
{
if(int(*s)>=65&&int(*s)<=90)
{
variables=(BNode)malloc(sizeof(tree));
if(!variables) exit(0);
variables->data=*s;
push(variable,variables);
}
else if(int(*s)>90||int(*s)<65)
{
gettop(logic,e);//优先级比较
switch(paixu(*s,e->data))
{
case '<':
logics=(BNode)malloc(sizeof(tree));
if(!logics) exit(0);
logics->data=*s;
push(logic,logics);
break;
case '=':
pop(logic,parentheses);break;
case '>':pop(logic,theta);
pop(variable,a);
b=NULL;
if(theta->data!='~')
pop(variable,b);
k=k+1;
create(theta,b,a);
push(variable,theta);
if(*s!='#'&&*s!=')')
{
logics=(BNode)malloc(sizeof(tree));
if(!logics) exit(0);
logics->data=*s;
push(logic,logics);
}
else s=s-1;
break;
}
}
s++;
}
tree=theta;
}
int value_tree(BNode tree)
{
if(!tree) return 0; //空结点;
else if(tree->data!='|'&&tree->data!='&'&&tree->data!='~')//变量;
return zuhe[int(tree->data)-65];
else if(int(tree->data)<65||int(tree->data)>90) //运算符;
switch(tree->data)
{
case '|': return(value_tree(tree->lch)||value_tree(tree->rch));
case '&': return(value_tree(tree->lch)&&value_tree(tree->rch));
case '~': return(!value_tree(tree->rch));
}
}
void main()
{
char str[60],string[60],*pstr;
int loop=20,i=0,sum;
BNode Tree;
while(loop)
{
pstr=str;
i=0;
int SUM=0,l;
cout<<"Please input the number of variables"<<endl;
cin>>N;
if(N<=1||N>20)
{
cout<<"Input error"<<endl;
break;
}
else{
sum=int(pow(2,N));
cout<<"Please input the logic expression(use'|',or '&' or '~')"<<endl;
cin>>str;
for(;*pstr!=NULL;pstr++)
if(*pstr!=' ') string[i++]=*pstr;
string[i]='#';
string[i+1]='\0';
creattree(string,Tree);//建立重言式二叉树;
cout<<" Result "<<endl;
printf("| ");
for(l=65;l<65+N;l++)
printf("%-4c",l);
printf(" expression");
printf(" |\n");
for(loop=0;loop<sum;loop++)
{
creatzuhe(loop);//产生变量取值组合;
SUM+=value_tree(Tree);
printf("| ");
for(int h=0;h<N;h++)
printf("%-4d",zuhe[h]);
printf("%8d",value_tree(Tree));
printf(" |\n");
}
string[i]='\0';
if(SUM==0) cout<<"logic expression: "<<string<<" is contradiction"<<endl;//矛盾式
else if(SUM==sum)
cout<<"logic expression: "<<string<<" is tautology"<<endl;//重言式
else
cout<<"logic expression: "<<string<<" is neither tautology,nor contradiction"<<endl;
break;
}
}
}
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