利用真值表法求主析取范式及主合取范式的实现参考word

1、实 验 报 告（ / 学年 第 一 学期）课程名称离散数学实验名称利用真值表法求主析取范式及主合取范式的实现实验时间年月日指导单位指导教师学生姓名班级学号学院(系)专 业 推荐精选推荐精选实 验 报 告实验名称利用真值表法求主析取范式及主合取范式的实现指导教师实验类型上机实验学时4实验时间一、 实验目的和要求能够列出合式公式的真值表并给出相应主析取范式和主合取范式。二、实验环境(实验设备)硬件:PC机。软件:Code:Blocks （C+ ）三、实验原理及内容 内容：编程实现用真值表法求任意含三个以内变量的合式公式的主析取范式和主合取范式。 原理：首先读入变元个数，然后读入合式公式，用堆栈的知

2、识将中缀表达式转化为后缀表达式，调用否定、析取、合取、条件、双条件的函数计算P、Q、R取不同真值时合式公式的真值，然后输出真值表，调用计算主析取范式和主合取范式的函数并输出。 推荐精选程序：#include<iostream>#include<string>#include<conio.h>#include<windows.h>using namespace std;string OriginalForm; /原式string Hequ; /主合取范式string Xiqu; /主析取范式class SeqStack/建立一个堆栈，利用将中缀表达

3、式转为后缀表达式public: SeqStack(int mSize); SeqStack(); char Top(); bool Push(char x); bool Pop();private: char *st; int top; int maxtop;推荐精选SeqStack:SeqStack(int mSize) maxtop = mSize - 1; top = -1; st = new charmSize;SeqStack:SeqStack() deletest;char SeqStack:Top() return sttop;bool SeqStack:Push(char x)

4、 if(top = maxtop) return false; st+top = x; return true;bool SeqStack:Pop()推荐精选 if(top = -1) return false; top-; return true;int p, q, r, s, t, u;int a, b, result;int v =0;int number;/用number表示变元的个数SeqStack stack(200);void Not() /否定 a = stack.Top(); stack.Pop(); result = a = 1 ? 0 : 1; stack.Push(re

5、sult);void Or() /析取 result = a + b;推荐精选 result = result > 1 ? 1 : result; stack.Push(result);void And() /合取 result = a * b; stack.Push(result);void If() /条件,b->a result = (b = 1 && a = 0) ? 0 : 1; stack.Push(result);void Doubleif() /双条件 result = (b = a) ? 1 : 0; stack.Push(result);bool

6、 CanIn(char out)/优先级的判断 char in = stack.Top(); int i, o;推荐精选 switch(in) case '#':i = 0; break; case '(':i = 1; break; case '-':i = 3; break; case '>':i = 5; break; case '|':i = 7; break; case '&':i = 9; break; case '!':i = 11; break; cas

7、e ')':i = 12; break; switch(out) case '#':o = 0; break; case '(':o = 12; break; case '-':o = 2; break; case '>':o = 4; break; case '|':o = 6; break; case '&':o = 8; break; case '!':o = 10; break; case ')':o = 1; break; 推荐

8、精选 if(i < o) return true; else return false;void InfixToPostfix()/中缀表达式转后缀表达式 string tmp = "" stack.Push('#'); for(int i = 0; (unsigned)i <OriginalForm.length(); i+) if(OriginalFormi = 'P' | OriginalFormi = 'Q' |OriginalFormi = 'R' | OriginalFormi = &#

9、39;S' | OriginalFormi = 'T' | OriginalFormi = 'U') tmp=tmp+OriginalFormi; continue; if(CanIn(OriginalFormi) stack.Push(OriginalFormi); else if(OriginalFormi = ')') 推荐精选 while(stack.Top() != '(') tmp = tmp + stack.Top(); stack.Pop(); stack.Pop(); else do tmp = tmp

10、+ stack.Top(); stack.Pop(); while(!CanIn(OriginalFormi); stack.Push(OriginalFormi); while(stack.Top() != '#') tmp = tmp + stack.Top(); stack.Pop(); 推荐精选 stack.Pop(); OriginalForm = tmp;void Calculate()/计算主析取范式和主合取范式的函数 if(number = 3) for(int i = 0; (unsigned)i <OriginalForm.length(); i+)

11、if(OriginalFormi = 'P' | OriginalFormi = 'Q'| OriginalFormi = 'R') v = OriginalFormi = 'P' ? p : OriginalFormi= 'Q' ? q : r; stack.Push(v); continue; if(OriginalFormi != '!') a = stack.Top(); stack.Pop(); b = stack.Top();推荐精选 stack.Pop(); switch(Origin

12、alFormi) case '-':Doubleif(); break; case '>':If(); break; case '|':Or(); break; case '&':And(); break; case '!':Not(); break; if(number = 2) for(int i = 0; (unsigned)i <OriginalForm.length(); i+) if(OriginalFormi = 'P' | OriginalFormi = '

13、;Q') v = OriginalFormi = 'P' ? p : q; stack.Push(v); continue; 推荐精选 if(OriginalFormi != '!') a = stack.Top(); stack.Pop(); b = stack.Top(); stack.Pop(); switch(OriginalFormi) case '-':Doubleif(); break; case '>':If(); break; case '|':Or(); break; case &

14、#39;&':And(); break; case '!':Not(); break; void Print() if(number = 3) 推荐精选 cout << "Pt" <<"Qt" << "Rt" << "Z" << endl; for(p = 1; p >= 0; p-) for(q = 1; q >= 0; q-) for(r = 1; r >= 0; r-) Calculate(); if

15、(result = 1) Xiqu =Xiqu +"(" + (p = 1 ? "P" : "!P") + "&" +(q = 1 ? "Q" : "!Q") + "&" + (r = 1 ?"R" : "!R") + ")" + " | " else Hequ = Hequ +"(" + (p = 0 ? "P" : &

16、quot;!P") + "|" + (q= 0 ? "Q" : "!Q") + "|" + (r = 0 ?"R" : "!R") + ")" + " & " cout << p <<"t" << q << "t" << r <<"t" << result << e

17、ndl; 推荐精选 if(number = 2) cout << "Pt" <<"Qt" << "Z" << endl; for(p = 1; p >= 0; p-) for(q = 1; q >= 0; q-) Calculate(); if(result = 1) Xiqu =Xiqu + "("+ (p = 1 ? "P" : "!P") + "&" + (q = 1 ?"

18、Q" : "!Q") + ")" + " | " else Hequ = Hequ + "(" + (p =0 ? "P" : "!P") + "|" + (q = 0 ? "Q" :"!Q") + ")" + " & " cout << p <<"t" << q << "t&quo

19、t; << result << endl; cout<<endl; if(Xiqu.length() != 0)推荐精选 Xiqu.erase(Xiqu.length() - 2); if(Hequ.length() != 0) Hequ.erase(Hequ.length() - 2);cout << "主析取范式："<<Xiqu << endl << endl; cout << "主合取范式：" << Hequ << endl &l

20、t;< endl;int main()int flag=1; while(flag=1) SetConsoleTextAttribute(GetStdHandle (STD_OUTPUT_HANDLE),FOREGROUND_INTENSITY |FOREGROUND_GREEN|FOREGROUND_BLUE); /设置绿色和蓝色相加(即青色) system("cls");/清屏 cout<<"-"<<endl; cout <<" 欢迎使用！"<<endl<<endl

21、; cout << " ! 表示否定" << endl<<endl; cout << " | 表示析取" << endl<<endl; cout << " & 表示合取" << endl<<endl;推荐精选 cout << " > 表示条件" << endl<<endl; cout << " - 表示双条件" << endl; cout<<"-"<<endl; cout <<endl<< " 输入你的变元数目:" cin >> number;cout<<" 请输入合式公式：" switch(number) case 2:cout << endl <<" 变元请用P、Q表示" << endl << endl; br