c++ - 将卡诺 map 简化程序从SOP更改为POS结果

Question

这是我简化卡诺 map 的程序。对于大学项目，我需要将其从SOP更改为POS。我已经将符号从+更改为*，并从*更改为+。我现在要做的就是使其从文本文件中读取零，而不是一个。我尽力解决这个问题，但我不知道。我有两个txt文件，一个带有输入，一个带有输出。此时，我需要输入0而不是1，并在另一侧输入它，以使其正常工作。请帮助我，我的生活取决于

#include <iostream>
#include <fstream>
#include <cmath>
#include <windows.h>



using namespace std;
//?????
int pow2(int n) // ??2?????,2^n
{
    int result=1
    ;
    while(n>0)
        result*=2,n--;
    return result;
}
int combination(int n,int r) // ?????C_n^r,? n! / ((n-r)!*(r)!)
{
    int fm=1,fz=1;
    for(int i=1;i<=r;i++,n--)
    {
        fm*=i;
        fz*=n;
    }
    return fz/fm;
}
void d2b(int d,char* b,int n) // ???????,??????d???????????b?
{
    for(n--;n>=0;n--)
    {
        b[n]='0'+d%2;
        d/=2;
    }
}
int ispair(char* a1,char* a2,int n) // ????????????????
{
    int x,y=0;
    for(int i=0;i<n;i++)
        if(a1[i]!=a2[i])
            x=i,y++;
    if(y==1) return x;
    else return-1;
}
bool issame(char* a1,char* a2,int n) // ????
{
    for(int i=0;i<n;i++)
        if(a1[i]!=a2[i])
            return 0;
    return 1;
}
int left1(char* a,int N) // ???????????,??????1,??????
{
    for(int i=0;i<N;i++)
        if(a[i]=='1')
            return i;
    return -1;
}
void copy(char* a1,char* a2,int n) // ??????
{
    for(int i=0;i<n;i++)
        a2[i]=a1[i];
}
bool isinside(int x,char* a,int n) // ???x????????a?
{
    for(n--;n>=0;n--)
    {
        if(a[n]!='x' && a[n]!=(char)('0'+x%2))
            return 0;
        x/=2;
    }
    return 1;
}
void output(fstream& file,char* a,int n) 
{


    for(int i=0;i<n;i++)

    {


        if(i==0)file<<'(';
        if(a[i]=='0'||a[i]=='1')

        {

            if(a[i]=='0') file<<'¬'<<(char)('a'+i);
            if(a[i]=='1') file<<(char)('a'+i);
            if(a[i+1]=='0'||a[i+1]=='1'||a[i+2]=='0'||a[i+2]=='1'||a[i+3]=='0'||a[i+3]=='1')
            {
            file<<'+';
            }
            else{
            file<<')';
            }
        }
        else{}


        ;
    }
    file<<'*';
}
int count(char* table,char* a,int n,int N) // ??,????????1???????????a?
{
    int counter=0;
    for(int i=0;i<N;i++)
    {
        if(table[i]!='1') continue;
        if(isinside(i,a,n)) counter++;
    }
    return counter;
}
void clean(char* table,char* a,int n,int N) // ??????????????????1,???x
{
    for(int j=0;j<N;j++)
        if(isinside(j,a,n))
            table[j]='x';
}
int main()
{
    SetConsoleTitle("???-???????????");
    system("color 06");
    //????
    fstream inputFile("input.txt",ios::in); //??????????input.txt
    int valNum; //???
    inputFile >> valNum; // ???????????
    cout<<"\n ????????: "<<valNum;
    cout<<"\n ?????????:\n";
    int minTermLength=pow2(valNum); // ??????????2^valNum?
    char* minTermExpression=new char[minTermLength]; // ???????????????
    int lineOff = pow2(ceil(double(valNum)/2));
    // ???????
    if (inputFile.is_open())
    {
        for(int i=0;i<minTermLength;i++) // ?????
        {
            inputFile>>minTermExpression[i];
            if(i%lineOff == 0&&(i!=0))
                cout<<"\n";
            cout<<"\t"<<minTermExpression[i];
        }
        inputFile.close();
    }
    // ??????
    else{
        cout<<"\n ??input.txt??,???????";
        return 0;
    }
    // ?implication????????
    char*** implication=new char**[valNum]; // ????

    int nonZeroNum=1;
    for(int i=0;i<minTermLength;i++)
        if(minTermExpression[i]!='0') // ?????ON???DC?????????
            nonZeroNum++;
    for(int i=0;i<valNum;i++) // i-??,???????3?,i=0???????,i=1???,i=2???
    {
        if(pow2(i)>nonZeroNum)break;
        int x=pow2(i-1)*combination(nonZeroNum,pow2(i));

        implication[i]=new char*[x];
        for(int j=0;j<x;j++)
            implication[i][j]=new char[valNum];
    }
    //?????????
    int* countNum=new int[valNum+1];
    countNum[0]=0;
    for(int i=0;i<minTermLength;i++)
        if(minTermExpression[i]!='0') // ???0?
        {
            d2b(i,implication[0][countNum[0]],valNum); // ??????????implication[0]???
            countNum[0]++;
        }
    int isOptimal=0;
    while(countNum[isOptimal]>0) // ??????????
    {
        countNum[isOptimal+1]=0;
        for(int i=0;i<countNum[isOptimal]-1;i++)
            for(int j=i+1;j<countNum[isOptimal];j++)
            {
                int x=ispair(implication[isOptimal][i],implication[isOptimal][j],valNum); // ??????????
                if(x==-1) continue;
                copy(implication[isOptimal][i],implication[isOptimal+1][countNum[isOptimal+1]],valNum);// ???implication??
                implication[isOptimal+1][countNum[isOptimal+1]][x]='x'; // ??????????,?????????x,???????????
                countNum[isOptimal+1]++;
            }
        for(int i=0;i<countNum[isOptimal+1]-1;i++)
            for(int j=i+1;j<countNum[isOptimal+1];j++)
                if(issame(implication[isOptimal+1][i],implication[isOptimal+1][j],valNum)) // ??????
                {
                    for(int k=j;k<countNum[isOptimal+1]-1;k++)
                        copy(implication[isOptimal+1][k+1],implication[isOptimal+1][k],valNum);
                    countNum[isOptimal+1]--;
                }
        isOptimal++;
    }
    isOptimal--;
    //???????
    fstream outputFile("output.txt",ios::out);
    outputFile<<"F=";
    while(left1(minTermExpression,minTermLength)>=0) //?minTermExpression???1
    {
        bool flag=0; // ???,????
        for(int i=0;i<minTermLength&&flag==0;i++)
        {
            if(minTermExpression[i]!='1') continue;
            int counter=0,recorder;
            for(int j=0;j<countNum[isOptimal];j++)
                if(isinside(i,implication[isOptimal][j],valNum))
                    counter++,recorder=j;
            if(counter!=1) continue;
            output(outputFile,implication[isOptimal][recorder],valNum);
            clean(minTermExpression,implication[isOptimal][recorder],valNum,minTermLength);
            flag=1;
        }
        if(flag==1) continue;
        int termMaxInclude=0;
        int recorder=0;
        for(int i=0;i<countNum[isOptimal];i++) // ??????????,???????????1???,??????
            if(count(minTermExpression,implication[isOptimal][i],valNum,minTermLength)>termMaxInclude)
                termMaxInclude=count(minTermExpression,implication[isOptimal][i],valNum,minTermLength),recorder=i;
        if(termMaxInclude==0) {isOptimal--; continue;}
        output(outputFile,implication[isOptimal][recorder],valNum);
        clean(minTermExpression,implication[isOptimal][recorder],valNum,minTermLength);
    }
    outputFile.close();
    // ?????
    outputFile.open("output.txt",ios::in);
    char finalExpression[201];
    outputFile.getline(finalExpression,200);
    outputFile.close();
    int termMaxInclude=0;
    for(;finalExpression[termMaxInclude]!='\0';termMaxInclude++);
        finalExpression[termMaxInclude-1]='\0';
    outputFile.open("output.txt",ios::out);
    outputFile<<finalExpression;
    outputFile.close();
    cout<<"\n??????????????";

    cout<<"\n ??????,??????!";
    cin.get();
    return 0;
}

Answer 1

零就是那些不在“1”列表中的零。