一次失败的课设——逻辑函数化简
写在前面
首先感谢我的舍友吴思滢同学，在设计之初我们进行了思路上的探讨，所以这个程序能得以实现也得益于她的启发和帮助。
这个大作业前前后后做了几个月，在这期间我对实现方式改了又改，string数组总是导致报错，即使到了课程快结束时，仍有一些问题没能顺利解决，所以为了不影响整体的功能，最后只能用最原始的枚举特殊情况的方法解决问题了，这也是一个遗憾。
通过这个大作业，也让我对”编程”二字有了新的认识，如何用编程实现一个实际的、具体的问题，与解决C++辅导书里的例题的方式真的相差甚远。
借此感谢老师这学期以来的教学和辅导，每次课后总是耐心解答我的许多问题，我也感受到了数电在逻辑上的魅力。
设计优缺点评价
优点：
①使用类，用户引导好，main函数简单
②输入出错时会产生提示，让用户重新输入
缺点：
①由于使用string数组过程中因内存问题频繁报错，始终没能解决，最后只能使用繁琐且笨的方法解决特殊情况
②用户需要使用最小项下标的二进制数进行输入，比较繁琐
代码部分
一坨答辩
//为了方便测试，设置成了四变量卡诺图化简的形式，实际可根据需要调整
#include <iostream>
#include <string>
using namespace std;
// Kmap类
class Kmap {
public:
int merged_count=0;
int count_1[16] = {0};
int count_2[16] = {0};
int index=0;
string original_part[17];
string merged_part[16][16];
string combined_part[16][16];
string simplified_part[16][16];
void input();
void count();
void merge();
void combine();
void simplify();
void combine1();
void printCombined();
};
// 原始数据的输入
void Kmap::input() {
cout << "你好！这是我的数电大作业：逻辑函数化简" << endl;
cout<< " 作者：杨语希 "<<endl;
cout<<"个人能力所限，做得比较粗糙，请多原谅" << endl;
cout << endl;
cout<<"***************************************"<<endl;
cout<<"* *"<<endl;
cout<<"* 输入最小项对应的二进制数,输入a结束 *" << endl;
cout << "* *" << endl;
cout<<"***************************************"<<endl;
cout << endl;
cout <<"请在此输入最小项对应的 4 位二进制数，以空格或回车分隔，如: 0011"<<endl;
int i = 0;
while (true) {
string str;
cin >> str;
if (str == "a") {
break;
}
if (str.length() != 4) {
cout<<"您的输入有误，请重新输入"<<endl;
continue;
}
original_part[i] = str;
i++;
}
}
// 统计original_part每个字符串中"1"的个数
void Kmap::count() {
int i = 0;
while (original_part[i] != "") {
int temp = 0;
for (int j = 0; j < original_part[i].length(); j++) {
if (original_part[i][j] == '1') {
temp++;
}
}
count_1[i] = temp;
i++;
}
}
// count_1数量相同的元素合并到merged_part中
void Kmap::merge() {
int i = 0;
while (original_part[i] != "") {
int temp = count_1[i];
int j = 0;
while (merged_part[temp][j] != "") {
j++;
}
merged_part[temp][j] = original_part[i];
i++;
}
merged_count = i;
}
// 合并相同数量的最小项
void Kmap::combine() {
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
int diffCount = 0;
string result = "";
int maxLength = min(merged_part[i][j].length(), merged_part[i + 1][j].length());
for (int k = 0; k < maxLength; k++) {
if (merged_part[i][j][k] == merged_part[i + 1][j][k]) {
result += merged_part[i][j][k];
}
else {
result += '-';
diffCount++;
}
}
if (diffCount <= 1) {
combined_part[i][j] = result;
}
}
}
}
// 合并相邻的最小项(函数汇总)
void Kmap::simplify() {
input();
count();
merge();
combine();
combine1();
}
// 打印输出
void Kmap::printCombined() {
cout << endl;
cout << "***************************************" << endl;
cout << "化简后的结果为：" << endl;
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
if (combined_part[i][j] != "") {
string term = "";
for (int k = 0; k < 4; k++) {
if (combined_part[i][j][k] == '1') {
term += 'A' + k;
}
else if (combined_part[i][j][k] == '0') {
term += 'a' + k;
}
}
if (term != "")
cout << term;
else if(term=="")
cout << "请检查您的输入是否有误，或者您的输入不足以化简";
if (combined_part[i+1][j]!="" || combined_part[i+2][j]!="" || combined_part[i+3][j] != "" || combined_part[i][j+1] != "" || combined_part[i][j+2] != "" || combined_part[i][j+3] != "" || combined_part[i+1][j+1] != "" || combined_part[i+2][j+1] != "" || combined_part[i+3][j+1] != "" || combined_part[i+1][j-1] != "" || combined_part[i+2][j-1] != "" || combined_part[i+3][j-1] != "") {
cout << "+";
}
}
}
}
cout << endl;
cout << "***************************************" << endl;
}
//主函数
int main() {
Kmap kmap;
kmap.simplify();
kmap.printCombined();
return 0;
}
//代码的不足之处，8个或16个相邻的部分的处理（琢磨了很久总是报错，最后只能用这种笨方法解决，希望老师能够原谅）
void Kmap::combine1() {
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
if (combined_part[i][j] == "000-" && combined_part[i + 1][j] == "00-1")
{
combined_part[i][j] = "00--";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "010-" && combined_part[i + 1][j] == "01-1")
{
combined_part[i][j] = "01--";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "100-" && combined_part[i + 1][j] == "10-1")
{
combined_part[i][j] = "10--";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "110-" && combined_part[i + 1][j] == "11-1")
{
combined_part[i][j] = "11--";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "0-00" && combined_part[i + 1][j] == "-100")
{
combined_part[i][j] = "--00";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "0-01" && combined_part[i + 1][j] == "-101")
{
combined_part[i][j] = "--01";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "0-10" && combined_part[i + 1][j] == "-110")
{
combined_part[i][j] = "--10";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "0-11" && combined_part[i + 1][j] == "-111")
{
combined_part[i][j] = "--11";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "000-" && combined_part[i + 1][j] == "--01")
{
combined_part[i][j] = "--0-";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "001-" && combined_part[i + 1][j] == "--11")
{
combined_part[i][j] = "--1-";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "00--" && combined_part[i + 1][j] == "0-11")
{
combined_part[i][j] = "0---";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "1-00" && combined_part[i + 1][j] == "11--")
{
combined_part[i][j] = "1---";
combined_part[i + 1][j] = "";
}
else if (combined_part[i][j] == "00--" && combined_part[i + 2][j] == "--11")
{
combined_part[i][j] = "----";
combined_part[i + 1][j] = "";
}
}
}
}
