C语言实现合式公式的判断示例

void Del_Negation(char* str,int n)
{

	assert(str);
	char* tmp = (char*)calloc(n+1, sizeof(char));//多一个1，是为了放置\0,避免strcpy越界拷贝
	assert(tmp);
	int cnt = 0;
	int i = 0;
	while (i < n)//将除！的字符赋值到tmp中
	{
		if (str[i] != '!')
		{
			tmp[cnt++] = str[i];
		}
		++i;
	}

	memset(str, 0, sizeof(char)*n);//对str那块内存重新赋值为\0,防止tmp拷贝到str中后，s扔有旧的数据
	strcpy(str, tmp);
	free(tmp);
	tmp = NULL;
}

void Del_Bracket( char* str, int left, int right)
{
	assert(str);
	char* tmp = (char*)calloc(right - left +1+1, sizeof(char));
	assert(tmp);
	str[right] = '\0';
	strcpy(tmp, str+left+1);
	strcpy(str+left, tmp);
}

int Find_Fist_operator( char* str, int left, int right)
{
	int ret = 0;
	while (left<right)
	{

		if (str[left + 1] == '+' || str[left + 1] == '*' || str[left + 1] == '>' || str[left + 1] == '=')
		{
			ret = left + 1;
			return ret;
		}

		++left;
	}
	return 0;//如果ret是0，说明是非法，反之就正确
}

bool Is_CombForm(char* str, int left, int right)
{

	if ((0 == (right - left))//区间是原子命题
		&& ('A' <= str[left] || 'Z' >= str[left]))
	{
		return true;
	}

	if (str[left] != '(')//第一个字符是字母： A>(B)
	{
		int keyi = Find_Fist_operator(str, left, right);//找寻第一个双目运算符
		if (keyi > 0)
		{
			if (str[keyi + 1] == '(')//A<(B)
			{
				Del_Bracket(str, keyi + 1, right);
				return Is_CombForm(str, left, keyi - 1) && Is_CombForm(str, keyi + 1, right - 2);
			}
			else//A<B
			{
				return Is_CombForm(str, left, keyi - 1) && Is_CombForm(str, keyi + 1, right);
			}
		}
	}
	else//第一个是（:(A)<B
	{
		int brackt = 0;//当brackt为0，说明将双目运算符的左操作数全体找到了
		int cnt = left;
		int flag = 0;
		while (cnt<right)
		{
			if (str[cnt] == '(')
			{
				brackt++;
			}
			if (str[cnt] == ')')
			{
				brackt--;
				flag = cnt;
			}
			++cnt;
			//[  left  , flag]  >  [flag+2,right]
			if (brackt == 0)
			{
				Del_Bracket(str, left, flag);
				if (str[flag + 2] == '(')
				{
					Del_Bracket(str, flag + 2, right);
					return Is_CombForm(str, left, flag - 2) && Is_CombForm(str, flag+ 2, right - 2);
				}
				else
				{
					return Is_CombForm(str, left, flag - 2) && Is_CombForm(str, flag + 2, right);
				}	
			}
		}
	}
	return false;
}

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include <stdlib.h> 
#include <string.h>
#include <assert.h>
#include<time.h>
#include<windows.h>

using namespace std;
//思路用一个辅助数组去占时存储非!的字符，之后拷贝到原区间，不过要对原区间进行赋值0
void Del_Negation(char* str,int n)
{

	assert(str);
	char* tmp = (char*)calloc(n+1, sizeof(char));//多一个1，是为了放置\0,避免strcpy越界拷贝
	assert(tmp);
	int cnt = 0;
	int i = 0;
	while (i < n)//将除！的字符赋值到tmp中
	{
		if (str[i] != '!')
		{
			tmp[cnt++] = str[i];
		}
		++i;
	}

	memset(str, 0, sizeof(char)*n);//对str那块内存重新赋值为\0,防止tmp拷贝到str中后，s扔有旧的数据
	strcpy(str, tmp);
	free(tmp);
	tmp = NULL;
}

//删除括号，因为是对首尾进行的删除，这里通过2次strcpy就可以完成
void Del_Bracket( char* str, int left, int right)
{
	assert(str);
	char* tmp = (char*)calloc(right - left +1+1, sizeof(char));
	assert(tmp);
	str[right] = '\0';
	strcpy(tmp, str+left+1);
	strcpy(str+left, tmp);
}

//判断是否为原子式
//因为去除括号的原因，当只有一个字母是原子式，否则不是
bool Is_operator(const char* str,int left,int right)
{
	assert(str);

	if ((0==(right-left))
		&&('A' <=str[left]||'Z'>=str[left]))
	{
	return true;
	}
	return false;
}

//找寻区间中第一双目运算符：找到就返回下标，否则就返回0.
int Find_Fist_operator( char* str, int left, int right)
{
	int ret = 0;
	while (left<right)
	{

		if (str[left + 1] == '+' || str[left + 1] == '*' || str[left + 1] == '>' || str[left + 1] == '=')
		{
			ret = left + 1;
			return ret;
		}

		++left;
	}
	return 0;//如果ret是0，说明是非法，反之就正确
}

bool Is_CombForm(char* str, int left, int right)
{

	if ((0 == (right - left))//区间是原子命题
		&& ('A' <= str[left] || 'Z' >= str[left]))
	{
		return true;
	}

	if (str[left] != '(')//第一个字符是字母： A>(B)
	{
		int keyi = Find_Fist_operator(str, left, right);//找寻第一个双目运算符
		if (keyi > 0)
		{
			if (str[keyi + 1] == '(')//A<(B)
			{
				Del_Bracket(str, keyi + 1, right);
				return Is_CombForm(str, left, keyi - 1) && Is_CombForm(str, keyi + 1, right - 2);
			}
			else//A<B
			{
				return Is_CombForm(str, left, keyi - 1) && Is_CombForm(str, keyi + 1, right);
			}
		}
	}
	else//第一个是（:(A)<B
	{
		int brackt = 0;//当brackt为0，说明将双目运算符的左操作数全体找到了
		int cnt = left;
		int flag = 0;
		while (cnt<right)
		{
			if (str[cnt] == '(')
			{
				brackt++;
			}
			if (str[cnt] == ')')
			{
				brackt--;
				flag = cnt;
			}
			++cnt;
			//[  left  , flag]  >  [flag+2,right]
			if (brackt == 0)
			{
				Del_Bracket(str, left, flag);
				if (str[flag + 2] == '(')
				{
					Del_Bracket(str, flag + 2, right);
					return Is_CombForm(str, left, flag - 2) && Is_CombForm(str, flag+ 2, right - 2);
				}
				else
				{
					return Is_CombForm(str, left, flag - 2) && Is_CombForm(str, flag + 2, right);
				}	
			}
		}
	}
	return false;
}



void Text(char *str)
{
	cout << str;
	int sz = strlen(str);
	Del_Negation(str, sz);
	sz = strlen(str);

	if (Is_CombForm(str, 0, sz-1))
	{
		printf("-------YES\n");
	}
	else
	{
		printf("----------NO\n");
	}
}
int main ()
{	
	


	char arr1[] = "P>!R";
	char  arr2[] = "!(P>Q)>!R";
	
	char arr3[] = "P>((P*R)>Q)";

	char arr4[] = "((P>R)*(Q*(P>R)))=R";
	char arr5[] = "((P>Q)>R)>Y";
	char arr6[] = "PQ";
	char arr7[] = "(P>RT)>Q";
	char arr8[] = "((P>Q)*(P>QT))>(R*T)";
	Text(arr1);
	Text(arr2);
	Text(arr3);
	Text(arr4);
	Text(arr5);
	Text(arr6);
	Text(arr7);
	Text(arr8);


	printf("-------------------------BY New Young\n");

	return 0;
}