POJ-3295 Tautology (构造法）

POJ-3295 Tautology (构造法）#

题目链接poj-3295
大意就是pqrst五个布尔变量，KCEAN五个方法
K是并，A是或，N是取反，C是蕴含，E是相等
构造的是这个变量的2^5=32种取法很有意思

int a[35][6];
for (int i = 0; i < 32; i++)
{
	a[i + 1][5] = i % 2;
	a[i + 1][4] = i / 2 % 2;
	a[i + 1][3] = i / 4 % 2;
	a[i + 1][2] = i / 8 % 2;
	a[i + 1][1] = i / 16 % 2;
}  

结果是这样的

0 0 0 0 0
0 0 0 0 1
0 0 0 1 0
0 0 0 1 1
0 0 1 0 0
0 0 1 0 1
0 0 1 1 0
0 0 1 1 1
0 1 0 0 0
0 1 0 0 1
0 1 0 1 0
0 1 0 1 1
0 1 1 0 0
0 1 1 0 1
0 1 1 1 0
0 1 1 1 1
1 0 0 0 0
1 0 0 0 1
1 0 0 1 0
1 0 0 1 1
1 0 1 0 0
1 0 1 0 1
1 0 1 1 0
1 0 1 1 1
1 1 0 0 0
1 1 0 0 1
1 1 0 1 0
1 1 0 1 1
1 1 1 0 0
1 1 1 0 1
1 1 1 1 0
1 1 1 1 1  

后面的算式用后缀表达式去模拟
代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<stack>
using namespace std;
string str;

int main()
{
	int a[35][6];
	for (int i = 0; i < 32; i++)
	{
		a[i + 1][5] = i % 2;
		a[i + 1][4] = i / 2 % 2;
		a[i + 1][3] = i / 4 % 2;
		a[i + 1][2] = i / 8 % 2;
		a[i + 1][1] = i / 16 % 2;
	}
	for (int i = 1; i <= 32; i++)
	{
		for (int j = 1; j <= 5; j++)
			cout << a[i][j] << " ";
		cout << endl;
	}
	while (cin >> str)
	{
		if (str == "0")
			break;
		bool flag = false;
		for (int i = 1; i < 33; i++)
		{
			stack<int>s;
			for (int j = str.size() - 1; j >= 0; j--) {
				if (str[j] >= 'p' && str[j] <= 't')
				{
					s.push(a[i][str[j] - 'o']);
				}
				else {
					int n1, n2;
					char now = str[j];
					if (now == 'K')
					{
						n1 = s.top();
						s.pop();
						n2 = s.top();
						s.pop();
						s.push(n1 && n2);
					}
					if (now == 'A')
					{
						n1 = s.top();
						s.pop();
						n2 = s.top();
						s.pop();
						s.push(n1 || n2);
					}
					if (now == 'C')
					{
						n1 = s.top();
						s.pop();
						n2 = s.top();
						s.pop();
						s.push((!n1) || n2);
					}
					if (now == 'E')
					{
						n1 = s.top();
						s.pop();
						n2 = s.top();
						s.pop();
						s.push(n1 == n2);
					}
					if (now == 'N')
					{
						n1 = s.top();
						s.pop();
						s.push(!n1);
					}
				}
			}
			flag = s.top();
			if (flag == 0)
			{
				cout << "not\n";
				break;
			}
		}
		if (flag)
			cout << "tautology\n";
	}
	return 0;
}