Description The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V )where V is for True, and F is for False. The expressions may include the following ope…

总时间限制: 1000ms  内存限制: 65536kB 描述 The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next:  Expression:( V | V ) & F & ( F | V ) where V is for True, and F is for False. The expressions may incl…

/* * POJ_2106.cpp * * Created on: 2013年10月30日 * Author: Administrator */ #include <iostream> #include <cstdio> using namespace std; const int maxn = 110; int op[maxn], otop; int val[maxn],vtop; void insert(int b){//将操作数b压入操作数栈val while(otop &a…

题意:关于!,&,| 的运算,表达式中V代表true,F代表false. 思路:见代码吧,很详细了. 要注意 !!!F,!(...) 的情况. #include <iostream> #include <stdio.h> #include <stack> #include <string.h> #include <map> using namespace std; ; stack<int> val; //存储操作数和中间运算结…

#include<cstdio> const int maxn=100 +10; int val[maxn],vtop; int op[maxn],otop; void insert(int b) { while(otop &&op[otop-1]==3) { b=!b; --otop; } val[vtop++]=b; } void calc(void) { int b=val[--vtop]; int a=val[--vtop]; int opr=op[--otop]; i…

The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V ) where V is for True, and F is for False. The expressions may include the following operators: ! f…

总时间限制: 1000ms  内存限制: 65536kB 描述The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V )where V is for True, and F is for False. The expressions may includ…

Boolean Expressions Time Limit: 1000MS   Memory Limit: 30000K       Description The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V )where V is for Tru…

题目链接: http://poj.org/problem?id=3295 题目描述: 给一个字符串,字符串所表示的表达式中p, q, r, s, t表示变量,取值可以为1或0.K, A, N, C, E 分别表示且,或,非,真蕴含,等值.问表达式是不是永真的,如果是输出“tautology”,否则输出“not”. 解题思路: 这里借用到了递归的本质,也就是对栈的模拟,用递归进行压栈,求表达式的值,再加上对变量状态压缩进行枚举. #include <cstdio>//本代码用G++交就ac,c+…

https://stackoverflow.com/questions/2802055/what-does-the-construct-x-x-y-mean -------------------------------------------------------- What is the double pipe operator (||)? The double pipe operator (||) is the logical OR operator . In most language…