POJ 2106 Boolean Expressions

Boolean Expressions

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3665		Accepted: 1104

Description

The objective of the program, going to produce was to evaluate Boolean expressions as the one shown next:
Expression: (V | V) & F & (F | V
Where V is for True, and F are for False. The expressions may include the following operators:! For no, & for and, | For or, the use of parenthesis for operations grouping is also allowed.

To perform the evaluation of an expression, it'll be considered the "the" the operators, the not have the highes T, and the or the lowest. The program must yield V or F, as the result of each expression in the input file.

Input

The expressions is of a variable length, although'll never exceed. Symbols May is separated by any number of spaces or no spaces at all, therefore, the total length of a expression, as a n Umber of characters, is unknown.

The number of expressions in the input file is variable and would never be greater than 20. Each expression was presented in a new line, as shown below.

Output

For each test expression, print "expression" followed by its sequence number, ":", and the resulting value of the Corres ponding test expression. Separate the output for consecutive test expressions with a new line.

Use the same format as, shown in the sample output shown below.

Sample Input

(V | V) & F & (f| V)! X 2 V & V &! F & (F | V) & (! F | F |! V & V) (f&f| v|! v&! f&! (f| F&V))

Sample Output

Expression 1:fexpression 2:vexpression 3:v

Source

México and central America 2004ACcode:

#include <iostream> #include <cstring> #include <cstdio>using namespace Std;int sum[110],poss;int op[        110],poso;void Insert (int t) {while (poso&&op[poso-1]==3) {t=!t;    --poso; } sum[poss++]=t;}    void Calc () {int B=sum[--poss];    int A=sum[--poss];    int O=op[--poso];    int c= (A&AMP;B);    if (o==1) c= (a|b); Insert (c);}    int main () {int loop=1;    char c;        while ((C=getchar ())!=eof) {poss=poso=0;            do{if (c== ' (') {op[poso++]=0;                    }else if (c== ') ') {while (poso&&op[poso-1]!=0) calc ();                --poso;            Insert (Sum[--poss]); }else if (c== '! ')            {op[poso++]=3;                }else if (c== ' & ') {while (poso&&op[poso-1]>=2) calc ();            op[poso++]=2; }else if (c== ' | ')   {while (poso&&op[poso-1]>=1) calc ();             Op[poso++]=1; } else if (c== ' V ' | |            c== ' F ') {insert (c== ' V '? 1:0);        }}while ((C=getchar ())! = ' \ n ' &&c!=eof);        while (Poso) calc (); printf ("Expression%d:%c\n", loop++, (sum[0]? ')    V ': ' F ')); } return 0;}

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POJ 2106 Boolean Expressions