POJ | Boolean Expressions
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 include the following operators: ! for not , & for and, | for or , the use of parenthesis for operations grouping is also allowed.
To perform the evaluation of an expression, it will be considered the priority of the operators, the not having the highest, and the or the lowest. The program must yield V or F , as the result for each expression in the input file.
输入
The expressions are of a variable length, although will never exceed 100 symbols. Symbols may be separated by any number of spaces or no spaces at all, therefore, the total length of an expression, as a number of characters, is unknown.
The number of expressions in the input file is variable and will never be greater than 20. Each expression is presented in a new line, as shown below.
输出
For each test expression, print "Expression " followed by its sequence number, ": ", and the resulting value of the corresponding test expression. Separate the output for consecutive test expressions with a new line.
Use the same format as that shown in the sample output shown below.
样例输入
( V | V ) & F & ( F| V)
!V | V & V & !F & (F | V ) & (!F | F | !V & V)
(F&F|V|!V&!F&!(F|F&V))
样例输出
Expression 1: F
Expression 2: V
Expression 3: V
解题思路
此题目可以通过递归的方式来做，下面是根据题意分析的Grammar解析过程，据此我们可以写实现代码。
1 #include <stdio.h> 2 #include <stdbool.h> 3 #include <stdlib.h> 4 5 char str[100] = {'\0'}; 6 int idx = 0; 7 bool expression(); 8 9 10 11 bool primary() 12 { 13 char op = str[idx]; 14 bool result; 15 if (op == '(') 16 { 17 idx++; //( 18 result = expression(); 19 idx++; //) 20 } 21 else if (op == 'V') 22 { 23 result = true; 24 idx++; 25 } 26 else if (op == 'F') 27 { 28 result = false; 29 idx++; 30 } 31 else if (op == '!') 32 { 33 idx++; 34 result = !primary(); 35 } 36 return result; 37 38 } 39 40 bool expression() 41 { 42 bool result = primary(); 43 while (true) 44 { 45 char op = str[idx]; 46 if (op == '&' || op == '|') 47 { 48 idx++; 49 bool value = primary(); 50 if (op == '&') 51 { 52 result &= value; 53 } 54 else 55 { 56 result |= value; 57 } 58 } 59 else 60 { 61 break; 62 } 63 } 64 return result; 65 } 66 67 bool getline_ns(char *str, int max) 68 { 69 bool ret = false; 70 char *s = (char *)malloc(sizeof(char) * max * 10); 71 if (fgets(s, max * 10, stdin)) 72 { 73 int i = 0, j = 0; 74 for (; s[i] != '\0'; i++) 75 { 76 if (s[i] != ' ') 77 str[j++] = s[i]; 78 } 79 str[j++] = '\0'; 80 ret = true; 81 } 82 free(s); 83 return ret; 84 } 85 int main() 86 { 87 int i = 0; 88 while (getline_ns(str, 100)) 89 { 90 idx = 0; 91 printf("Expression %d: %c\n", ++i, expression() ? 'V' : 'F'); 92 } 93 return 0; 94 }
