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ECNUOJ 2574 Principles of Compiler

of compiler
2023-09-11 14:15:28 时间

Principles of Compiler

Time Limit:1000MS Memory Limit:65536KB
Total Submit:473 Accepted:106

Description 

After learnt the Principles of Compiler,partychen thought that he can solve a simple expression problem.So he give you strings of less than 100 characters which strictly adhere to the following grammar (given in EBNF):
    A:= '(' B')'|'x'.
    B:=AC.
    C:={'+'A}.
Can you solve them too?

Input 

The first line of input gives the number of cases, N(1 ≤ N ≤ 100). N test cases follow.
The next N lines will each contain a string as described above.

Output 

For each test case,if the expression is adapt to the EBNF above output “Good”,else output “Bad”.

Sample Input 

3
(x)
(x+(x+x))
()(x)

Sample Output 

Good
Good
Bad

Source

解题：几十万只草泥马呼啸而过，一直没搞懂{+A}的含义

 

重复 { ... } 也就是说这个是表示+A可以出现0次或多次

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 210;
 4 char str[maxn];
 5 int cur;
 6 bool A();
 7 bool B();
 8 bool C();
 9 bool A() {
10     if(str[cur] == 'x') {
11         cur++;
12         return true;
13     }
14      if(str[cur] == '(') {
15         cur++;
16         if(B() && str[cur] == ')') {
17             cur++;
18             return true;
19         }
20     }
21     return false;
22 }
23 bool B() {
24     return A() && C();
25 }
26 bool C() {
27     while(str[cur] == '+'){
28         cur++;
29         if(!A()) return false;
30     }
31     return true;
32 }
33 int main() {
34     int kase;
35     scanf("%d",&kase);
36     getchar();
37     while(kase--) {
38         gets(str);
39         cur = 0;
40         puts(A() && str[cur] == '\0'?"Good":"Bad");
41     }
42     return 0;
43 }
View Code

 

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