sicily1000. 输入输出LL(1)语法分析程序_语法分析输入输出

作者：笔触狂放9 | 2024-05-30 00:36:42

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语法分析输入输出

1000. 输入输出LL(1)语法分析程序

题目：

Time Limit: 1sec Memory Limit:256MB

Description

输入开始符号，非终结符，终结符，产生式，LL(1)分析表
输出LL(1)分析表

G[E]:E →E+T | E-T | T

T →T*F | T/F | F

F →(E) | D

D →x | y | z

消除左递归G1[E]:

E →TA

A →+TA | -TA | e

T →FB

B →*FB | /FB | e

F →(E) | D

D →x | y | z

Input

输入开始符号；
非终结符个数，非终结符，空格符分隔；
终结符个数，终结符，空格符分隔；
产生式的个数，各产生式的序号，产生式的左边和右边符号，空格符分隔；
LL(1)分析表中的产生式个数，序号，行符号，列符号，产生式编号，空格符分隔；

Output

第一行：空，安终结符循序输出终结符，结束符‘#’，每个符号占5格；
其余行：非终结符符号，各对应终结符的产生式的右边，每个符号占5格；

Sample Input

Copy sample input to clipboard

E
6  E A T B F D
9 + - * / ( ) x y z 
13
1  E TA
2  A +TA
3  A -TA
4  A k
5  T FB
6  B *FB
7  B /FB
8  B k
9  F (E)
10 F D
11 D x
12 D y
13 D z
25
1  E ( 1
2  E x 1
3  E y 1
4  E z 1
5  A + 2
6  A - 3
7  A ) 4
8  A # 4
9  T ( 5
10 T x 5
11 T y 5
12 T z 5
13 B + 8
14 B - 8
15 B * 6
16 B / 7
17 B ) 8
18 B # 8
19 F ( 9
20 F x 10
21 F y 10
22 F z 10
23 D x 11
24 D y 12
25 D z 13

Sample Output

         +    -    *    /    (    )    x    y    z    #
    E                       TA        TA   TA   TA     
    A  +TA  -TA                   k                   k
    T                       FB        FB   FB   FB     
    B    k    k  *FB  /FB         k                   k
    F                      (E)         D    D    D     
    D                                  x    y    z

代码：


#include <cstring>
#include <string.h>
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
 
struct v {
    char vn;
    string vt;
    //v(string vn1, string vt1) {vn = vn1, vt = vt1;}
};
 
struct sel {
    char v, t;
    int ch;
};
 
vector<char> s1,s2; //s1非终结符， s2终结符
vector<v> s3;
vector<sel> s4;
 
string find(char v, char t) {
    for (auto i : s4) {
        if (i.v == v&&i.t == t) return s3[i.ch-1].vt;
    }
    return "";
}
 
int main() {
    string start;
    cin >> start;
    int s1num, s2num, s3num,s4num;
    cin >> s1num;
    for (int i = 0; i < s1num; i++) {
        char tmp;
        cin >> tmp;
        s1.push_back(tmp);
    }
    cin >> s2num;
    for (int i = 1; i <= s2num; i++) {
        char tmp;
        cin >> tmp;
        s2.push_back(tmp);
    }
    s2num++;
    s2.push_back('#');
    cin >> s3num;
    for (int i = 1; i <= s3num; i++) {
        char vn;
        string vt;
        int n;
        cin >> n;
        cin >> vn;
        cin >> vt;
        v tmp;
        tmp.vn = vn, tmp.vt = vt;
        s3.push_back(tmp);
    }
    cin >> s4num;
    for (int i = 1; i <= s4num; i++) {
        char v, t;
        int num, n;
        cin >> n,cin >> v >> t, cin >> num;
        sel tmp;
        tmp.v = v, tmp.t = t, tmp.ch = num;
        s4.push_back(tmp);
    }
    int widthOccupy = 5;
    cout << setw(widthOccupy) << ' ';
    for (int i = 0; i < s2num; ++i)
        cout << setw(widthOccupy) << s2[i];
    cout << endl;
    for (int i = 0; i < s1num; ++i) {
        cout << setw(widthOccupy) << s1[i];
        for (int j = 0; j < s2num; ++j) {
            cout << setw(widthOccupy) << find(s1[i], s2[j]);
        }
        cout << endl;
    }
    return 0;
}

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