标签：
In this problem, we consider a simple programming language that has only declarations of one-dimensional integer arrays and assignment statements. The problem is to find a bug in the given program.
The syntax of this language is given in BNF as follows:
| <program> | ::= | <declaration> | <program><declaration> | <program><assignment> |
| <declaration> | ::= | <array_name>[<number>]<new_line> |
| <assignment> | ::= | <array_name>[<expression>]=<expression><new_line> |
| <expression> | ::= | <number> | <array_name>[<expression>] |
| <number> | ::= | <digit> | <digit_positive><digit_string> |
| <digit_string> | ::= | <digit> | <digit><digit_string> |
| <digit_positive> | ::= | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| <digit> | ::= | 0 | <digit_positive> |
| <array_name> | ::= | a | b | c | d | e | f | g | h | i | j | k | l | m |n | o | p | q | r | s | t | u | v | w | x | y | z |A | B | C | D | E | F | G | H | I | J | K | L | M |N | O | P | Q | R | S | T | U | V | W | X | Y | Z |
where <new_line> denotes a new line character (LF).
Characters used in a program are alphabetical letters, decimal digits, =, [, ] and new line characters. No other characters appear in a program.
A declaration declares an array and specifies its length. Valid indices of an array of length n are integers between 0 and n - 1, inclusive. Note that the array names are case sensitive, i.e. array a and array A are different arrays. The initial value of each element in the declared array is undefined.
For example, array a of length 10 and array b of length 5 are declared respectively as follows.
a[10] b[5]
An expression evaluates to a non-negative integer. A <number> is interpreted as a decimal integer. An <array_name>[<expression>] evaluates to the value of the <expression>-th element of the array. An assignment assigns the value denoted by the right hand side to the array element specified by the left hand side.
Examples of assignments are as follows.
a[0]=3 a[1]=0 a[2]=a[a[1]] a[a[0]]=a[1]
A program is executed from the first line, line by line. You can assume that an array is declared once and only once before any of its element is assigned or referred to.
Given a program, you are requested to find the following bugs.
An index of an array is invalid.
An array element that has not been assigned before is referred to in an assignment as an index of array or as the value to be assigned.
You can assume that other bugs, such as syntax errors, do not appear. You can also assume that integers represented by <number>s are between 0 and 231 - 1 (= 2147483647), inclusive.
The input consists of multiple datasets followed by a line which contains only a single ‘.’ (period). Each dataset consists of a program also followed by a line which contains only a single ‘.’ (period). A program does not exceed 1000 lines. Any line does not exceed 80 characters excluding a new line character.
For each program in the input, you should answer the line number of the assignment in which the first bug appears. The line numbers start with 1 for each program. If the program does not have a bug, you should answer zero. The output should not contain extra characters such as spaces.
a[3] a[0]=a[1] . x[1] x[0]=x[0] . a[0] a[0]=1 . b[2] b[0]=2 b[1]=b[b[0]] b[0]=b[1] . g[2] G[10] g[0]=0 g[1]=G[0] . a[2147483647] a[0]=1 B[2] B[a[0]]=2 a[B[a[0]]]=3 a[2147483646]=a[2] . .
2 2 2 3
4 0
玩的又是字符串！！！要疯！！！
写了一半，实在是不会用stack啊啊啊啊啊
#include <algorithm> #include <cstdio> #include <iostream> #include <string> #include <map> #include <stack> #include <sstream> using namespace std; map<string,int> zhi; stack<char>s; string change(char t) { char ss[2]; ss[0] = t; ss[1] = ‘\0‘; return ss; } void deal_dingyi(string exp) { int n = exp.length(); string t1; stringstream ss; if(isalpha(exp[0]))t1 = change(exp[0]); int range,x1 = 0,x2; for(int i = 0;i < n; i++){ if(exp[i] <‘9‘&& exp[i] > ‘0‘){ if(!x1)x1 = i; if(exp[i+1] == ‘]‘){ x2 = i - 1; break; } } } //cout<<"x1 "<<x1<<"x2 "<<x2<<endl; string t2 = exp.substr(x1,x2); //cout<<t2<<endl; ss.str(t2); ss >> range; zhi[t1] = range; } bool deal_fuzhi(string exp) { int n = exp.length(); stringstream ss; if(!(exp.find(‘=‘) < n &&exp.find(‘=‘) >= 0))deal_dingyi(exp); //定义数组的存储 else{ for(int i = 0;i <= n; i++){ } } return false; } int main() { string exp; while(cin >> exp){ if(exp == ".")break; int n = 1; //计数第几行 deal_fuzhi(exp);//if(!)cout << n++ <<endl; while(cin>>exp){ if(exp == ".")break; if(!deal_fuzhi(exp))cout << n++ <<endl; } } system("pause"); return 0; }
那就不用stack，想想其他方法。
结果想了想其他，应该是要用递归做的，然而递归也不太会，好难过......
最后看别人的代码，好不容易把递归整了出来。但感觉还是不能完全理解递归，需要好好琢磨
#include <algorithm> #include <cstdio> #include <iostream> #include <string> #include <map> #include <sstream> using namespace std; map<string,long long> zhi; map<string,string>val; bool bug; string change(char t) { char ss[2]; ss[0] = t; ss[1] = ‘\0‘; return ss; } void deal_arry(string exp) //存储数组定义 { int n = exp.length(); string t1; stringstream ss; long long range; t1 = exp.substr(0,exp.find(‘[‘)); string t2 = exp.substr(exp.find(‘[‘)+1,exp.find_last_of(‘]‘)-2); ss.str(t2); ss >> range; //cout<<range<<"///"<<endl; zhi[t1] = range; } string deal_index(string exp,string arry) { long long t = 0; string y, z; //cout<<"!"<<exp<<endl; if(exp.find("[") == string::npos){ //exp为数字 stringstream ss(exp); ss >> t; //cout<<endl<<arry<<"+"<<zhi[arry]<<"+"<<t<<endl; if( arry != "\0" &&t >= zhi[arry]){ //cout<<"*1*"<<endl; bug = true; } return exp; } y = exp.substr(0, exp.find_first_of("[")); z = exp.substr(exp.find("[")+1, exp.find_last_of("]") - 2); //cout<<"z "<<z <<endl; //cout<<" y "<<y<<endl; z = deal_index(z, y); if(bug) return "/0" ; exp = y + "[" + z + "]"; if( !val.count(exp) ) bug = 1; return val[exp]; } int main() { string exp; int n = 1; bool mark = false; while(cin >> exp){ if(exp == "."){ val.clear(); zhi.clear(); if(mark)break; else{ if(!bug)cout<<0<<endl; n = 1; //计数第几行 bug = false; mark = true; } } else{ mark = false; if(bug)continue; if(exp.find(‘=‘) == string::npos){ //find函数找不到字符返回" string::npos " deal_arry(exp); } else{ string index, arry1, value, arry2,s; s = exp.substr(0,exp.find(‘=‘)); //cout << s<<"%"<<endl; index = s.substr(s.find(‘[‘)+1,s.find_last_of(‘]‘)-2); //数组第几个数 arry1 = s.substr(0,s.find(‘[‘)); //数组 value = exp.substr(exp.find(‘=‘)+1); //数组存储的值 //cout << index <<"@"<<endl; index = deal_index(index,arry1); stringstream ss(index); long long t; ss >> t; if( t>= zhi[arry1]){ bug = true; //数组越界 } string left_value = arry1+‘[‘+index+‘]‘; value = deal_index(value, "\0"); if(bug)cout <<n <<endl; else { val[left_value] = value; } } n++; } } // system("pause"); return 0; }
以下是别人的代码，看了看改了改，是用stack做的
话说写的真的很棒
#include <iostream> #include <string> #include <vector> #include <set> #include <map> #include <sstream> #include <fstream> #include <stack> using namespace std; map<string, long long> array_table; map<string,map< long long, long long>> array_value_table; //string数组的下标为int的位置的值为int #define FILE void partition(string origin,string &left,string &right) //把式子等号两边分开 { int index = origin.find_first_of(‘=‘,0); cout << "index "<<index<<endl; if(index != -1){ //有等号 赋值 left = origin.substr(0,index); right = origin.substr(index+1); } else{ //无等号 定义 left = origin; right = ""; } } void getArray(string str,string &name, string &value) //把数组字母与[]内数字分开 { int begin = str.find_first_of(‘[‘,0); int end = str.find_last_of(‘]‘); if(begin != -1) //有[ 是数组 { name = str.substr(0,begin); value = str.substr(begin+1,end-begin-1); } else //无[ 不是数组—常量或者字母 { name = ""; value = str; } } long long calculateArray(string str) { string name,value; getArray(str,name,value); stack<string> s; long long ans; while(name!=""){ //只剩下[]内的数字时 停止 string left,right; getArray(value,left,right); if (array_table.count(name) == 0)return -1; //无此数组的定义 value = right; s.push(name); name = left; } ans = atoi(value.c_str()); if(ans<0) return -1; //数组下标小于0 while (!s.empty()) { string left = s.top(); long long num = array_table[left]; if(ans >= num) return -1; //数组下标超出数组范围 if(array_value_table[left].count(ans) == 0)return -1; //数组下标为 ans 的数未定义 ans = array_value_table[left][ans]; s.pop(); } return ans; } int main(int argc, char* argv[]) { #ifdef FILE ifstream in("data.txt"); ofstream out("output.txt"); cin.rdbuf(in.rdbuf()); cout.rdbuf(out.rdbuf()); #endif string str; bool mark = false,isfirst = true; int num = 0; while(cin>>str) { if(str!=".") { mark = false; num++; string left,right; partition(str,left,right); if(right=="") //定义数组 { string name,value; getArray(left,name,value); array_table[name] = atoi(value.c_str()); //atoi将字符串转化为整型 } else //赋值语句 { string name, value; long long ans,temp; getArray(left,name,value); temp = calculateArray(value); //value为数组下标 ans = calculateArray(right); //赋给数组的值 if(temp >= 0&& temp < array_table[name]&& ans != -1){ array_value_table[name][temp] = ans; } else{ if(isfirst){ cout<<num<<endl; isfirst = false; } } } } else { array_table.clear(); array_value_table.clear(); if(mark) break; else { if(isfirst) cout<<0<<endl; mark = true; isfirst = true; num = 0; } } } //system("pause"); return 0; }
标签：
原文地址：http://www.cnblogs.com/farewell-farewell/p/5477457.html
