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.

 

Input

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.

 

Output

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.

 

Sample Input

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]
.
.

Sample Output

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[];
ss[] = t;
ss[] = '\0';
return ss;
} void deal_dingyi(string exp)
{
int n = exp.length();
string t1;
stringstream ss;
if(isalpha(exp[]))t1 = change(exp[]);
int range,x1 = ,x2;
for(int i = ;i < n; i++){
if(exp[i] <''&& exp[i] > ''){
if(!x1)x1 = i;
if(exp[i+] == ']'){
x2 = i - ;
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('=') >= ))deal_dingyi(exp); //定义数组的存储
else{
for(int i = ;i <= n; i++){ }
} return false;
} int main()
{
string exp;
while(cin >> exp){
if(exp == ".")break;
int n = ; //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 ;
}

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[];
ss[] = t;
ss[] = '\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(,exp.find('['));
string t2 = exp.substr(exp.find('[')+,exp.find_last_of(']')-);
ss.str(t2);
ss >> range;
//cout<<range<<"///"<<endl;
zhi[t1] = range;
} string deal_index(string exp,string arry)
{
long long t = ;
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(, exp.find_first_of("["));
z = exp.substr(exp.find("[")+, exp.find_last_of("]") - );
//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 = ;
return val[exp]; } int main()
{
string exp;
int n = ;
bool mark = false;
while(cin >> exp){
if(exp == "."){
val.clear();
zhi.clear();
if(mark)break;
else{
if(!bug)cout<<<<endl;
n = ; //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(,exp.find('='));
//cout << s<<"%"<<endl;
index = s.substr(s.find('[')+,s.find_last_of(']')-); //数组第几个数
arry1 = s.substr(,s.find('[')); //数组
value = exp.substr(exp.find('=')+); //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 ;
}

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('=',);
cout << "index "<<index<<endl;
if(index != -){ //有等号 赋值
left = origin.substr(,index);
right = origin.substr(index+);
}
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('[',);
int end = str.find_last_of(']');
if(begin != -) //有[ 是数组
{
name = str.substr(,begin);
value = str.substr(begin+,end-begin-);
}
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) == )return -; //无此数组的定义 value = right;
s.push(name);
name = left; }
ans = atoi(value.c_str());
if(ans<) return -; //数组下标小于0
while (!s.empty())
{
string left = s.top();
long long num = array_table[left];
if(ans >= num) return -; //数组下标超出数组范围
if(array_value_table[left].count(ans) == )return -; //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 = ; 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 >= && temp < array_table[name]&& ans != -){
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<<<<endl;
mark = true;
isfirst = true;
num = ;
}
}
}
//system("pause");
return ;
}
												

uva 1596 Bug Hunt的更多相关文章

  1. 【技巧性(+递归运用)】UVa 1596 - Bug Hunt

    In this problem, we consider a simple programming language that has only declarations of onedimensio ...

  2. UVA 1596 Bug Hunt (大模拟 栈)

    题意: 输入并模拟执行一段程序,输出第一个bug所在的行. 每行程序有两种可能: 数组定义: 格式为arr[size]. 例如a[10]或者b[5],可用下标分别是0-9和0-4.定义之后所有元素均为 ...

  3. UVa 1596 Bug Hunt (string::find && map && 模拟)

    题意 : 给出几组由数组定义与赋值构成的编程语句, 有可能有两种BUG, 第一种为数组下标越界, 第二种为使用尚未定义的数组元素, 叫你找出最早出现BUG的一行并输出, 每组以' . '号分隔, 当有 ...

  4. UVa 1596 Bug Hunt (STL栈)

    题意:给定两种操作,一种是定义一个数组,另一种是赋值,让你找出哪一步时出错了,出错只有两种,一种是数组越界,另一种是访问未定义变量. 析:当初看到这个题时,感觉好麻烦啊,然后就放过去了,而现在要重新回 ...

  5. 【UVA】1596 Bug Hunt(模拟)

    题目 题目     分析 算是个模拟吧     代码 #include <bits/stdc++.h> using namespace std; map<int,int> a[ ...

  6. [刷题]算法竞赛入门经典(第2版) 5-9/UVa1596 - Bug Hunt

    //开学了,好烦啊啊啊啊啊!怎么开个学那么多破事情!!都俩星期了,终于有时间写出来一道题 题意:不难理解,不写了.这几天忙的心累. 代码:(Accepted, 0.010s) //UVa1596 - ...

  7. Bug Hunt UVA - 1596

      In this problem, we consider a simple programming language that has only declarations of onedimens ...

  8. 【习题 5-9 UVA - 1596】Bug Hunt

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] map模拟 map<string,int>记录每个数组的大小 map <pair<string, int&g ...

  9. 在iOS上使用ffmpeg播放视频

    国外靠谱的有这几个:1.Mooncatventures group https://github.com/mooncatventures-group 2.KxMoviePlayer (use Open ...

随机推荐

  1. windows中java读目录空格变成%20 处理方法

    URL url = Thread.currentThread().getContextClassLoader().getResource(""); String path = ur ...

  2. uva Fire!

    算法:BFS Joe works in a maze. Unfortunately, portions of the maze have caught on fire, and the owner o ...

  3. Session与Cookie间不得不说的一些事

    在很久很久以前,刚有浏览器和网页的时候,web开发者发现了一个问题,我必须要在客户端这边保存一些东西才能实现某些功能,比如大家喜闻乐见的购物车.用户登录.自动登陆等.但是客户端只有一个浏览器,怎么在用 ...

  4. HTML5视频

    <video>标签用于定义视频. 案例1: <!DOCTYPE html><html><head lang="en"> <me ...

  5. [JS]九种网页弹窗代码

    [1.最基本的弹出窗口代码] 其实代码非常简单: <SCRIPT LANGUAGE="javascript"><!--window.open ("pag ...

  6. [算法导论]练习2-4.d求排列中逆序对的数量

    转载请注明:http://www.cnblogs.com/StartoverX/p/4283186.html 题目:给出一个确定在n个不同元素的任何排列中逆序对数量的算法,最坏情况需要Θ(nlgn)时 ...

  7. Linux下安装memcached图文版

    所用到的工具: putty小软件  可以远程连接主机的小软件     然后在屏幕上输入用户名root  密码:******即可连接 由于安装这两个工具需要编译所以:第一步要进行安装编译: 习惯于将下载 ...

  8. [android]-如何在向服务器发送request时附加已保存的cookie数据

    [android]-如何在向服务器发送request时附加已保存的cookie数据 应用场景:在开发android基于手机端+服务器端的应用时,登陆->获取用户信息->获取授权用户相关业务 ...

  9. SIM卡基础,各管脚意义,封装定义

    1. SIM简介 SIM卡(Subscriber Identity Module).即用户识别模块,是一张符合GSM规范的"智慧卡".SIM卡可以插入任何一部符合GSM规范的移动电 ...

  10. 关于Microsoft app下同义词的整理

    Windows os 以下词表达的是同一个概念 windows store app windows metro app windows modern app windows runtime app w ...