M. Variable Shadowing

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100513/problem/M

Description

In computer programming, variable shadowing occurs when a variable declared within a certain scope has the same name as a variable declared in an outer scope. The outer variable is said to be shadowed by the inner variable, and this can lead to a confusion. If multiple outer scopes contain variables with the same name, the variable in the nearest scope will be shadowed.

Formally, a declared variable shadows another declared variable if the following conditions are met simultaneously:

  • the other variable is declared in outer scope and before (in terms of position in program source code) the declaration of the first variable,
  • the other variable is nearest among all variables satisfying the condition above.

Here is an example containing exactly one variable shadowing:

/* Prints a+max(b,c) */
int main() {
int a, b, c;
cin » a » b » c;
if (b > c) {
int a = b; // <– variable 'a' shadows outer 'a'
int x = c;
b = x;
c = a;
}
int x = a + c; // <– no shadowing here
cout « x « endl;
}

Variable shadowing is permitted in many modern programming languages including C++, but compilers can warn a programmer about variable shadowing to avoid possible mistakes in a code.

Consider a trivial programming language that consists only of scopes and variable declarations. The program consists of lines, each line contains only characters '{', '}', 'a' ... 'z' separated by one or more spaces.

  • Scopes. A scope (excluding global) is bounded with a pair of matching curly brackets '{' and '}'. A scope is an inner scope relative to another scope if brackets of the first scope are enclosed by brackets of the second scope.
  • Variables. A variable declaration in this language is written just as a name of the variable. In addition all variables are lowercase Latin letters from 'a' to 'z' inclusive (so there are at most 26 variable names). A variable is declared in each scope at most once.

Given a syntactically correct program (i.e. curly brackets form a regular bracket sequence), write an analyzer to warn about each fact of variable shadowing. Warnings should include exact positions of shadowing and shadowed variables. Your output should follow the format shown in the examples below.

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of lines in the program. The following n lines contain the program. Each program line consists of tokens '{', '}', 'a' ... 'z' separated by one or more spaces. The length of each line is between 1 and 50 characters. Each program line contains at least one non-space character.

The curly brackets in the program form a regular bracket sequence, so each opening bracket '{' has uniquely defined matching closing bracket '}' and vice versa. A variable is declared in a scope at most once. Any scope (including global) can be empty, i.e. can contain no variable declarations.

Output

For each fact of shadowing write a line in form "r1:c1: warning: shadowed declaration of ?, the shadowed position is r2:c2", where "r1:c1" is the number of line and position in line of shadowing declaration and "r2:c2" is the number of line and position in line of shadowed declaration. Replace '?' with the letter 'a' ... 'z' — the name of shadowing/shadowed variable. If multiple outer scopes have variables named as the shadowing variable, the variable in the nearest outer scope is shadowed.

Print warnings in increasing order of r1, or in increasing order of c1 if values r1 are equal. Leave the output empty if there are no variable shadowings.

Sample Input

1
{ a { b { a } } } b

Sample Output

1:11: warning: shadowed declaration of a, the shadowed position is 1:3

HINT

题意

给你一堆字符串,让你找到变量的父亲是啥

题解:

暴力找就好了……

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** struct node
{
int x,y;
};
node a[maxn];
char ss[][];
string s;
int main()
{
int n=read();
for(int i=;i<n;i++)
gets(ss[i]);
int num=;
for(int i=;i<n;i++)
for(int j=;j<strlen(ss[i]);j++)
{
s+=ss[i][j];
a[num].x=i,a[num++].y=j;
}
for(int i=;i<s.size();i++)
{
if(s[i]<='z'&&s[i]>='a')
{
int flag=;
int ans=;
for(int j=i-;j>=;j--)
{
if(ans==&&s[j]==s[i])
{
printf("%d:%d: warning: shadowed declaration of %c, the shadowed position is %d:%d\n",a[i].x+,a[i].y+,s[i],a[j].x+,a[j].y+);
flag=;
}
if(flag)
break;
if(s[j]=='}')
ans++;
if(s[j]=='{'&&ans>)
ans--;
} }
}
}

Codeforces Gym 100513M M. Variable Shadowing 暴力的更多相关文章

  1. Codeforces Gym 100513G G. FacePalm Accounting 暴力

    G. FacePalm Accounting Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100513 ...

  2. Codeforces Gym 100002 C "Cricket Field" 暴力

    "Cricket Field" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1000 ...

  3. Codeforces Gym 100342E Problem E. Minima 暴力

    Problem E. MinimaTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/attac ...

  4. Codeforces Gym 100203G G - Good elements 暴力

    G - Good elementsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...

  5. Codeforces Gym 101190M Mole Tunnels - 费用流

    题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...

  6. Codeforces Gym 101252D&&floyd判圈算法学习笔记

    一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈 ...

  7. Codeforces Gym 101623A - 动态规划

    题目传送门 传送门 题目大意 给定一个长度为$n$的序列,要求划分成最少的段数,然后将这些段排序使得新序列单调不减. 考虑将相邻的相等的数缩成一个数. 假设没有分成了$n$段,考虑最少能够减少多少划分 ...

  8. 【Codeforces Gym 100725K】Key Insertion

    Codeforces Gym 100725K 题意:给定一个初始全0的序列,然后给\(n\)个查询,每一次调用\(Insert(L_i,i)\),其中\(Insert(L,K)\)表示在第L位插入K, ...

  9. Codeforces gym 101343 J.Husam and the Broken Present 2【状压dp】

     2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Hu ...

随机推荐

  1. Mac终端编译运行C++

    1.在编辑器中写好C++代码 2.打开终端打开文件对应的地址 3.用g++命令来编译.cpp文件 4.用./文件名来运行 观察文件的目录可发现 g++ 源文件名 编译源文件,产生a.out ./文件名 ...

  2. 【大数模板】C++大数类 大数模板

    分别使用C++中的运算符重载的方法来实现大数之间的数学运算,包括加法.减法.乘法.除法.n次方.取模.大小比较.赋值以及输入流.输出流的重载. 感觉很麻烦... [代码] #include<io ...

  3. JAVA数据库处理(连接,数据查询,结果集返回)

    package john import java.io.IOException; import java.util.*; public class QueryDataRow { public Hash ...

  4. POJ 1423 Big Number

    题意:求n阶乘的位数. 解法:斯特林公式,,然后取log10就是位数了,因为精度问题需要化简这个式子,特判1. 代码: #include<stdio.h> #include<iost ...

  5. How to Calculate difference between two dates in C# z

    Do you need to find the difference in number of days, hours or even minute between the two date rang ...

  6. [Irving] SQL 2005/SQL 2008 备份数据库并自动删除N天前备份的脚本

    以下为SQL脚本,本人以执行计划来调用,所以改成了执行命令,大家可根据自己需要改为存储过程使用 )='E:\MsBackUp\SqlAutoBackup\' --备份路径; --备份类型为全备,1为差 ...

  7. STM32查看系统时钟

    调用库函数RCC_GetClocksFreq,该函数可以返回片上的各种时钟的频率 函数原形  void  RCC_GetClocksFreq(RCC_ClocksTypeDef*  RCC_Clock ...

  8. PHP 正则表达式总结

    可以用字符作为一个通配符来代替除换行符(\n)之外的任一个字符.例如,正则表达式:.at可以与"cat"."sat"."#at"和" ...

  9. python中struct模块及packet和unpacket

    转自:http://www.cnblogs.com/gala/archive/2011/09/22/2184801.html 我们知道python只定义了6种数据类型,字符串,整数,浮点数,列表,元组 ...

  10. Python闭包与javascript闭包比较

    实例一 python def line_conf(): def line(x): return 2*x+1 print(line(5)) # within the scope     line_con ...