Codeforces Gym 100513M M. Variable Shadowing 暴力
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
- 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 ...
- 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 ...
- 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 ...
- 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/ ...
- 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$\left\lfloor \frac{i}{2}\rig ...
- Codeforces Gym 101252D&&floyd判圈算法学习笔记
一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈 ...
- Codeforces Gym 101623A - 动态规划
题目传送门 传送门 题目大意 给定一个长度为$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 ...
- 【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, ...
- 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
- Spring学习之声明式事物管理
public List<Student> selectStudent() { Student s = new Student(); s.setName("zhengbin&quo ...
- 【RMQ问题】求数组区间最大值,NYOJ-1185-最大最小值
转自:http://blog.csdn.net/lilongherolilong/article/details/6624390 先挖好坑,明天该去郑轻找虐 RMQ(Range Minimum/Max ...
- andorid 下拉刷新
1:android的下拉刷新操作是需要一个ListView,通过onTouchEvent来判断用户的手势操作,用户触摸屏幕并且下拉时,当下拉超过指定的设定高度时就提示用户进行刷新. 2:当进行刷新时, ...
- golang windows程序获取管理员权限(UAC ) via gocn
golang windows程序获取管理员权限(UAC ) 在windows上执行有关系统设置命令的时候需要管理员权限才能操作,比如修改网卡的禁用.启用状态.双击执行是不能正确执行命令的,只有右键以管 ...
- [Everyday Mathematics]20150111
设 $n$ 阶方阵 $A=(\al_1,\cdots,\al_n)$ 非奇异, $B=(0,\al_2,\cdots,\al_n)$. 试证: $BA^{-1}$, $A^{-1}B$ 的秩均为 $n ...
- CodeForce---Educational Codeforces Round 3 D. Gadgets for dollars and pounds 正题
对于这题笔者无解,只有手抄一份正解过来了: 基本思想就是 : 二分答案,对于第x天,计算它最少的花费f(x),<=s就是可行的,这是一个单调的函数,所以可以二分. 对于f(x)的计算,我用了nl ...
- C语言-简单哈希表(hash table)
腾讯三面的时候,叫我写了个哈希表,当时紧张没写好···结果跪了··· 回来后粪发涂墙,赶紧写了一个! 什么都不说了···先让我到厕所里面哭一会··· %>_<% 果然现场发挥,以及基础扎实 ...
- res/raw和assets的 区别
res/raw和assets的相同点: 两者目录下的文件在打包后会原封不动的保存在apk包中,不会被编译成二进制. res/raw和assets的不同点: 1.res/raw中的文件会被映射到R.ja ...
- 内核源码分析之linux内核栈(基于3.16-rc4)
在3.16-rc4内核源码中,内核给每个进程分配的内核栈大小为8KB.这个内核栈被称为异常栈,在进程的内核空间运行时或者执行异常处理程序时,使用的都是异常栈,看下异常栈的代码(include/linu ...
- 【sgu282】Isomorphism
题意: 给出n(n<=53)点的无向完全图 要将每条边染上m(m<=1000)种颜色的一种 只改变顶点编号的图视为同种方案 求本质不同方案数%p(p>n且为质树)的值 题解: 这题貌 ...