Detect the Virus


Time Limit: 2 Seconds      Memory Limit: 65536 KB

One day, Nobita found that his computer is extremely slow. After several hours' work, he finally found that it was a virus that made his poor computer slow and the virus was activated by a misoperation of opening an attachment of an email.

Nobita did use an outstanding anti-virus software, however, for some strange reason, this software did not check email attachments. Now Nobita decide to detect viruses in emails by himself.

To detect an virus, a virus sample (several binary bytes) is needed. If these binary bytes can be found in the email attachment (binary data), then the attachment contains the virus.

Note that attachments (binary data) in emails are usually encoded in base64. To encode a binary stream in base64, first write the binary stream into bits. Then take 6 bits from the stream in turn, encode these 6 bits into a base64 character according the following table:

That is, translate every 3 bytes into 4 base64 characters. If the original binary stream contains 3k + 1 bytes, where k is an integer, fill last bits using zero when encoding and append '==' as padding. If the original binary stream contains 3k + 2 bytes, fill last bits using zero when encoding and append '=' as padding. No padding is needed when the original binary stream contains 3k bytes.

Value 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
Encoding 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
Value 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63
Encoding g h i j k l m n o p q r s t u v w x y z 0 1 2 3 4 5 6 7 8 9 + /

For example, to encode 'hello' into base64, first write 'hello' as binary bits, that is: 01101000 01100101 01101100 01101100 01101111
Then, take 6 bits in turn and fill last bits as zero as padding (zero padding bits are marked in bold): 011010 000110 010101 101100 011011 000110 111100
They are 26 6 21 44 27 6 60 in decimal. Look up the table above and use corresponding characters: aGVsbG8
Since original binary data contains 1 * 3 + 2 bytes, padding is needed, append '=' and 'hello' is finally encoded in base64: aGVsbG8=

Section 5.2 of RFC 1521 describes how to encode a binary stream in base64 much more detailedly:

Click here to see Section 5.2 of RFC 1521 if you have interest

Here is a piece of ANSI C code that can encode binary data in base64. It contains a function, encode (infile, outfile), to encode binary file infile in base64 and output result to outfile.

Click here to see the reference C code if you have interest

Input

Input contains multiple cases (about 15, of which most are small ones). The first line of each case contains an integer N (0 <= N <= 512). In the next N distinct lines, each line contains a sample of a kind of virus, which is not empty, has not more than 64 bytes in binary and is encoded in base64. Then, the next line contains an integer M (1 <= M <= 128). In the following M lines, each line contains the content of a file to be detected, which is not empty, has no more than 2048 bytes in binary and is encoded in base64.

There is a blank line after each case.

Output

For each case, output M lines. The ith line contains the number of kinds of virus detected in the ith file.

Output a blank line after each case.

Sample Input

3
YmFzZTY0
dmlydXM=
dDog
1
dGVzdDogdmlydXMu 1
QA==
2
QA==
ICAgICAgICA=

Sample Output

2

1
0

Hint

In the first sample case, there are three virus samples: base64, virus and t: , the data to be checked is test: virus., which contains the second and the third, two virus samples.


Author: WU, Jun
Contest: ZOJ Monthly, November 2010

AC自动机的题目。

就是编码转换有点麻烦。

而且这个要用一个字节256个点来表示状态

//============================================================================
// Name : ZOJ.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================ #include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std; struct Trie
{
int next[*][],fail[*],end[*];
int root,L;
int newnode()
{
for(int i = ;i < ;i++)
next[L][i] = -;
end[L++] = -;
return L-;
}
void init()
{
L = ;
root = newnode();
}
void insert(unsigned char buf[],int len,int id)
{
int now = root;
for(int i = ;i < len;i++)
{
if(next[now][buf[i]] == -)
next[now][buf[i]] = newnode();
now = next[now][buf[i]];
}
end[now] = id;
}
void build()
{
queue<int>Q;
fail[root] = root;
for(int i = ;i < ;i++)
if(next[root][i] == -)
next[root][i] = root;
else
{
fail[next[root][i]]=root;
Q.push(next[root][i]);
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
for(int i = ;i < ;i++)
if(next[now][i] == -)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
bool used[];
int query(unsigned char buf[],int len,int n)
{
memset(used,false,sizeof(used));
int now = root;
for(int i = ;i < len;i++)
{
now = next[now][buf[i]];
int temp = now;
while( temp!=root )
{
if(end[temp] != -)
used[end[temp]]=true;
temp = fail[temp];
}
}
int res = ;
for(int i = ;i < n;i++)
if(used[i])
res++;
return res;
}
}; unsigned char buf[];
int tot;
char str[];
unsigned char s[];
unsigned char Get(char ch)
{
if( ch>='A'&&ch<='Z' )return ch-'A';
if( ch>='a'&&ch<='z' )return ch-'a'+;
if( ch>=''&&ch<='' )return ch-''+;
if( ch=='+' )return ;
else return ;
}
void change(unsigned char str[],int len)
{
int t=;
for(int i=;i<len;i+=)
{
buf[t++]=((str[i]<<)|(str[i+]>>));
if(i+ < len)
buf[t++]=( (str[i+]<<)|(str[i+]>>) );
if(i+ < len)
buf[t++]= ( (str[i+]<<)|str[i+] );
}
tot=t;
}
Trie ac;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n,m;
while(scanf("%d",&n) == )
{
ac.init();
for(int i = ;i < n;i++)
{
scanf("%s",str);
int len = strlen(str);
while(str[len-]=='=')len--;
for(int j = ;j < len;j++)
{
s[j] = Get(str[j]);
}
change(s,len);
ac.insert(buf,tot,i);
}
ac.build();
scanf("%d",&m);
while(m--)
{
scanf("%s",str);
int len=strlen(str);
while(str[len-]=='=')len--;
for(int j = ;j < len;j++)
s[j] = Get(str[j]);
change(s,len);
printf("%d\n",ac.query(buf,tot,n));
}
printf("\n");
}
return ;
}

ZOJ 4114 Detect the Virus(AC自动机)的更多相关文章

  1. ZOJ - 3430 Detect the Virus —— AC自动机、解码

    题目链接:https://vjudge.net/problem/ZOJ-3430 Detect the Virus Time Limit: 2 Seconds      Memory Limit: 6 ...

  2. zoj 3430 Detect the Virus(AC自己主动机)

    题目连接:zoj 3430 Detect the Virus 题目大意:给定一个编码完的串,将每个字符相应着表的数值转换成6位二进制.然后以8为一个数值,又一次形成字符 串,推断给定询问串是否含有字符 ...

  3. ZOJ 3228 Searching the String(AC自动机)

    Searching the String Time Limit: 7 Seconds      Memory Limit: 129872 KB Little jay really hates to d ...

  4. ZOJ 3430 Detect the Virus(AC自动机)

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3430 题意:给你n个编码后的模式串,和m个编码后的主串,求原来主 ...

  5. zoj 3430 Detect the Virus(AC自己主动机)

    Detect the Virus Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita found that his co ...

  6. ZOJ 3430 Detect the Virus

    传送门: Detect the Virus                                                                                ...

  7. ZOJ 3494 BCD Code(AC自动机 + 数位DP)题解

    题意:每位十进制数都能转化为4位二进制数,比如9是1001,127是 000100100111,现在问你,在L到R(R <= $10^{200}$)范围内,有多少数字的二进制表达式不包含模式串. ...

  8. ZOJ 3430 Detect the Virus 【AC自动机+解码】

    解码的那些事儿,不多说. 注意解码后的结果各种情况都有,用整数数组存储,char数组会超char类型的范围(这个事最蛋疼的啊)建立自动机的时候不能用0来判断结束. #include <cstdi ...

  9. ZOJ 3430 Detect the Virus(AC自动机 + 模拟)题解

    题意:问你主串有几种模式串.但是所有串都是加密的,先解码.解码过程为:先把串按照他给的映射表变成6位数二进制数,然后首尾衔接变成二进制长串,再8位8位取变成新的数,不够的补0.因为最多可能到255,所 ...

随机推荐

  1. C#中控件的CheckState和Checked属性区别?

    Checked 和CheckState都是检查控件选中状态,都能判断是否选中控件. 只是Checked 通过布尔判断(true & false): CheckState 通过枚举判断. che ...

  2. [POJ2398]Toy Storage(计算几何,二分,判断点在线段的哪一侧)

    题目链接:http://poj.org/problem?id=2398 思路RT,和POJ2318一样,就是需要排序,输出也不一样.手工画一下就明白了.注意叉乘的时候a×b是判断a在b的顺时针还是逆时 ...

  3. C# 按拼音/笔划 排序的简单示例(转)

    class Program { static void Main(string[] args) { string[] arr = { "趙(ZHAO)", "錢(QIAN ...

  4. useradd和adduser的区别

    1. 在root权限下,useradd只是创建了一个用户名,如 (useradd  +用户名 ),它并没有在/home目录下创建同名文件夹,也没有创建密码,因此利用这个用户登录系统,是登录不了的,为了 ...

  5. HDU 4893 线段树

    比赛时太大意,斐波拉契数列开小了. 题目大意:1个序列,3种操作,改变序列某个数大小,将序列中连续的一段每个数都变成其最近的斐波拉契数,以及查询序列中某一段的数之和. 解题思路:维护add[]数组表示 ...

  6. GBDT(Gradient Boosting Decision Tree)算法&协同过滤算法

    GBDT(Gradient Boosting Decision Tree)算法参考:http://blog.csdn.net/dark_scope/article/details/24863289 理 ...

  7. noip2001提高组题解

    今天继续感动滚粗.第一次提交170分,不能多说. 第一题:一元三次方程 明明是寒假讲分治的时候做过的题居然还是WA而且只拿了60分,说明知识掌握实在不够牢固. 寒假做的是保留4位小数,原题只保留2位, ...

  8. 【Java学习笔记】Hello world

    package aaa; public class aaa { public static void main(String args[]){ System.out.println("hel ...

  9. ffmpeg+rtsp+dss

    1. push stream to dss ffmpeg -f mpegts -re -i film.v -c:v libx264 -s 352x288 -aspect 4:3 -b:v 300k - ...

  10. PNG文件结构分析 ---Png解析

    PNG文件结构分析 ---Png解析   为了实现更高级的应用,我们必须充分挖掘PNG的潜力. PNG的文件结构 根据PNG文件的定义来说,其文件头位置总是由位固定的字节来描述的:   十进制数 13 ...