ZOJ-3430
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.
/**
题意:给你n个模版,然后再给出一个字符串,问该字符串包含几个模版
做法:AC自动机 字符串的转换有些麻烦,先将字符串转换成ASCII 然后根据相应的ASCII 转换成二进制,每一个是6位,不够加0,然后取8位为一个字符,求得的字符串为要的字符串
**/
#include <iostream>
#include <cmath>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 520*64
using namespace std;
int n;
int tot = ;
unsigned char ch1[maxn];
struct Tire
{
int next[maxn][],end[maxn],fail[maxn];
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)
{
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 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 solve(char ch[],int len)
{
string str;
string str1;
str1 = "";
for(int i=; i<len; i++)
{
str = "";
int res = int(ch[i]);
while(res)
{
str += res% + '';
res /= ;
}
while(str.length() != ) str += "";
reverse(str.begin(),str.end());
str1 += str;
}
int len1 = str1.length();
while(len1% !=) len1--;
int mm = ;
memset(ch1,'\0',sizeof(ch1));
for(int i=; i<len1; i+=)
{
int res = ;
for(int j=; j<; j++)
{
res += (str1[i+j] -'') * pow(2.0,-j);
}
ch1[mm++] = char(res);
}
//cout<<ch1<<"\n";
tot = mm;
return ;
} Tire ac;
char buf[maxn];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int m;
while(~scanf("%d",&n))
{
ac.init();
char xx[maxn];
for(int i=; i<n; i++)
{
memset(xx,'\0',sizeof(xx));
scanf("%s",buf);
int len = strlen(buf);
while(buf[len-]=='=')len--;
for(int j=; j<len; j++)
{
xx[j] = Get(buf[j]);
}
solve(xx,len);
ac.insert(ch1,tot,i);
}
ac.build();
scanf("%d",&m);
while(m--)
{
scanf("%s",buf);
int len = strlen(buf);
memset(xx,'\0',sizeof(xx));
while(buf[len-] == '=')
{
--len;
}
for(int j=; j<len; j++)
{
xx[j] = Get(buf[j]);
}
solve(xx,len);
printf("%d\n",ac.query(ch1,tot));
}
printf("\n");
}
return ;
}
ZOJ-3430的更多相关文章
- 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 ...
- HDU - 2222,HDU - 2896,HDU - 3065,ZOJ - 3430 AC自动机求文本串和模式串信息(模板题)
最近正在学AC自动机,按照惯例需要刷一套kuangbin的AC自动机专题巩固 在网上看过很多模板,感觉kuangbin大神的模板最为简洁,于是就选择了用kuangbin大神的模板. AC自动机其实就是 ...
- Detect the Virus ZOJ - 3430 AC自动机
One day, Nobita found that his computer is extremely slow. After several hours' work, he finally fou ...
- ZOJ 3430 Detect the Virus
传送门: Detect the Virus ...
- 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 ...
- 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 ...
- Detect the Virus - ZOJ 3430(恶心的自动机)
题目大意:给你一些病毒的特征码,然后再给一些文本,判断每个文本有多少种病毒,不过给的字符串都是加密处理过的,给的每个字符串都有对应一个64以内的一个数(题目里面那个表就是),然后可以把这个64以内的这 ...
- 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 ...
- 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 ...
- ZOJ - 3430 ac自动机
这题主要就是解码过程很恶心,不能用char存,一共wa了20发 题意:先给n串加密后的字符,然后m串加密后的字符,解码之后求n对应每个m的匹配数,很显然的ac自动机 加密过程是先用对应ascii表的标 ...
随机推荐
- BZOJ3224:普通平衡树——题解
http://www.lydsy.com/JudgeOnline/problem.php?id=3224 题面源于洛谷 题目描述 您需要写一种数据结构(可参考题目标题),来维护一些数,其中需要提供以下 ...
- Numpy模块(数值计算)
Numpy是高性能科学计算和数据分析的基础包.它是pandas等其他各种工具的基础. NumPy的主要功能: ndarray,一个多维数组结构,高效且节省空间 无需循环对整组数据进行快速运算的数学函数 ...
- Apple Tree POJ - 3321 dfs序列构造树状数组(好题)
There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. ...
- 解决oracle数据库 ora-00054:resource busy and acquire with NOWAIT specified 错误
解决oracle数据库 ora-00054:resource busy and acquire with NOWAIT specified 错误 本人在使用pl/sql developer 客户端调用 ...
- ZooKeeper文档(二)
ZooKeeper:因为协调的分布式系统是一个动物园 ZooKeeper对分布式应用来说是一个高性能的协调服务.它暴露通常的服务-比如命名,配置管理,同步,和组服务-用一种简单的接口,所以你不用从头开 ...
- [洛谷P2596] [ZJOI2006]书架
洛谷题目链接:书架 题目描述 小T有一个很大的书柜.这个书柜的构造有些独特,即书柜里的书是从上至下堆放成一列.她用1到n的正整数给每本书都编了号. 小T在看书的时候,每次取出一本书,看完后放回书柜然后 ...
- 「模板」 FHQ_Treap 区间翻转
「模板」 FHQ_Treap 区间翻转 没有旋转的 Treap 实现区间操作的功能,很好理解,也很好写,只是速度不算太快. 对于要翻转的区间,把整棵 Treap(存有区间 \([1,n]\) 的信息) ...
- jquery checkbox选中状态以及实现全选反选
jquery1.6以下版本获取checkbox的选中状态: $('.ck').attr('checked'); $('.ck').attr('checked',true);//全选 $('.ck'). ...
- 省队集训 Day3 杨北大
[题目大意] 给出平面上$n$个点$(x_i, y_i)$,请选择一个不在这$n$个点之内的点$(X, Y)$,定义$(X, Y)$的价值为往上下左右四个方向射出去直线,经过$n$个点中的数量的最小值 ...
- 数据分析之CE找数据大法
一.基本介绍 CE的全称为Cheat Engine,是一款内存修改编辑工具,其官网是http://www.cheatengine.org,可以在其官网下载到最新的CE工具,目前最新版本是Cheat E ...