ZOJ 3430 Detect the Virus
传送门: 3430 Detect the Virus 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
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Solution: A-C Automation
注意:
1.弄清题意,“If these binary bytes can be found in the email attachment (binary data), then the attachment contains the virus.“题意是,将base64编码的模式串与文本串翻译成ASCII编码的字符串,再进行匹配。并不是将两者翻译成二进制(0-1串)进行匹配。
2.也可将模式串与文本串翻译成0-1串做匹配,但注意匹配的位置必须是整字节处。这样做代码更简洁,内存与时间都更优(Maybe, 因为字符集是{0, 1})。
#include<bits/stdc++.h>
using namespace std;
const char *m1="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
int m2[];
void build_map(){
for(int i=; m1[i]; i++){
m2[m1[i]]=i;
}
}
const int MAX_N=1e6+, ls=, lt=2e4, MAX_M=;
bool s[ls], t[lt], used[MAX_M];
char v[ls], f[lt];
int q[MAX_N], head, tail;
struct node
{
int pre, id, pos[], last;
void init(){
id=;
memset(pos, , sizeof(pos));
}
};
node trie[MAX_N];
void b2b(char ch, bool *b, int &i){
if(ch=='=') i-=;
else{
int p=;
while(p){
b[i++]=m2[ch]&p;
p>>=;
}
}
}
int trans(char *s, bool *b){
int i=, j=;
for(; s[i]; i++){
b2b(s[i], b, j);
}
//for(int i=0; i<j; i++) printf("%d", b[i]);
//puts("");
return j;
}
int build_trie(int N){
int tot=, now, len;
trie[tot].init();
for(int id=; id<=N; id++){
scanf("%s", v);
len=trans(v, s);
now=;
for(int i=; i<len; i++){
int &nt=trie[now].pos[s[i]]; //error-prone
now=nt?nt:(trie[++tot].init(), nt=tot);
}
trie[now].id=id;
}
return tot;
}
void build_ac(){
head=tail=;
trie[].last=;
for(int i=; i<; i++){
int &nt=trie[].pos[i];
if(nt){
trie[nt].pre=;
trie[nt].last=;
q[tail++]=nt;
}
}
int pre;
while(head!=tail){
int &top=q[head++];
for(int i=; i<; i++){
int &nt=trie[top].pos[i];
pre=trie[top].pre;
if(!nt) nt=trie[pre].pos[i];
else{
q[tail++]=nt;
while(!trie[pre].pos[i]&&pre) pre=trie[pre].pre;
pre=trie[nt].pre=trie[pre].pos[i];
int &last=trie[nt].last;
last=trie[pre].id?pre:trie[pre].last; //error-prone
}
}
}
}
void get_ans(int pre, int &cnt){ //error-prone
pre=trie[pre].id?pre:trie[pre].last;
while(pre && !used[trie[pre].id]){
used[trie[pre].id]=true;
cnt++;
pre=trie[pre].last;
}
}
void match(int N){
while(N--){
scanf("%s", f);
int len=trans(f, t);
memset(used, , sizeof(used));
int cnt=;
for(int i=, pre=; i<len; i++){
pre=trie[pre].pos[t[i]];
if(i%==){
get_ans(pre, cnt);
}
}
printf("%d\n", cnt);
}
} int main(){
//freopen("in", "r", stdin);
build_map();
int N, M;
while(~scanf("%d", &N)){
build_trie(N);
build_ac();
scanf("%d", &M);
match(M);
puts("");
}
return ;
}
ZOJ 3430 Detect the Virus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 ...
- 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 ...
- 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 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】
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 ...
- ZOJ 4114 Detect the Virus(AC自动机)
Detect the Virus Time Limit: 2 Seconds Memory Limit: 65536 KB One day, Nobita found that his co ...
- 【ZOJ】3430 Detect the Virus
动态建树MLE.模仿别人的代码模板各种原因wa后,终于AC. #include <iostream> #include <cstdio> #include <cstrin ...
- 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 ...
随机推荐
- Beaufort密码
博福特密码,是一种类似于维吉尼亚密码的替代密码,由弗朗西斯·蒲福(Francis Beaufort)发明.它最知名的应用是M-209密码机.博福特密码属于对等加密,即加密演算法与解密演算法相同 博福特 ...
- usb驱动开发2之代码地图
USB只是Linux庞大家族里的一个小部落,host controller是它们的族长,族里的每个USB设备都需要被系统识别.下图显示包含一个USB接口的USB鼠标导出的结果. USB系统中的第一个U ...
- 携手K2 BPM,华住酒店完美实现“互联网+”转型
华住酒店集团,旗下6大品牌酒店,包括商旅品牌—禧玥酒店.全季酒店.星程酒店.汉庭酒店.海友酒店,以及度假品牌—漫心度假酒店.高端大气上档次一气呵成,2013年签约K2,携手成就美好生活. 演讲人:宋洪 ...
- zabbix架构
监控系统: 数据的采集---存储---展示和分析---报警 snmp cacti:rrd agent nagios:不保存数据,mysql icmp/ssh ...
- 解决centos7重启后出现ata bus error
昨天把centos7装在电脑上了,还把win7系统格掉了,从此电脑上只装centos,有一种弃暗投明的感觉. 装完重启后欣赏了一番成果,一个halt命令想把系统关掉,却发现屏幕没黑,机器不转了,电源灯 ...
- C#基础系列:实现自己的ORM(反射以及Attribute在ORM中的应用)
反射以及Attribute在ORM中的应用 一. 反射什么是反射?简单点吧,反射就是在运行时动态获取对象信息的方法,比如运行时知道对象有哪些属性,方法,委托等等等等.反射有什么用呢?反射不但让你在运行 ...
- [转]基于四叉树(QuadTree)的LOD地形实现
实现基于四叉树的LOD地形时,我遇到的主要问题是如何修补地形裂缝. 本段我将描述使用LOD地形的优势,我实现LOD地形的思路,实现LOD地形核心模块的详细过程,以及修补地形裂缝的思路. 首先,LOD地 ...
- DIV相关的操作总结
由于有时候需要做网站项目,遇到CSS的问题总是需要百度或者google一下,比较麻烦,索性今天就来总结一下,这里就拿div开刀先. DIV在HTML前端页面布局中,非常重要,我们经常遇到的问题是:DI ...
- Django添加Last-Modified和ETag
用Django REST Framework做的REST API,其中有个API有这样的需求: APP端请求这个API,服务器端从数据库读数据,返回json.返回的数据量稍微有些大,但是可能一年才修改 ...
- Bootstrap Paginator 分页插件参数介绍及使用
Bootstrap Paginator是一款基于Bootstrap的js分页插件,功能很丰富,个人觉得这款插件已经无可挑剔了.它提供了一系列的参数用来支持用户的定制,提供了公共的方法可随时获得插件状态 ...