poj 3073 Spam

最新推荐文章于 2017-07-20 22:41:14 发布

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最新推荐文章于 2017-07-20 22:41:14 发布

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本文介绍了一种垃圾邮件编码算法，该算法通过替换特定字符来规避过滤器。文章详细解释了编码过程，并提供了一个示例程序，展示了如何计算特定字符串对应的垃圾邮件编码有多少种可能的原始形式。

http://poj.org/problem?id=3073

Spam

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1033		Accepted: 593

Description

To thwart content-based spam filters, spammers often modify the text of a spam email to prevent its recognition by automatic filtering programs. For any plain text string s (containing only upper-case letters), let Φ(s) denote the string obtained by substituting each letter with its “spam alphabet” equivalent:

`A`	`4`	(four)	`N`	`\|\\|`	(pipe backslash pipe)
`B`	`\|3`	(pipe three)	`O`	`0`	(zero)
`C`	`(`	(left-parenthesis)	`P`	`\|0`	(pipe zero)
`D`	`\|)`	(pipe right-parenthesis)	`Q`	`(,)`	(left-parenthesis comma right-parenthesis)
`E`	`3`	(three)	`R`	`\|?`	(pipe question-mark)
`F`	`\|=`	(pipe equals)	`S`	`5`	(five)
`G`	`6`	(six)	`T`	`7`	(seven)
`H`	`#`	(pound)	`U`	`\|_\|`	(pipe underscore pipe)
`I`	`\|`	(pipe)	`V`	`\/`	(backslash forward-slash)
`J`	`_\|`	(underscore pipe)	`W`	`\/\/`	(backslash forward-slash backslash forward-slash)
`K`	`\|<`	(pipe less-than)	`X`	`><`	(greater-than less-than)
`L`	`\|_`	(pipe underscore)	`Y`	`-/`	(minus forward-slash)
`M`	`\|\/\|`	(pipe backslash forward-slash pipe)	`Z`	`2`	(two)

In this scheme, any plain text message s corresponds to exactly one spam-encoded message Φ(s). The reverse, however, is not necessarily true: a spam-encoded message may correspond to more than one plain text message.

Given a plain text message s, your goal is to determine the number of unique plain text messages whose spam encoding is Φ(s).

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing a plain text string s containing from 1 to 100 upper-case letters. The end-of-file is denoted by a single line containing the word “end”.

Output

For each test case, print the number of unique plain text messages (including the original message) whose spam encoding is Φ(s). The number of unique plain text messages is guaranteed to be no greater than 1,000,000,000.

Sample Input

BU
UJ
THEQUICKBROWNFOXJUMPEDOVERTHELAZYDOGS
end

Sample Output

6
5
144

Hint

In the first test case, the spam encoding of ‘BU’ is ‘|3|_|’. The 6 plain text messages with this spam encoding are ‘BU’, ‘IEU’, ‘BIJ’, ‘IEIJ’, ‘BLI’, and ‘IELI’. In the second test case, the spam encoding of ‘UJ’ is ‘|_|_|’. The 5 plain text messages with this spam encoding are ‘UJ’, ‘LU’, ‘IJJ’, ‘LLI’, and ‘LIJ’.

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=65535;
char rp[26][6]=
{
	"4","|3","(","|)","3","|=","6","#","|",
	"_|","|<","|_","|\\/|","|\\|","0","|0",
	"(,)", "|?", "5", "7", "|_|", "\\/", "\\/\\/",
    "><", "-/", "2"	
};
int dp[maxn];
char buffer[maxn];
char s[maxn];
int main()
{
//freopen("C:\\Users\\5201\\Desktop\\1.txt","r",stdin);
	
	while(scanf("%s",buffer)&&buffer[0]!='e')
	{
		s[0]=0;
		int len=strlen(buffer);
		int i,j;
		for(i=0;i<len;i++)
		{
			strcat(s,rp[buffer[i]-'A']);
		}
		len=strlen(s);
		memset(dp,0,sizeof(dp));
		for(i=0;i<len;i++)
		{
			char ch=s[i+1];
			s[i+1]=0;
			for(j=0;j<26;j++)
			{
				if(strcmp(s,rp[j])==0){ dp[i]++;break; }
			}
			for(j=1;j<=i;j++)
			{
				for(int t=0;t<26;t++)
				{
					if(strcmp(rp[t],s+j)==0) dp[i]+=dp[j-1];
				}
			}
			s[i+1]=ch;
		}
		cout<<dp[len-1]<<endl;
	}
	return 0;
}

1. 动态规划，i由dp[j-1]累加.