## 【模拟/字符串】DRM Messages

DRM Encryption is a new kind of encryption. Given an encrypted string (which we’ll call a DRM
message), the decryption process involves three steps: Divide, Rotate and Merge. This process is
described in the following example with the DRM message “EWPGAJRB”:
Divide — First, divide the message in half to “EWPG” and “AJRB”.
Rotate — For each half, calculate its rotation value by summing up the values of each character
(A = 0, B = 1, . . . Z = 25). The rotation value of “EWPG” is 4 + 22 + 15 + 6 = 47. Rotate each
character in “EWPG” 47 positions forward (wrapping from Z to A when necessary) to obtain the
new string “ZRKB”. Following the same process on “AJRB” results in “BKSC”.
Merge — The last step is to combine these new strings (“ZRKB” and “BKSC”) by rotating each character
in the first string by the value of the corresponding character in the second string. For the first
position, rotating ‘Z’ by ‘B’ means moving it forward 1 character, which wraps it around to ‘A’.
Continuing this process for every character results in the final decrypted message, “ABCD”.
Input
The input file contains several test cases, each of them as described below.
The input contains a single DRM message to be decrypted. All characters in the string are uppercase
letters and the string’s length is even and ≤ 15000.
Output
For each test case, display the decrypted DRM message.
Sample Input
EWPGAJRB
UEQBJPJCBUDGBNKCAHXCVERXUCVK
Sample Output
ABCD
ACMECNACONTEST

## 【KMP】Blue Jeans

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 21885 Accepted: 9712

Description

The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated.

As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.

A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.

Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.

Input

Input to this problem will begin with a line containing a single integer n indicating the number of datasets. Each dataset consists of the following components:

• A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
• m lines each containing a single base sequence consisting of 60 bases.

Output

For each dataset in the input, output the longest base subsequence common to all of the given base sequences. If the longest common subsequence is less than three bases in length, display the string “no significant commonalities” instead. If multiple subsequences of the same longest length exist, output only the subsequence that comes first in alphabetical order.

Sample Input

3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalities
AGATAC
CATCATCAT

1、  最长公共串长度小于3输出   no significant commonalities

2、  若出现等长的最长的子串，则输出字典序最小的串

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
const int N=10+5;
const int M=60+5;

string s[N];
int nxt[M];

void getnext(string str,int len){
int i=0,j=-1;
nxt[0]=-1;
while(i<len){
if(j==-1||str[i]==str[j])
nxt[++i]=++j;
else j=nxt[j];
}
}
bool kmp(string s1,int len1,string s2,int len2){
int i=0,j=0;
while(i<len1){
if(j==-1||s1[i]==s2[j])
++i,++j;
else j=nxt[j];
if(j==len2)return true;
}
return false;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int T,n;
cin>>T;
while(T--){
cin>>n;
for(int i=0;i<n;i++)cin>>s[i];
string ans="";
for(int i=0;i<s[0].size();i++) //起点
{
for(int j=1;i+j<=s[0].size();j++) //长度
{
string res=s[0].substr(i,j);
getnext(res,res.size());
bool flag=0;
for(int k=1;k<n;k++)
if(!kmp(s[k],s[k].size(),res,res.size()))
flag=1;
if(!flag)
{
if(ans.size()<res.size())ans=res;  //优先长度更长的公共子串
else if(ans.size()==res.size())ans=min(ans,res); //长度相同，按字典序排序
}
}
}
if(ans.size()<3)cout<<"no significant commonalities"<<endl;
else cout<<ans<<endl;
}
return 0;
}

## Codeforces：Oh Those Palindromes

A non-empty string is called palindrome, if it reads the same from the left to the right and from the right to the left. For example, “abcba“, “a“, and “abba” are palindromes, while “abab” and “xy” are not.

A string is called a substring of another string, if it can be obtained from that string by dropping some (possibly zero) number of characters from the beginning and from the end of it. For example, “abc“, “ab“, and “c” are substrings of the string “abc“, while “ac” and “d” are not. 继续阅读Codeforces：Oh Those Palindromes

## 优美字符串(字符串拼接|java String::endsWith的用法)

ABB ABB

ABB

import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
String str1=cin.next(),str2=cin.next();
int len=str2.length();
for(;len>0;len--) {
if(str1.endsWith(str2.substring(0,len))) {
break;
}
}
System.out.println(str1+str2.substring(len));

}
}