TwiceString

Problem Statement
	You are given a string s. Return the shortest string which contains s as a contiguous substring twice. Note that two occurrences of s may overlap. For example, "ababa" contains "aba" twice.
Definition
	Class: TwiceString Method: getShortest Parameters: string Returns: string Method signature: string getShortest(string s) (be sure your method is public)
Notes
-	The shortest string which contains s as a contiguous substring twice is always unique.
Constraints
-	s will contain between 1 and 50 characters, inclusive.
-	Each character in s will be a lowercase letter ('a'-'z').
Examples
0)
	"aba" Returns: "ababa" This is the example shown in the problem statement.
1)
	"xxxxx" Returns: "xxxxxx"
2)
	"topcoder" Returns: "topcodertopcoder"
3)
	"abracadabra" Returns: "abracadabracadabra"

     题意：就是给你个字符串，然后求包含该字符串的最短字符串。

             

// BEGIN CUT HERE

// END CUT HERE
#line 5 "TwiceString.cpp"
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,n) for(i=0;i<(n);++i)
#define FOR(i,l,h) for(i=(l);i<=(h);++i)
#define FORD(i,h,l) for(i=(h);i>=(l);--i)

typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;


class TwiceString
{
        public:
        string getShortest(string s)
        {      
                //int j;
                  // int len=s.size();
               int i=1,j;//j在这里用来标记重复字符串的长度。
               for(j=0;j<s.size();)
                    {
                        if(s1[i]=='\0') break;
                      if(s1[i]==s[j]) {i++;j++;}
                      else {i++;j=0;}
                    }
                     return s+s.substr(j);
                     
           法二：从第二个字符开始依次截到最后，然后和从第一个开始截到和刚才相同字符长度进行比较，一旦有相同就跳出。
                   /* for(j=1;j<len;j++)
                      if(s.substr(j)==s.substr(0,len-j))
                           return s+s.substr(len-j);
                           return s+s;
                           */
                    
        }
        
};

// BEGIN CUT HERE
int main()
{
        TwiceString ___test;
           string s;
            while(cin>>s)
        cout<<___test.getShortest(s)<<endl;
        return 0;
}
// END CUT HERE