Cyclic Nacklace
Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=3746
Mean:
给你一个字符串，让你在后面加尽量少的字符，使得这个字符串成为一个重复串。
例：
abca---添加bc，成为abcabc
abcd---添加abcd，成为abcdabcd
aa---无需添加
analyse:
经典的求最小循环节。
首先给出结论：一个字符串的最小循环节为：len-next[len]。
证明：
举个例子：abcabc的最小循环节是abc，abcda的最小循环节是abcd，abbab的最小循环节是abb。
看出点什么端倪没？
证明开始：
-----------------------
-----------------------
k m x j i
由上，next[i]=j，两段红色的字符串相等（两个字符串完全相等），s[k....j]==s[m....i]
设s[x...j]=s[j....i] ，记为:(xj=ji)
则可得，以下简写字符串表达方式:
kj=kx+xj;
mi=mj+ji;
因为xj=ji，所以kx=mj，如下图所示
-------------
-------------
k m x j
看到了没，此时又重复上面的模型了，kx=mj，所以可以一直这样递推下去。
所以可以推出一个重要的性质len-next[len]为此字符串的最小循环节。
另外如果len%(len-next[len])==0,此字符串的最小周期就为len/(len-next[i])。
有了这个结论，这题就好做多了。注意判断一下是否原串就是一个重复串。
Time complexity: O(N)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-07-27-21.10
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 100010;
char s [ MAXN ];
int Next [ MAXN ];
void getNext()
{
Next [ 0 ] = 0;
int s_len = strlen(s);
for( int i = 1 , k = 0; i < s_len; ++ i)
{
while(s [ i ] !=s [ k ] && k) k = Next [ k - 1 ];
if(s [ i ] ==s [ k ]) ++ k;
Next [ i ] = k;
}
}
int main()
{
ios_base :: sync_with_stdio( false);
cin . tie( 0);
int t;
scanf( "%d" , & t);
while( t --)
{
scanf( "%s" ,s);
getNext();
int s_len = strlen(s);
int min_cycle = s_len - Next [ s_len - 1 ];
if( min_cycle != s_len && s_len % min_cycle == 0)
{
puts( "0");
}
else
{
int need_add = min_cycle - Next [ s_len - 1 ] % min_cycle;
printf( "%d \n " , need_add);
}
}
return 0;
}
/*
*/
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-07-27-21.10
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 100010;
char s [ MAXN ];
int Next [ MAXN ];
void getNext()
{
Next [ 0 ] = 0;
int s_len = strlen(s);
for( int i = 1 , k = 0; i < s_len; ++ i)
{
while(s [ i ] !=s [ k ] && k) k = Next [ k - 1 ];
if(s [ i ] ==s [ k ]) ++ k;
Next [ i ] = k;
}
}
int main()
{
ios_base :: sync_with_stdio( false);
cin . tie( 0);
int t;
scanf( "%d" , & t);
while( t --)
{
scanf( "%s" ,s);
getNext();
int s_len = strlen(s);
int min_cycle = s_len - Next [ s_len - 1 ];
if( min_cycle != s_len && s_len % min_cycle == 0)
{
puts( "0");
}
else
{
int need_add = min_cycle - Next [ s_len - 1 ] % min_cycle;
printf( "%d \n " , need_add);
}
}
return 0;
}
/*
*/
