Cyclic Nacklace(KMP求循环节)

Cyclic Nacklace

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3746

Description

CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of "HDU CakeMan", he wants to sell some little things to make money. Of course, this is not an easy task. 

As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl's fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls' lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 and its cyclic count is 2: 

Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden. 
CC is satisfied with his ideas and ask you for help.

 

Input

The first line of the input is a single integer T ( 0 < T <= 100 ) which means the number of test cases. 
Each test case contains only one line describe the original ordinary chain to be remade. Each character in the string stands for one pearl and there are 26 kinds of pearls being described by 'a' ~'z' characters. The length of the string Len: ( 3 <= Len <= 100000 ).

 

Output

For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.

 

Sample Input

3
aaa
abca
abcde 

 

Sample Output

0
2
5 

运用next数组求循环节。。

相关知识：（http://www.cnblogs.com/wuyiqi/archive/2012/01/06/2314078.html）

-----------------------

-----------------------

 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](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[i]为此字符串的最小循环节(i为字符串的结尾)，另外如果len%(len-next[i])==0,此字符串的最小周期就为len/(len-next[i]);

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int MAXN=100010;
char str[MAXN];
int nnext[MAXN];

void getNext(char *p)
{
    int j,k;
    j=0;
    k=-1;
    int len=strlen(p);
    nnext[0]=-1;
    while(j<len)
    {
        if(k==-1||p[j]==p[k])
        {
            j++;
            k++;
            nnext[j]=k;
        }
        else k=nnext[k];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",&str);
        getNext(str);
        int len=strlen(str);
        if(nnext[len]==0)
        {
            printf("%d\n",len);
            continue;
        }
        int t=len-nnext[len]; ///最小循环节
        if(len%t==0)printf("0\n");
        else
        {
            printf("%d\n",t-len%t);
        }
    }
    return 0;
}