8.11 E

                                    E - Cyclic Nacklace

CC always becomes very depressed at the end of thismonth, he has checked his credit card yesterday, without any surprise, thereare only 99.9 yuan left. he is too distressed and thinking about how to tideover the last days. Being inspired by the entrepreneurial spirit of "HDUCakeMan", 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 presentsto send to their girlfriends. It is believed that chain bracelet is a goodchoice. However, Things are not always so simple, as is known to everyone,girl's fond of the colorful decoration to make bracelet appears vivid andlively, meanwhile they want to display their mature side as college students.after CC understands the girls demands, he intends to sell the chain braceletcalled CharmBracelet. The CharmBracelet is made up with colorful pearls to showgirls' lively, and the most important thing is that it must be connected by acyclic chain which means the color of pearls are cyclic connected from the leftto right. And the cyclic count must be more than one. If you connect theleftmost pearl and the rightmost pearl of such chain, you can make aCharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 andits cyclic count is 2: 

Now CC has brought in some ordinary bracelet chains, he wants to buy minimumnumber of pearls to make CharmBracelets so that he can save more money. butwhen remaking the bracelet, he can only add color pearls to the left end andright 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 tobe remade. Each character in the string stands for one pearl and there are 26kinds of pearls being described by 'a' ~'z' characters. The length of thestring Len: ( 3 <= Len <= 100000 ).

Output

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

Sample Input

3

aaa

abca

abcde

Sample Output

0

2

5

题意：输入N组样例，对于每一组样例判断输入的样例中含有的最小循环节，如果不含有2个或2个以上的循环节时，补全字符串使字符串至少含有2个循环节，输出需要补全的字符个数。

思路：首先计算出next数组，用字符串长度减去next数组最后一位存储的数字，得出字符串中最小单个循环节长度，然后用next数组最后一位存储的数字对单个循环节长度取余，得出含有几个循环节，如果含有余数，则用单个循环节长度减去余数，如果不含余数同时循环节长度大于等于2，输出0；

 

#include<stdio.h>#include<string.h>int next[151000];char s[151000];int main(){int i,j,k,m,n,w;while(scanf("%d",&n)!=EOF){while(n--){memset(next,0,sizeof(next));scanf("%s",s);m=strlen(s);i=0;j=-1;next[0]=-1;while(i<m){if(j==-1||s[i]==s[j]){j++;i++; next[i]=j;}elsej=next[j];}/*for(i=0;i<=m;i++)printf("%d ",next[i]);printf("\n");*/w=m-next[m];//计算单个循环节长度if(w!=m&&m%w==0)//判断字符串字符是否全部为一样的字符{printf("0\n");}else{printf("%d\n",w-(next[m]%w));//输出补全字符串的长度}//printf("w=====%d\n",w);}}return 0;}