D - Cyclic Nacklace （kmp的nex数组）

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

可以发现一个规律，从第二个循环节的最后一个字符开始，所有的Next值都是连续的，并且到最后一个最后一个字符的Next的值，与从第二个循环节从1开始计数的结果是一样的，那么我们就可以得到这样一个计算字符串中的方式，使用Next数组！

假设图中S段是该串的最小循环节，S‘是末尾没有循环完的一小段，Next[m]是将最后一个字符进行匹配的位置，那么若S’不是第一个循环节，那么我们可以知道Next[m]会指向它前面那段S的相同位置。

从这个图我们可以得出循环节为   strlen(s)-Next[strlen(s)]，需要添加的为   循环节-Next[strlen(s)]%循环节。

需要注意的是，当前循环整好结束，也就是不需要添加就已经是一个循环了的，满足之前讨论的最后一个字符的next值是有规律的，也就是Next[m]%循环节=0（循环次数大于等于2，如果是第一次循环的，那么我们可以通过判断循环节==strlen(s）来确定)
代码：

//Full of love and hope for life

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <queue>
#define inf 0x3f3f3f3f
//https://paste.ubuntu.com/

using namespace std;

int nex[100010];
char n[100010];
int ans;

void getnext(char a[],int x){
    int j=-1;
    nex[0]=-1;
    int i=0;
    while(i<x){
        if(j==-1||a[i]==a[j]){
            i++;
            j++;
            nex[i]=j;
        }
        else{
            j=nex[j];
        }
    }
}
/*
int kmp(char a[],char b[],int x,int y){
     getnext(b,y);
     int j=-1;
     for(int i=0;i<x;i++){
        while(j!=-1&&a[i]!=b[j+1]){
             j=nex[j];
        }
        if(a[i]==b[j+1]){
            j++;
        }
        if(j==y-1){
            ans++;
        }
    }
    return ans;
}
*/
int main(){
    //ios::sync_with_stdio(false);
    int a;
    scanf("%d",&a);
    while(a--){
            ans=0;
        scanf("%s",n);
        int len1=strlen(n);
        getnext(n,len1);
        int t=len1-nex[len1];
        if(t!=len1&&len1%t==0){
            cout << 0 << endl;
            continue;
        }
        cout << t-(nex[len1]%t) << endl;
    }
    return 0;
}