題目：
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

分析：
題意是差幾個字元可以構成迴圈串。
1.掌握Next陣列的意義：當前字首和字尾最大匹配數。如abcab，為2.
2.掌握迴圈節：int len=tlen-Next[tlen];
剩下的就很簡單了。

程式碼：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define exp 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
#define lie() {ans++;continue;}
using namespace std;
const int N = 1000005;
int Next[N];
char S[N], T[N];//母串 子串
int slen,tlen;

void getNext()  //求next陣列
{
    int j, k;
    j = 0; k = -1; Next[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
            Next[++j] = ++k;
        else
            k = Next[k];
}


int KMP_Count()    //子串數目
{
    int ans = 0;
    int i, j = 0;

    if(slen == 1 && tlen == 1)
    {
        if(S[0] == T[0])
            return 1;
        else
            return 0;
    }
    getNext();
    for(i = 0; i < slen; i++)
    {
        while(j>=0 && S[i] != T[j])
            j = Next[j];
        if(j==-1||S[i] == T[j])
            j++;
        if(j == tlen)
        {
            ans++;
            j = Next[j];
        }
    }
    return ans;
}

int KMP_Index()    //子串第一次出現的位置
{
    int i = 0, j = 0;
    getNext();
    while(i < slen && j < tlen)
    {
        if(j == -1 || S[i] == T[j])
        {
            i++; j++;
        }
        else
            j = Next[j];
    }
    if(j == tlen)
        return i - tlen+1;
    else
        return -1;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",T);
        tlen=strlen(T);
        getNext();
        int len=tlen-Next[tlen];
        //迴圈節的長度
        if(len!=tlen && tlen%len==0)
            printf("0\n");
        else
        {
            int ans=len-Next[tlen]%len;
            //取餘的作用：abcab，去掉abc
            printf("%d\n",ans);
        }
    }
    return 0;
}