HDU 3746 Cyclic Nacklace

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

 

题意：要制作一个幸运手链（字符串），制造手链的每个珠子分别用a~z之间的字母代表，手链开始已经有了一些珠子，现在要制作幸运手链（就是这个字符串至少要有两次循环，注意是循环，那么ababababa就不能算作两次循环啦，这时还需要加1），问最少还需要加几个珠子（字母）。

#include<stdio.h>
#include<string.h>

const int N=1e5+10;

char s[N];
int Next[N], n;

void Getnext()
{
    int i = 0, j = -1;

    Next[0] = -1; ///这里千万别忘啦，不过调试也是能很快找到这个错误的

    while (i <= n)
    {
        if (j == -1 || s[i] == s[j])
        {
            i++;
            j++;

            Next[i] = j;
        }
        else j = Next[j];
    }
}

int main ()
{
    int T, x, k;

    scanf("%d", &T);

    while (T--)
    {
        scanf("%s", s);

        n = strlen(s);

        Getnext();

        x = n - Next[n]; ///x是字符串的循环节
        k = x - Next[n] % x; ///Next[n]%x是多个循环节后多出的部分，那么我们只需要用循环节减去这个部分就是我们要求的最少珠子数啦

        if (x != n && n % x == 0) printf("0
"); ///注意这个特判，万一整个字符串就是多个循环节组成的，那么肯定不需要再添加啦（但是这种情况如果不特判的话就会输出n啦）
        else printf("%d
", k);
    }

    return 0;
}