HDU 3746 Cyclic Nacklace (用kmp求循环节）

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11516    Accepted Submission(s):
4913

Problem 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

 

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
using namespace std;
char a[];
int nxt[];
void get_next(int l)
{
    int i = , j = -;
    nxt[] = -;
    while (i != l)
    {
        if (j == - || a[i] == a[j])
        {
            i++;
            j++;
            nxt[i] = j;
        }
        else
            j = nxt[j];
    }
}
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        scanf("%s", a);//居然用cin>>a就wa了
        int m = strlen(a);
        get_next(m);
        int n = m - nxt[m];//代表循环节的长度
        if (n != m && m%n == )printf("0\n");//如果可以多次循环
        else printf("%d\n", n - nxt[m] % n);//取余的作用：abcab，去掉abc
    }
    return ;
}