HSU 3746 Cyclic Nacklace

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4786    Accepted Submission(s): 2179

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

 

这个题目也是完全适合KMP算法的一个题目，题意是给定一个字符串，为了让其首尾连接后可以组成一个有着循环子串的主串需要添加几个字符。

一般的KMP是保存到模式串的长度为止，但是对于循环这种情况，我们需要把next延长到模式串长度+1，然后根据这个位求出的next的值，计算循环子串还需要几个字符。

分以下几种情况，第一种就是这个位的next值为0，则显然这个只能将整个串作为循环子串，添加整个串长长度的字符串。

第二种是计算出temp为串长减去这个值，求得的就是循环子串的长度，显然如果循环子串长度为主串的因子的话则无需添加任何字符，否则添加的字符长度就是循环子串长度减去主串余循环子串的长度。得解

代码如下：

/*************************************************************************
	> File Name: Cyclic_Nacklace.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Tue 24 Nov 2015 05:25:45 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

int T;

void preKMP(char x[], int m, int kmpNext[]){
    int i, j;
    j = kmpNext[0] = -1;
    i = 0;
    while(i <= m){
        while(j != -1 && x[i] != x[j])
            j = kmpNext[j];
        if(x[++ i] == x[++ j])
            kmpNext[i] = kmpNext[j];
        else 
            kmpNext[i] = j;
    }
}


int nexti[100010];
void KMP_Count(char x[], int m){
    int i, j;
    int ans = 0;
    preKMP(x, m, nexti);
    int temp = m - nexti[m];
    if(nexti[m]){
        if(m % temp == 0)
            puts("0");
        else
            printf("%d
",temp - m % temp);
    }
    else 
        cout << m << endl;
}
char a[100010];
int main(void){
    cin >> T;
    getchar();
    while(T --){
        gets(a);
        KMP_Count(a, strlen(a));
    }
}