Cyclic Nacklace （kmp next数组的应用）

   

Cyclic Nacklace

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

3aaaabcaabcde

Sample Output

025

题意：给定一个字符串最少添加多少个字符使这串字符变成循环的，至少2次循环

思路：类似于之前做过的一个求周期字符串的题目，同样是只需要求出字符串的next数组来就可以，然后如果字符长度减去next[字符长度]就是循环节的长度，如果能被长度整除，并且商大于二，说明本身就是循环的，所需增加的个数为0，否则每次加一个，每次判断能否被整除并且商大于2，就可以了、

code：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int Next[100100];void getNext(char s[]){    int i = -1,j = 0;    memset(Next,0,sizeof(Next));    Next[0] = -1;    int lens = strlen(s);    while(j < lens){        if(i==-1||s[i]==s[j]){            i++,j++;            Next[j] = i;//不能优化        }        else i = Next[i];    }}int main(){    int t;    char s[100100];    scanf("%d",&t);    while(t--){        scanf("%s",s);        getNext(s);        int len = strlen(s);        int temp = len-Next[len];//循环节长度        if(len%temp==0&&len/temp>1)printf("0\n");//是循环的直接输出0        else{            int i;            for(i = 1; ;i++){//否则判断加多少能变成循环的                if((len+i)%temp==0&&(len+i)/temp>1){                    printf("%d\n",i);                    break;                }            }        }    }    return 0;}