Cyclic Nacklace（KMP之找寻环节）

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10306    Accepted Submission(s): 4412

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

3aaaabcaabcde

 

Sample Output

025

题意：题目要求的是给定一个字符串，问我们还需要添加几个字符可以构成一个由 若干个循环节 组成的字符串。巧妙利用next[]数组即可

心得：next数组表示的是一个字符串前缀和后缀最大的公共长度，比如字符串s=  abcabca  这个字符串 ,其next[0]=-1,next[1]=0,next[2]=0,next[3]=0,next[4]=1,

next[5]=2,next[6]=3,next[7]=4,我们发现next[0]=-1，可是长度是不会为负数的,所以next[0]不用来表示s[0],而是用next[1]来表示s[0],那next[2]自然而然就是表示s[1]的,以此类推......所以这里求next数组的那个while()循环是i<len，而不是i<len-1,这和求子串不一样,求子串的话随便哪个都行,求循环节的话必须用i<len,所以最好用i<len

#include <cstdio>#include <cstring>using namespace std;char str[100005];int next[100005];void getnext(int len){    int i = 0,k=-1;    next[0] = -1;    while(i<len)    {        if(k== -1 || str[i] == str[k])        {            i++;            k++;            next[i] = k;        }        else            k = next[k];    }}int main(){    int n;    scanf("%d",&n);    while(n--)    {        scanf("%s",str);        memset(next,0,sizeof(next));        int len = strlen(str);        int ans;        getnext(len);//        for(int i=0;i<=len;i++)//            printf("%c:%d\n",str[i],next[i]);        ans = len - next[len];//ans就是循环节长度        if(ans!=len && len%ans == 0)//如果循环节长度不等于字符串长度，并且字符串长度是循环节长度的倍数,就不用添加了            printf("0\n");        else           printf("%d\n",ans-next[len]%ans);//最后少了几个字符就添加几个    }    return 0;}