HDU - 3746 - Cyclic Nacklace

题目：

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1892    Accepted Submission(s): 830

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

 

 

　　题意：给你一个字符串，问还需要在后面添加多少个字符才可以使字符串出现两个循环节。

　　解法是这样的，先求出这个字符串的next数组，这里我是从第0好元素开始，所以next[i]的意义是0~i-1的字符串中最长的字符串的长度是多少。设整个字符串的长度为len，则分析如图：

　　

　　len-next[len]的意思是最小循环节的长度。

　　如果len可以整除len-next[len]且next[len]不等于0，就说明当前的字符串里面已经有循环节了，所以不需要再添加字符。

　　否则，答案就是(len-next[len])-next[len]%(len-next[len])，这条式子中next[len]%(len-next[len])的意思是当前已经有了多少个字符，则结果就是还剩多少个字符了。

 

代码：

#include <stdio.h>
#include <string.h>
#define MAX 200010
using namespace std;

char s[MAX];
int next[MAX];

int get_next()
{
    int i,j,len;
    len=strlen(s);
    memset(next,-1,sizeof(next));
    i=0;
    j=-1;
    while(i<len)
    {
        if(j==-1 || s[i]==s[j]){i++; j++; next[i]=j;}
        else j=next[j];
    }
    return len;
}

int main()
{
    int t,len,ans;
    //freopen("data.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        len=get_next();
        ans=len-next[len];
        if(next[len] && len%ans==0) printf("0\n");
        else
        {
            printf("%d\n",ans-next[len]%ans);
        }
    }
    return 0;
}