HDU - 3746 Cyclic Nacklace (最小循环节)

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 11473    Accepted Submission(s): 4890

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

结论：最小循环节长度xl = len-next[len]

证明：

假设图中的黑色是原来的字符串，现在要求最小循环节，对于nxt[len]来说指的是图中蓝色和黄色的长度，而且蓝色和黄色是相等的，
那么绿色和紫色也是相等的，，对比原串可知紫色跟粉色是相同的子串，那么绿色跟粉色相同，然后对比蓝色跟黄色可知红色跟粉色相同，对比原串，红色跟棕色相同，那么棕色跟粉色相同，不断重复此过程可知若此字符串有最小循环节，那么循环次数一定为len/（len-nxt[len]）。

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<cstring>
#include<string>
#include<vector>
#include<cmath> 
#include<map>
#define mem(a,b) memset(a,b,sizeof(a))
#define mod 1000000007
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
const double esp = 1e-7;
const int ff = 0x3f3f3f3f;

char s[maxn];
int ne[maxn];

void get_next(char *s,int len)//求next数组
{
	ne[0] = -1;
	int i = 0,j = -1;
	while(i< len)
	{
		if(j == -1||s[i] == s[j])
		{
			i++;j++;
			ne[i] = j;
		}
		else
			j = ne[j];
	}
}

int main()
{
	int tt;
	cin>>tt;
	while(tt--)
	{
		mem(ne,0);
		scanf(" %s",s);
		int len = strlen(s);
		get_next(s,len);
		int xl = len-ne[len];
		if(len!= xl&&len%xl == 0)
			cout<<0<<endl;
		else
			cout<<xl-(len%xl)<<endl;
	}
	
	return 0;
}