# [HDU3746]Cyclic Nacklace（KMP）

### 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.

HDU

KMP

## 代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

#define ll long long
#define mem(Arr,x) memset(Arr,x,sizeof(Arr))

const int maxN=100100;
const int inf=2147483647;

int n;
char str[maxN];
int Next[maxN];
bool Pre[maxN];

int main()
{
ios::sync_with_stdio(false);
int T;cin>>T;
while (T--)
{
cin>>(str+1);n=strlen(str+1);
Next[0]=-1;Next[1]=0;
for (int i=2;i<=n;i++)//构造Next
{
int j=Next[i-1];
while ((j!=0)&&(str[j+1]!=str[i])) j=Next[j];
if (str[j+1]==str[i]) Next[i]=j+1;
else Next[i]=0;
}
mem(Pre,0);Pre[0]=1;//标记出所有合法前缀
for (int i=n;i!=0;i=Next[i]) Pre[i]=1;
int mn=n;//记录最短补全长度
for (int i=1;i<=n;i++)
{
int cnt=n/i,pos=cnt*i;//cnt求出循环多少次，pos求出最后一个完整的循环节结束的位置
if ((pos%(pos-Next[pos])!=0)||(pos-Next[pos]!=i)) continue;//判断这个循环节是否真的是循环的
if (Pre[n-pos]==0) continue;//判断后面剩下的几位是否能匹配
cnt=pos/(pos-Next[pos]);//求出循环次数
if ((pos==n)&&(cnt>1)) mn=0;//特判原串就是
mn=min(mn,pos+i-n);
}
printf("%d\n",mn);
}
return 0;
}

HNCJ OIer 一枚