HDU 3746 Cyclic Nacklace(KMP)

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

题解：

题意：

给你一个串，问你最少加多少个字符可以形成一个循环的串

思路：

第一次写KMP的题，还只是会套模板，只能看题解了。。。

首先根据Next数组的性质，那么最小的循环节应该是字符串长度len-Next[len]，设为ans，这里可以理解为模板串从第一位到第ans位和要匹配字符串相同

那么分情况讨论这个ans，如果ans==len，那么说明最小循环节是自己，那么就只能添加len个字符了

如果len%ans==0，那么说明整个字符串已经是循环了也不用添加

除了以上的情况，剩下的情况就要添加:ans-Next[len]%ans个字符

代码：

#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2#define INF 1008611111#define ll long long#define eps 1e-15int Next[100005];char T[100005];int tlen,slen;void init(){    int j, k;    j=0;k=-1;Next[0]=-1;    while(j<tlen)        if(k ==-1||T[j]==T[k])            Next[++j]=++k;        else            k=Next[k];}int main(){    int i,j,m,n,ans;    while(scanf("%d",&n)!=EOF)    {        while(n--)        {            scanf("%s",T);            tlen=slen=strlen(T);            init();            tlen=strlen(T);            ans=tlen-Next[tlen];            if(ans!=tlen&&tlen%ans==0)                printf("0\n");            else                printf("%d\n",ans-Next[tlen]%ans);        }    }    return 0;}