hdu3746 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.

题意：

给出一个字符串，将它补成有完整循环节的串至少需要加多少个字符。

完整循环节即至少要出现两次，整个字符串必须仅由若干个完整的循环节构成。

用KMP的失配数组的性质，可以直接求出循环节，那么字符串长度对循环节取模就是最后一个循环节的已有长度，那么循环节长度减去取模后的值就是要补的长度。

 #include<stdio.h>
 #include<string.h>

 const int maxn=1e6+;
 const int maxm=1e5+;

 char t[maxm];
 int p[maxm];

 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         scanf("%s",t);
         int i,j,ans=;
         int m=strlen(t);
         p[]=p[]=;
         for(i=;i<m;i++){
             j=p[i];
             while(j&&t[i]!=t[j])j=p[j];
             p[i+]=t[i]==t[j]?j+:;
         }
         if(p[m])printf("%d\n",(m-p[m]-m%(m-p[m]))%(m-p[m]));
         else printf("%d\n",m);
     }
     return ;
 }