HDU 3746 Cyclic Nacklace(KMP:补齐循环节)

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:

此题补齐循环节的几种判断方法

如果f[m]=0那么要补m个

如果m%f[m]==0那么已经存在循环节

否则需要补齐： 

分析：

最小循环节长度是m-f[m],那么还要补充m-m%f[m];
1 题目要求的是给定一个字符串，问我们还需要添加几个字符可以构成一个由n个循环节组成的字符串。
2 可知我们应该先求出字符串的最小循环节的长度：假设字符串的长度为len，那么最小的循环节就是cir = len-next[len] ;
如果有len%cir == 0,那么这个字符串就是已经是完美的字符串，不用添加任何字符；
如果不是完美的那么需要添加的字符数就是cir - (len-(len/cir)*cir))，相当与需要在最后一个循环节上面添加几个。
3 如果cir = 1，说明字符串只有一种字符例如“aaa” ;
 如果cir = m说明最小的循环节长度为m，那么至少还需m个；
 如果m%cir == 0，说明已经不用添加了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXM=100000+100;
int f[MAXM],m;
char P[MAXM];
void getFail(char *P,int *f)
{
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=f[i];
        while(j && P[i]!=P[j]) j=f[j];
        f[i+1]= (P[i]==P[j])?j+1:0;
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",P);
        m=strlen(P);
        getFail(P,f);//求next数组；
        if(f[m]==0)
        {
            printf("%d\n",m);
            continue;
        }//当f[m]=0时,输出它的长度;
        if(m%(m-f[m])==0) printf("0\n");//如果存在循环节
        else
        {
            int len1=m-f[m];//
            printf("%d\n",len1-m%len1);
        }
    }
    return 0;
}