HUST1010——The Minimum Length（KMP）

Description

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A. For example, A="abcdefg". I got abcd efgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

Input

Multiply Test Cases. For each line there is a string B which contains only lowercase and uppercase charactors. The length of B is no more than 1,000,000.

Output

For each line, output an integer, as described above.

Sample Input

bcabcabefgabcdefgabcde

Sample Output

37

用KMP求最小循环点。因为KMP是求一段字符中某一段与前段相等的长度，所以与前一段相等长度最大的数，再用总长度减去这个数，就是最小循环点

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN = 1000005;char T[MAXN];int  f[MAXN];void getFail(char *p,int *f){    int n=strlen(p);    f[0]=f[1]=0;    for(int i=1; i<n; ++i)    {        int j=f[i];        while(j && p[i]!=p[j])            j=f[j];        if(p[i]==p[j])            f[i+1]=1+j;        else            f[i+1]=0;    }}int main(){    while(gets(T))    {        getFail(T,f);        int n=strlen(T);        printf("%d\n", n-f[n]);    }    return 0;}