Poj2406 -- power strings (KMP minimum cycle)

Power strings

Time limit:3000 Ms		Memory limit:65536 K
Total submissions:33178		Accepted:13792

Description

Given two strings A and B we define a * B to be their concatenation. for example, if a = "ABC" and B = "def" Then a * B = "abcdef ". if we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a ^ 0 = "" (the empty string) and a ^ (n + 1) = A * (a ^ N ).

Input

Each test case is a line of input representing S, a string of printable characters. the length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you shoshould print the largest N such that S = a ^ N for some string.

Sample Input

abcdaaaaababab.

Sample output

143

Hint

This problem has huge input, use scanf instead of CIN to avoid time limit exceed.

Source

Waterloo local 2002.07.01

Use the next array to obtain the maximum prefix of the entire array. If the entire string is composed of Loop Segments, then n-next [N] is the smallest loop section, verify that the minimum cycle is rounded out by N.

 

#include <cstdio>#include <cstring>#include <algorithm>int next[1100000] ;char str[1100000] ;void getnext(int l){    int j = 0 , k = -1 ;    next[0] = -1 ;    while(j < l)    {        if( k == -1 || str[j] == str[k] )        {            j++ ;            k++ ;            next[j] = k ;        }        else            k = next[k] ;    }}int main(){    int l , m ;    while(scanf("%s", str)!=EOF)    {        if( str[0] == '.' ) break;        l = strlen(str);        getnext(l) ;        m = next[l];        if( l % (l-m) != 0 )            printf("1\n");        else        {            m = l / ( l-m );            printf("%d\n", m);        }        memset(str,0,sizeof(str));    }    return 0;}

 

Poj2406 -- power strings (KMP minimum cycle)