poj 2406 Power Strings【kmp循环节】

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 50448 Accepted: 21038

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 should print the largest n such that s = a^n for some string a.

Sample Input

abcdaaaaababab.

Sample Output

143

思路：KMP，next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。所以思路和上面一样，如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

例如：a    b    a    b    a    b

next:-1   0    0    1    2    3    4

next[n]==4,代表着，前缀abab与后缀abab相等的最长长度，这说明，ab这两个字母为一个循环节，长度=n-next[n];

/* *********************************************** ┆  ┏┓　　　┏┓ ┆ ┆┏┛┻━━━┛┻┓ ┆ ┆┃　　　　　　　┃ ┆ ┆┃　　　━　　　┃ ┆ ┆┃　┳┛　┗┳　┃ ┆ ┆┃　　　　　　　┃ ┆ ┆┃　　　┻　　　┃ ┆ ┆┗━┓　马　┏━┛ ┆ ┆　　┃　勒　┃　　┆　　　　　　 ┆　　┃　戈　┗━━━┓ ┆ ┆　　┃　壁　　　　　┣┓┆ ┆　　┃　的草泥马　　┏┛┆ ┆　　┗┓┓┏━┳┓┏┛ ┆ ┆　　　┃┫┫　┃┫┫ ┆ ┆　　　┗┻┛　┗┻┛ ┆ ************************************************ */  #include<stdio.h>  #include<string.h>char S[1000005];int next[1000005],l;void GetNext(){int i=0,j=-1;next[0]=-1;j=next[i];while(i<l){if(j==-1||S[i]==S[j]){next[i+1]=j+1;i++; j++;}else{j=next[j];}} }int main(){while( ~scanf("%s",S) ){if(S[0]=='.')break;l=strlen(S);GetNext();if(l%(l-next[l])==0)            printf("%d\n",l/(l-next[l]));        else            printf("1\n");}return 0;}