POJ

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 51263 Accepted: 21405

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

Hint

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

问 把所给字符串写成  s^n 的模式 ，n最大是多少

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;const int len=1e6+8;char a[len];int kmp_next[len];void KMP_next(){    int i=0,j=-1;    kmp_next[0]=-1;    while(a[i])        if(j==-1||a[i]==a[j])            kmp_next[++i]=++j;        else            j=kmp_next[j];}int main(){    while(scanf("%s",a)&&a[0]!='.')    {        KMP_next();        int n=strlen(a);        if(n%(n-kmp_next[n]))            printf("1\n");        else            printf("%d\n",n/(n-kmp_next[n]));    }    return 0;}