G - Power Strings

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

abcd

aaaa

ababab

.

Sample Output

Hint

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

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<vector>

#include<string>

#include<cstring>

using namespace std;

#define MAXN 1000001

typedef long long LL;

/*

给定一个串，找最短循环节

*/

char s[MAXN];

int Next[MAXN];

void kmp_pre(int m)

{

    int j,k;

    j = ;k = Next[] = -;

    while(j<m)

    {

        if(k==-||s[j]==s[k])

            Next[++j] = ++k;

        else

            k = Next[k];

    }

}

int main()

{

    while(scanf("%s",s))

    {

        if(s[]=='.') break;

        int l = strlen(s);

        kmp_pre(l);

        int ans = l - Next[l];

        if(l%ans==)

            printf("%d\n",l/ans);

        else

            printf("1\n");

    }

}

巴特西

G - Power Strings

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