Power Strings

题目描述

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).

输入

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.

输出

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

样例输入

abcd
aaaa
ababab
.

样例输出

1
4
3

#define UnQuestion4
#ifdef Question4

#define UnDebug

#ifdef Debug
#include <ctime>
#include <cstdio>
#endif

#include <iostream>
#include <cstring>

using namespace std;

const int Max = 1e6;
int Next[Max + 10];

void makeNext(char str[], int len);

int main()
{
#ifdef Debug
    freopen("in.txt", "r", stdin);
    clock_t start, finish;
    double totaltime;
    start = clock();
#endif

    char str[Max + 10];
    while (cin >> str && str[0] != '.')
    {
        int len = strlen(str);
        makeNext(str, len);
        if (len % (len - Next[len]) == 0)
            cout << len / (len - Next[len]) << endl;
        else
            cout << 1 << endl;
    }

#ifdef Debug
    finish = clock();
    totaltime = (double)(finish - start) / CLOCKS_PER_SEC;
    cout << "Time : " << totaltime * 1000 << "ms" << endl;
    system("pause");
#endif
    return 0;
}

void makeNext(char str[], int len)
{
    int i = 0, j = -1;
    memset(Next, -1, sizeof(Next));
    Next[0] = -1;
    while (i < len)
    {
        if (j == -1 || str[i] == str[j])
        {
            j++;
            i++;
            Next[i] = j;
        }
        else
            j = Next[j];
    }
}
#endif