POJ2406(SummerTrainingDay10-I KMP)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50036		Accepted: 20858

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

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

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

Source

Waterloo local 2002.07.01

 

KMP找不可重叠的最小循环节

//2017-08-10
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1000010;
char str[N];
int nex[N];

void getNext(int n)
{
    nex[0] = -1;
    for(int i = 0, fail = -1; i < n;)
    {
        if(fail==-1||str[i]==str[fail])
        {
            i++, fail++;
            nex[i] = fail;
        }else fail = nex[fail];
    }
}

int main()
{
    while(scanf("%s", str)!=EOF)
    {
        if(str[0] == '.')break;
        int n = strlen(str);
        getNext(n);
        int ans = n-nex[n];
        if(n%ans == 0)
            printf("%d
", n/ans);
        else printf("1
");
    }

    return 0;
}