Poj 2406--Power Strings(KMP)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38544		Accepted: 16001

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,   扩展数组 ;

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char str[1000010]; int len, NeXt[1000010];
void GetP()
{
    int i = 0, j = -1;
    NeXt[i] = j;
    while(i < len)
    {
        if(j == -1 || str[i] == str[j])
        {
            i++; j++;
            if(str[i] != str[j])
                NeXt[i] = j;
            else
                NeXt[i] = NeXt[j];
        }
        else
            j = NeXt[j];
    }
        
} 
int main()
{
    while(gets(str), str[0] != '.')
    {
        
        len = strlen(str);
        GetP();
        if(len%(len-NeXt[len])==0)
            printf("%d
", len/(len-NeXt[len]));
        else
            printf("1
");
    }
    return 0;    
}