Power Strings POJ - 2406 (KMP)

                                                                                                               

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.

 

分析： 题目求字符串的最小循环节，可以利用KMP的next数组性质

next[len]表示字符串前缀和后缀的最大公共长度。

比如长度为8的字符串 abababab   next[len]=6 即最长公共长度为6

进而推出最小循环节即为2

代码如下

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 1000002;
int next[N];
char S[N], T[N];
int slen, tlen;

void getNext()
{
    int j, k;
    j = 0; k = -1; next[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
            next[++j] = ++k;
        else
            k = next[k];

}
int main()
{

    int TT;
    int i, cc,ans;
    while(scanf("%s",T)!=EOF)
    {
        if(T[0]=='.')
            break;
       tlen=strlen(T);
        getNext();
        ans=tlen-next[tlen];
        if(tlen%ans==0)
        printf("%d\n",tlen/ans);
        else
        printf("1\n");
    }
    return 0;
}