（ KMP 求循环节的个数）Power Strings -- poj -- 2406

链接：

http://poj.org/problem?id=2406

Power Strings

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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.

 

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 1000007

char S[N];
int Next[N];  

/// Next中存的是前缀和后缀的最大相似度

void FindNext(int Slen, int Next[]) ///Next[i] 代表前 i 个字符的最大匹配度
{
    int i=0, j=-1;
    Next[0] = -1;

    while(i<Slen)
    {
        if(j==-1 || S[i]==S[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}

int main()
{
    while(scanf("%s", S), strcmp(S, "."))
    {
        int Slen=strlen(S);

        FindNext(Slen, Next);

        if(Slen%(Slen-Next[Slen]))
            printf("1
");
        else
            printf("%d
", Slen/(Slen-Next[Slen]));
    }

    return 0;
}