POJ：2406-Power Strings（寻找字符串循环节）

Power Strings

Time Limit: 3000MS Memory Limit: 65536K

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.

就是一个寻找字符串的循环节，用KMP的思想，其实很简单就是len/(len-next[n])。（当然是在没有余数的情况下，不然就是1）

#include<stdio.h>

#include<cstring>

#include<set>

using namespace std;

const int maxn = 1e6+100;

char s[maxn];

int next[maxn];

void cal_next()

{

    int k;

    next[0] = k = -1;

    int len = strlen(s);

    for(int i=1;i<len;i++)

    {

        while(k>-1 && s[i] != s[k+1])

            k = next[k];

        if(s[i] == s[k+1])

            k++;

        next[i] = k;

    }

}

int main()

{

    while(scanf("%s",s))

    {

        if(s[0] == '.' && s[1] == 0)

            break;

        cal_next();

        int len = strlen(s);

        if(len%(len - next[len-1] - 1) == 0)

            printf("%d\n",len/(len - next[len-1] - 1));

        else

            printf("1\n");

        memset(s,0,sizeof(s));

    }

    return 0;

}

巴特西