POJ - 2406 Power Strings(字符串周期)

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.

i-next[i]:表示已i结尾的前缀的循环节的长度

#include 
#include 
#include 

char s[1000005];
int nex[1000005],len;
void get_next();
int main()
{
    int a;
    while(scanf("%s",s)!=EOF)
    {
        if(s[0]=='.')
            break;
        get_next();
        a=len-nex[len-1]; //循环节长度
        if(len%a==0)
            printf("%d\n",len/a);
        else
            printf("1\n");
    }
    return 0;
}
void get_next()
{
    int i,j;
    len=strlen(s);
    i=0;
    j=0;
    nex[0]=0;
    for(i=1;i1];
        while(j!=0&&s[i]!=s[j])
        {
            j=nex[j-1];
        }
        if(s[i]==s[j])
            nex[i]=j+1;
        else
            nex[i]=0;
    }
}