POJ2406

我还不是很懂的一道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
题意：给出一个字符串 问它最多由多少相同的字串组成

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
using namespace std;
int next[1000005],slen,ans; 
char s[1000005];
void getNext()
{
    int j, k;
    j = 0; k = -1; next[0] = -1;
    while(j < slen)
        if(k == -1 || s[j] == s[k])
        {
            next[++j] = ++k;
        }

        else
            k = next[k];

}

int main()
{
    while(~scanf("%s",s))
    {
        if(s[0]=='.') return 0;
        slen=strlen(s);
        getNext();
        ans=1;
        if(slen%(slen-next[slen])==0)//求最小循环节
        ans=slen/(slen-next[slen]);
        printf("%d\n",ans);
    }
    return 0;
}