G - Power Strings

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

題解：

KMP求最小迴圈節應用。
注意判斷條件。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 1000000+100;

char str[MAXN];
int len;
int nt[MAXN];

void getNext()
{
    nt[0] = -1;
    int i = 0, j = -1;
    while (i <= len)
    {
        if
 
 (j == -1 || str[i] == str[j])
        {
            nt[++i] = ++j;
        }
        else
        {
            j = nt[j];
        }
    }
}

int main()
{
    while(scanf("%s",str)!=EOF)
    {
        if(str[0]=='.') break;
        len = strlen(str);
        getNext();
        if(nt[len]==0)
        {
            cout
 
<<1<<endl;
            continue;
        }

        int t = len-nt[len];

        if(len%t)
        {
            cout<<1<<endl;
        }
        else
            cout<<len/t<<endl;
    }
    return 0;
}