#POJ 2406 Power Strings (KMP)

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.

题目大意 ： 输入一个字符串， 输出该字符串可以最多由多少个字符串拼得

思路 ：  没啥难度，  KMP中getnext表示的就是该下标之前， 最大前缀与后缀相同的长度， 那么运行到i的时候， i - nxt[i]就是循环体的长度了， 所以最大的一定是看最后一个字符。

AC代码 ：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 1e7 + 5;

char str[maxn];
int nxt[maxn], len_;
void getnext() {
    nxt[0] = -1;
    int i = 0, j = -1;
    while (i < len_) {
        if (j == -1 || str[i] == str[j])
            nxt[++i] = ++j;
        else
            j = nxt[j];
    }
}

int main()
{
    while (scanf("%s", str) && strcmp(str, ".") != 0) {
        memset(nxt, 0, sizeof(nxt));
        len_ = strlen(str);
        getnext();
        if (len_ % (len_ - nxt[len_]) == 0) cout << len_ / (len_ - nxt[len_]) <<endl;
        else cout << 1 << endl;
    }
    return 0;
}