POJ - 2406 Power Strings

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Power Strings

Time Limit: 3000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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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.

Source

Waterloo local 2002.07.01

字符串的自身匹配kmp算法，找出一个字符串中最短的连续串。简单模板题，代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000000+10;
int next[maxn];
char s[maxn];
int n;
int getnext(char *s)//kmp不仅出现在kmp函数，在其他函数也都用到了kmp
{
int j=-1,i=0;
next[0]=-1;
while(i<n)
{
if(j==-1||s[i]==s[j])
{
i++,j++;
next[i]=j;
}
//else j=next[j];
else j=-1;//还是这句简洁高效
}
if(n%(n-j)==0) return n/(n-j);//写几个字符串手动模拟一下就知道这句话的含义了
else return 1;
}

int main()
{
while(scanf("%s",s)&&s[0]!='.')
{
n=strlen(s);
printf("%d\n",getnext(s));
}
}