POJ2406Power Strings

易水人去，明月如霜。

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

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

大意：给出一个字符串 问它最多由多少相同的字串组成 

如  abababab由4个ab组成

 

分析：

kmp中的next数组求最小循环节的应用

例如 

ababab  next[6] = 4; 即

 

ababab

   ababab

1~4位  与2~6位是相同的

 

 

那么前两位

就等于3、4位

3、4位就等于5、6位

……

所以 如果 能整除  也就循环到最后了

 

如果不能整除  

就最后余下的几位不在循环内

 

例如

1212121

　 1212121

最后剩余1不能等于循环节

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;char a[1000005];int d[1000005];int main(){while(1){    scanf("%s",a);    if(a[0]=='.') break;    int len=strlen(a);    for(int i=1;i<=len;i++)        d[i]=a[i-1];        int nxt[1000005];       nxt[1]=0;    int j=0;    for(int i=2;i<=len;i++)    {        if(j!=0&&d[i]!=d[j+1]) j=nxt[j];        if(d[i]==d[j+1]) j++;        nxt[i]=j;    }    int ans=1;   if(len%(len-nxt[len])==0)    ans=len/(len-nxt[len]);   printf("%d\n",ans);} return 0;}