POJ 2406 Power Strings（kmp） - c++编程基础

Language: Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33335		Accepted: 13852

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

求一个串的最小周期

对于一个串，如果abcdabc 那么next[len]=3,那么len-next【len】就大于len/2,那么len%(len-next[len])!=0;而对于一个周期串ababab next[len]=4,此时len-next[len]应该等于

最小串的长度，所以是不是有最小周期就可以用len%(len-next[len])是否为0来判断，（个人理解，如有错，请想告）

#include
  
   
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     #include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
         
           #define L(x) (x<<1) #define R(x) (x<<1|1) #define MID(x,y) ((x+y)>>1) #define eps 1e-8 using namespace std; #define N 100005 char a[N]; int len,next[N]; void getfail(char *a) { int i,j; len=strlen(a); i=0;j=-1; next[0]=-1; while(i