Poj2406 power strings

Power strings

Time limit:3000 Ms		Memory limit:65536 K
Total submissions:33273		Accepted:13825

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 shoshould print the largest N such that S = a ^ N for some string.

Sample Input

abcdaaaaababab.

Sample output

143

Hint

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

Source

Waterloo local 2002.07.01

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6  7 int next[1000010],m; 8 char s[1000010]; 9 10 void get_next()11 {12     int i=0,j=-1;13     next[0]=-1;14     while(i<m)15     {16         if(j==-1||s[i]==s[j])17         {18             i++;19             j++;20             next[i]=j;21         }22         else j=next[j];23     }24 }25 int main()26 {27     while(scanf("%s",s)!=EOF)28     {29         if(s[0]==‘.‘)break;30         m=strlen(s);31         get_next();32         if(m%(m-next[m])==0)printf("%d\n",m/(m-next[m]));33         else printf("1\n");34     }35     return 0;36 }

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Poj2406 power strings