Poj2406 power strings

Portal

Power strings

Time limit:3000 Ms		Memory limit:65536 K

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

 

The first KMP question .. The question is to find the loop section of a string, and then check the length of the loop section.

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

Codes

 

Poj2406 power strings