POJ2406 Power Strings

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K

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.

Source

Waterloo local 2002.07.01

 

第一道kmp題。。題目意思是求一個字串的迴圈節，求出迴圈節長度之後檢驗即可

 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