kuangbin专题十六 KMP&&扩展KMP POJ2406 Power Strings

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

Hint

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

prekmp写残了，导致以为EOF不会写，后来才发现Next[j]=-1了。。。。

其实就是求最小循环节，然后循环次数n就最大了。

 #include<stdio.h>

 #include<string.h>

 #include<iostream>

 #include<string>

 using namespace std;

 int Next[],n;

 char p[];

 void prekmp() {

     int i,j;

     j=Next[]=-;

     i=;

     while(i<n) {

         while(j!=-&&p[i]!=p[j]) j=Next[j];

         if(p[++i]==p[++j]) Next[i]=Next[j];

         else Next[i]=j;

     }

 }

 int main() {

     //freopen("in","r",stdin);

     while(~scanf("%s",p)) {

         if(p[]=='.') break;

         n=strlen(p);

         prekmp();

         int l=n-Next[n];

         if(n%l==) printf("%d\n",n/l);

         else printf("%d\n",);

     }

 }

巴特西

kuangbin专题十六 KMP&&扩展KMP POJ2406 Power Strings

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