【POJ2406】【KMP】Power Strings

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

【分析】

其实我是来贴诗句的。。

 /*
 唐代李白
 《三五七言 / 秋风词》

 秋风清，秋月明，
 落叶聚还散，寒鸦栖复惊。
 相思相见知何日？此时此夜难为情！
 入我相思门，知我相思苦。
 长相思兮长相忆，短相思兮无穷极。
 早知如此绊人心，何如当初莫相识。
 */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <assert.h>
 #include <map>
 #include <ctime>
 #include <cstdlib>
 #include <stack>
 #define LOCAL
 const int MAXN =  + ;
 const int INF = ;
 const int SIZE = ;
 const int MAXM =  + ;
 const int maxnode =  0x7fffffff + ;
 using namespace std;
 int l1, l2;
 char a[MAXN], b[MAXN];
 int next[];//不用开太大了..
 void getNext(){
      //初始化next数组
      next[] = ;
      int j = ;
      for (int i = ; i <= l1; i++){
          while (j >  && a[j + ] != a[i]) j = next[j];
          if (a[j + ] == a[i]) j++;
          next[i] = j;
      }
      return;
 }
 int kmp(){
     int j = , cnt = ;
     for (int i = ; i <= l2; i++){
         while (j >  && a[j + ] != b[i]) j = next[j];
         if (a[j + ] == b[i]) j++;
         if (j == l1){
            cnt++;
            j = next[j];//回到上一个匹配点
         }
     }
     return cnt;
 }

 void init(){
      scanf("%s", a + );
      scanf("%s", b + );
      l1 = strlen(a + );
      l2 = strlen(b + );
 }

 int main(){
     int T;

     while (scanf("%s", a + )){
           if (a[] == '.') break;
           l1 = strlen(a + );
           getNext();
           if (l1%(l1 - next[l1]) == ) printf("%d\n", l1/(l1 - next[l1]));
           else printf("1\n");
     }
     return ;
 }