【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。。

 1 /*
 2 唐代李白
 3 《三五七言 / 秋风词》
 4
 5 秋风清，秋月明，
 6 落叶聚还散，寒鸦栖复惊。
 7 相思相见知何日？此时此夜难为情！
 8 入我相思门，知我相思苦。
 9 长相思兮长相忆，短相思兮无穷极。
10 早知如此绊人心，何如当初莫相识。
11 */
12 #include <iostream>
13 #include <cstdio>
14 #include <algorithm>
15 #include <cstring>
16 #include <vector>
17 #include <utility>
18 #include <iomanip>
19 #include <string>
20 #include <cmath>
21 #include <queue>
22 #include <assert.h>
23 #include <map>
24 #include <ctime>
25 #include <cstdlib>
26 #include <stack>
27 #define LOCAL
28 const int MAXN = 1000000 + 10;
29 const int INF = 100000000;
30 const int SIZE = 450;
31 const int MAXM = 1000000 + 10;
32 const int maxnode =  0x7fffffff + 10;
33 using namespace std;
34 int l1, l2;
35 char a[MAXN], b[MAXN];
36 int next[1000000];//不用开太大了..
37 void getNext(){
38      //初始化next数组
39      next[1] = 0;
40      int j = 0;
41      for (int i = 2; i <= l1; i++){
42          while (j > 0 && a[j + 1] != a[i]) j = next[j];
43          if (a[j + 1] == a[i]) j++;
44          next[i] = j;
45      }
46      return;
47 }
48 int kmp(){
49     int j = 0, cnt = 0;
50     for (int i = 1; i <= l2; i++){
51         while (j > 0 && a[j + 1] != b[i]) j = next[j];
52         if (a[j + 1] == b[i]) j++;
53         if (j == l1){
54            cnt++;
55            j = next[j];//回到上一个匹配点
56         }
57     }
58     return cnt;
59 }
60
61 void init(){
62      scanf("%s", a + 1);
63      scanf("%s", b + 1);
64      l1 = strlen(a + 1);
65      l2 = strlen(b + 1);
66 }
67
68 int main(){
69     int T;
70
71     while (scanf("%s", a + 1)){
72           if (a[1] == ‘.‘) break;
73           l1 = strlen(a + 1);
74           getNext();
75           if (l1%(l1 - next[l1]) == 0) printf("%d\n", l1/(l1 - next[l1]));
76           else printf("1\n");
77     }
78     return 0;
79 }