【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 }

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