Codeforces - ZeptoLab Code Rush 2015 - D. Om Nom and Necklace：字符串

D. Om Nom and Necklace

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Om Nom found a thread with n beads of different colors. He decided to cut the first several beads from this thread to make a bead necklace and present it to his girlfriend Om Nelly.

Om Nom knows that his girlfriend loves beautiful patterns. That's why he wants the beads on the necklace to form a regular pattern. A sequence of beads S is regular if it can be represented as S = A + B + A + B + A + ... + A + B + A, where A and B are some bead sequences, " + " is the concatenation of sequences, there are exactly 2k + 1 summands in this sum, among which there are k + 1 "A" summands and k "B" summands that follow in alternating order. Om Nelly knows that her friend is an eager mathematician, so she doesn't mind if A or B is an empty sequence.

Help Om Nom determine in which ways he can cut off the first several beads from the found thread (at least one; probably, all) so that they form a regular pattern. When Om Nom cuts off the beads, he doesn't change their order.

Input

The first line contains two integers n, k (1 ≤ n, k ≤ 1 000 000) — the number of beads on the thread that Om Nom found and number k from the definition of the regular sequence above.

The second line contains the sequence of n lowercase Latin letters that represent the colors of the beads. Each color corresponds to a single letter.

Output

Print a string consisting of n zeroes and ones. Position i (1 ≤ i ≤ n) must contain either number one if the first i beads on the thread form a regular sequence, or a zero otherwise.

Sample test(s)

Input

7 2
bcabcab

Output

0000011

Input

21 2
ababaababaababaababaa

Output

000110000111111000011

Note

In the first sample test a regular sequence is both a sequence of the first 6 beads (we can take A = "", B = "bca"), and a sequence of the first 7 beads (we can take A = "b", B = "ca").

In the second sample test, for example, a sequence of the first 13 beads is regular, if we take A = "aba", B = "ba".

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思路：可以将AB看成一个串（设为C），然后问题就等价为：判断字符串是否由k个连续的C串以及C串的一个前缀组成（前缀可以为空）。

然后用Z算法求出z数组。

然后从1到n枚举C串的长度（设为len），判断z[0],z[len],z[len*2],....,z[len*(k-1)]的值是否都不小于len，如果都符合就标记最后一个匹配位置为true。

输出的时候维护一个last指针，表示C串的前缀最多能扩展到哪个位置。

复杂度o(n).

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 1 #include <iostream>

 2 #include <stdio.h>

 3 using namespace std;

 4 #define MAXN 1000010

 5 

 6 char s[MAXN];

 7 bool ans[MAXN] = {0};

 8 int z[MAXN] = {0};

 9 

10 int n, k;

11 bool check(int len)

12 {

13     for(int i = 0, cnt = 0; i < n && cnt < k; i += len, cnt++)

14     {

15         if(z[i] < len)

16             return false;

17     }

18     return true;

19 }

20 

21 int main()

22 {

23     freopen("in.txt", "r", stdin);

24     cin >> n >> k;

25 

26     scanf("%s", s);

27 

28     // Z[i] is the length of the longest substring starting from S[i] which is also a prefix of S

29     // s[0,n-1]

30     int L = 0, R = 0;

31     for(int i = 1; i < n; i++)

32     {

33         if(i > R)

34         {

35             L = R = i;

36             while(R < n && s[R - L] == s[R]) R++;

37             z[i] = R - L;

38             R--;

39         }

40         else

41         {

42             int k = i - L;

43             if(z[k] < R - i + 1) z[i] = z[k];

44             else

45             {

46                 L = i;

47                 while(R < n && s[R - L] == s[R]) R++;

48                 z[i] = R - L;

49                 R--;

50             }

51         }

52     }

53     z[0] = n;

54 

55 

56     // 枚举AB串的长度

57     for(int len = 1; len <= n; len++)

58     {

59         if(len * k > n)

60             break;

61         // 判断长度为len的AB串是否符合

62         if(check(len))

63         {

64             int last = len * k;

65             ans[last - 1] = true;

66         }

67     }

68 

69     int last = -1;

70     for(int i = 0; i < n; i++)

71     {

72         if(ans[i] || i<=last)

73             printf("1");

74         else

75             printf("0");

76         if(ans[i] && i < n - 1)

77         {

78             int len = (i+1)/k;

79             last = max(last, min(i+len, i+z[i+1]));

80         }

81     }

82 

83     return 0;

84 }