HDU1358（KMP）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1358

主要考查KMP算法next数组本质的理解。len - len[len]为循环节的大小~~~

代码如下：

 1 #include <cstdio>

 2 #include <cstring>

 3 using namespace std;

 4 

 5 const int N = 1000002;

 6 char str[N];

 7 int next[N];

 8 

 9 void get_next(int len)

10 {

11     int i = 0;

12     int j = -1;

13     next[i] = -1;

14     while(i < len)

15     {

16         if(j == -1 || str[i] == str[j])

17         {

18             i++;

19             j++;

20             next[i] = j;

21         }

22         else

23         {

24             j = next[j];

25         }

26     }

27 }

28 

29 int main()

30 {

31     int n;

32     int k = 0;

33     while(scanf("%d", &n) != EOF && n != 0)

34     {

35         scanf("%s", str);

36         get_next(n);

37         k++;

38         printf("Test case #%d\n", k);

39         for(int i = 2; i <= n; i++)

40         {

41             int x = i - next[i];

42             if((i % x == 0) && i / x != 1)

43             {

44                 printf("%d %d\n", i, i / x);

45             }

46         }

47         printf("\n");

48     }

49     return 0;

50 }

 

参考别人的代码：

 1 #include <cstdio>

 2 #include <cstring>

 3 using namespace std;

 4 

 5 const int N = 1000002;

 6 char str[N];

 7 int  next[N];

 8 

 9 void get_next(int len)

10 {

11     int i;

12     int j = 0;

13     for(next[1] = 0, i = 2; i <= len; i++)

14     {

15         while(j && str[i] != str[j + 1])

16         {

17             j = next[j];

18         }

19         if(str[i] == str[j + 1])

20         {

21             j++;

22         }

23         next[i] = j;

24     }

25 }

26 

27 int main()

28 {

29     int n;

30     int k = 1;

31     while(scanf("%d", &n) != EOF && n != 0)

32     {

33         scanf("%s", str + 1);

34         get_next(n);

35         printf("Test case #%d\n", k++);

36         for(int i = 2; i <= n; i++)

37         {

38             if(i % (i - next[i]) == 0 && i / (i - next[i]) != 1)

39             {

40                 printf("%d %d\n", i, i / (i - next[i]));

41             }

42         }

43         printf("\n");

44     }

45     return 0;

46 }