[数组hash] PKU 3274 Gold Balanced Lineup

题解；

数组hash：hash的是排列，不能用相加、相乘取模，常用hash函数的资料；

 1 # include <stdio.h>

 2 # include <string.h>

 3 

 4 # define get(x, i) ((((x)>>(i))&0x1) ? 1:0)

 5 

 6 # define N 100005

 7 # define K 35

 8 # define MOD 100007

 9 

10 int sum[N][K], c[N][K], a[N], h[MOD << 3];

11 

12 int Max(int x, int y)

13 {

14     return x > y ? x : y;

15 }

16 

17 int eql(int *s, int *t, int k)

18 {

19     int i;

20     for (i = 0; i < k; ++i)

21         if (s[i] != t[i]) return 0;

22     return 1;

23 }

24 

25 int hash(int *s, int k)

26 {

27     int i;

28     unsigned int h = 0, g;

29     for (i = 0; i < k; ++i)

30     {

31         h = (h<<4)+s[i];

32         g = h&0xF0000000;

33         if (g) h^=g>>24;

34         h &= ~g;

35     }

36     return (h%MOD+MOD)%MOD;

37 }

38 

39 int main()

40 {

41     int n, k, i, j, ans = 0, t, x;

42     scanf("%d%d", &n, &k);

43     memset(sum[0], 0, sizeof(sum[0]));

44     memset(h, -1, sizeof(h));

45     h[0] = 0;

46     for (i = 1; i <= n; ++i)

47     {

48         scanf("%d", &a[i]);

49         for (j = 0; j < k; ++j)

50         {

51             sum[i][j] += (sum[i-1][j]+get(a[i], j));

52             c[i][j] = sum[i][j] - sum[i][0];

53         }

54         t = hash(c[i], k);

55         while (h[t] != -1)

56         {

57             x = h[t];

58             for (j = 0; j < k; ++j)

59                 if (c[i][j] != c[x][j]) break;

60             if (j >= k)    {ans = Max(ans, i-x);break;}

61             ++t;

62         }

63         if (h[t] == -1) h[t] = i;

64     }

65     printf("%d\n", ans);

66     

67     return 0;

68 }