POJ 3274 Gold Balanced Lineup(哈希）

题目链接

很难想。会哈希，但是想不出。需要一个转化，本来是求某一段上的二进制上每一位的1数目相等，转化为找两段相等的，换元可推出公式。最后注意特判。。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <string>

 5 using namespace std;

 6 #define MOD 97777

 7 struct node

 8 {

 9     int data;

10     struct node *next;

11 }*head[MOD],hash[60000];

12 int sum[100001][31];

13 int c[100001][31];

14 int n,m;

15 int judge(int x,int y)

16 {

17     int i;

18     for(i = 0; i < m; i ++)

19     {

20         if(c[x][i] !=  c[y][i])

21             return 0;

22     }

23     return 1;

24 }

25 int main()

26 {

27     int i,j,a,key,ans = 0,num = 0;

28     node *q,*t;

29     scanf("%d %d",&n,&m);

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

31     {

32         scanf("%d",&a);

33         for(j = 0; j < m; j ++)

34         {

35             if(a&(1<<j))

36                 sum[i][j] = sum[i-1][j] + 1;

37             else

38                 sum[i][j] = sum[i-1][j];

39         }

40     }

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

42     {

43         key = 0;

44         for(j = 1; j < m; j ++)

45         {

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

47             key += c[i][j]*j;

48         }

49         for(j = 1;j < m;j ++)

50         {

51             if(c[i][j] != 0)

52             break;

53         }

54         if(j == m&&ans < i) ans = i;//特判是否全是0

55         key = key%MOD;

56         if(key < 0)

57             key = key + MOD;

58         for(t = head[key]; t != NULL; t = t->next)

59         {

60             if(ans < (i - t->data)&&judge(i,t->data))

61             {

62                 ans = i - t->data;

63             }

64         }

65         if(t == NULL)

66         {

67             q = &hash[num++];

68             q -> data = i;

69             q -> next = head[key];

70             head[key] = q;

71         }

72     }

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

74     return 0;

75 }