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 }

 

posted @ 2013-01-14 19:09  Naix_x  阅读(161)  评论(0)  编辑  收藏  举报