Description

Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might have spots, cows exhibiting feature #2 might prefer C to Pascal, and so on.

FJ has even devised a concise way to describe each cow in terms of its "feature ID", a single K-bit integer whose binary representation tells us the set of features exhibited by the cow. As an example, suppose a cow has feature ID = 13. Since 13 written in binary is 1101, this means our cow exhibits features 1, 3, and 4 (reading right to left), but not feature 2. More generally, we find a 1 in the 2^(i-1) place if a cow exhibits feature i.

Always the sensitive fellow, FJ lined up cows 1..N in a long row and noticed that certain ranges of cows are somewhat "balanced" in terms of the features the exhibit. A contiguous range of cows i..j is balanced if each of the K possible features is exhibited by the same number of cows in the range. FJ is curious as to the size of the largest balanced range of cows. See if you can determine it.

Input

Output

Sample Input

7 3
7
6
7
2
1
4
2

Sample Output

4

Hint

题目大意，按二进制给出一些特征，求最长的一段内的所有特征相同

首先用p[i][j]存储从第一个牛到底i头牛，关于第j种特征的总数，如果某一段中的[i,j]符合，那么

p[j][0] - p[i-1][0] = p[j][1] = p[i-1][1] = 。。 = p[j][k-1] - p[i-1][k-1]；从第0到k-1所有的差都相同，一定是p[j][0...k] - p[i-1][0...k] 才对应的是区间[i,j]；

那么，转化后的p[j][1] - p[j][0] = p[i-1][1] - p[i-1][0] , p[j][2] - p[j][0] = p[i-1][2] - p[i-1][0] 。。。。一直k-1

所以最后是p[i][j]转化为p[i][j] = p[i][j] - p[i][0] ；那么如果区间[i,j]相同，那么p[i-1]和p[j]应该完全相同，使用hash找到最长的一段

注意，存在可能区间[1,n]是最长的区间，所以在vec[0] 中加入0

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       using namespace std ; #define mod 999997 int p[110000][32] ; vector 
      
        vec[100000] ; int main() { int i , j , n , k , a , c[32] , num , flag , l , h , max1 = 0 ; __int64 sum ; scanf("%d %d", &n, &k); for(i = 0 , num = 1; i < k ; i++) { p[0][i] = 0 ; num *= 2 ; } sum = 0; vec[0].push_back(0) ; for(i = 1 ; i <= n ; i++) { scanf("%d", &a); if(a == num-1) max1 = 1 ; j = k-1 ; memset(c,0,sizeof(c)); while(a) { c[j--] = a%2 ; a /= 2 ; } for(j = 0 ; j < k ; j++) { p[i][j] = p[i-1][j] + c[j] ; } } for(i = 1 ; i <= n ; i++) { sum = 0 ; for(j = 1 ; j < k ; j++) { p[i][j] = p[i][j] - p[i][0] ; sum += p[i][j] ; } p[i][0] = 0 ; if(sum < 0) sum += 100000 ; sum %= mod ; num = vec[sum].size(); flag = 0 ; for(j = 0 ; j < num ; j++) { l = vec[sum][j] ; for(h = 0 ; h < k ; h++) if( p[i][h] != p[l][h] ) break; if(h == k) { flag = 1 ; if( max1 < i-l ) max1 = i-l ; } } if(!flag) vec[sum].push_back(i) ; } printf("%d\n", max1); }