这个是哈希的折叠法，，这样的题，发现用这样的方法可以大大的提高计算效率。


其中关键代码p值的计算是像代码中写的一样，都可以，只要找得比较乱就行。


#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<set>
#include<cstring>

using namespace std;

#define maxn 100010

const int prime=99983;

int n,k;

int a[maxn][31],sum[maxn][31],C[maxn][31];

int hash[1000000];

int ans;

int hashcode(int *v,int k)   

{

       int i,p=0;

	   int cur=0;

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

			p=p^(v[i]<<9+p+v[i]>>2);		

              //p=((p<<2)+(v[i]>>4))^(v[i]<<10);

       p = p%prime;

       if(p<0)   p=p+prime;

       return p;

}

int main()

{

       int i,j,x,p;

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

       memset(sum,0,sizeof(sum));

       memset(hash,-1,sizeof(hash));

       memset(C,0,sizeof(C));

       hash[hashcode(C[0],k)]=0;

       ans=0;

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

       {

              scanf("%d",&x);

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

              {

                     a[i][j]=x%2;

                     x>>=1;

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

                     C[i][j]=sum[i][j]-sum[i][0];

              }

              p=hashcode(C[i],k);

              while(hash[p]!=-1)

              {

                     for(j=1;j<k;j++)

                     {

                            if(C[i][j]!=C[hash[p]][j])

                                   break;

                     }

                     if(j==k)

                     {

                            if(i-hash[p]>ans)

                                   ans=i-hash[p];

                            break;

                     }

                     p++;

              }

              if(hash[p]==-1)

                     hash[p]=i;

       }

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

       return 0;

}





Gold Balanced Lineup


Time Limit: 2000MS

 

Memory Limit: 65536K

Total Submissions: 8376

 

Accepted: 2472


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 onlyK 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 featurei.

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 cowsi..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


Line 1: Two space-separated integers, N and K.
Lines 2.. N+1: Line i+1 contains a single K-bit integer specifying the features present in cow i. The least-significant bit of this integer is 1 if the cow exhibits feature #1, and the most-significant bit is 1 if the cow exhibits feature # K.


Output


Line 1: A single integer giving the size of the largest contiguous balanced group of cows.


Sample Input


7 3 7 6 7 2 1 4 2


Sample Output


4


Hint


In the range from cow #3 to cow #6 (of size 4), each feature appears in exactly 2 cows in this range