这题，看到别人的解题报告做出来的，分析：

大概意思就是：

数组sum[i][j]表示从第1到第i头cow属性j的出现次数。

所以题目要求等价为:

求满足

sum[i][0]-sum[j][0]=sum[i][1]-sum[j][1]=.....=sum[i][k-1]-sum[j][k-1] (j<i)

中最大的i-j

将上式变换可得到

sum[i][1]-sum[i][0] = sum[j][1]-sum[j][0]

sum[i][2]-sum[i][0] = sum[j][2]-sum[j][0]

......

sum[i][k-1]-sum[i][0] = sum[j][k-1]-sum[j][0]

令C[i][y]=sum[i][y]-sum[i][0] (0<y<k)

初始条件C[0][0~k-1]=0

所以只需求满足C[i][]==C[j][] 中最大的i-j，其中0<=j<i<=n。

C[i][]==C[j][] 即二维数组C[][]第i行与第j行对应列的值相等，

那么原题就转化为求C数组中 相等且相隔最远的两行的距离i-j

大概意思就是：

数组sum[i][j]表示从第1到第i头cow属性j的出现次数。

所以题目要求等价为:

求满足

sum[i][0]-sum[j][0]=sum[i][1]-sum[j][1]=.....=sum[i][k-1]-sum[j][k-1] (j<i)

中最大的i-j

将上式变换可得到

sum[i][1]-sum[i][0] = sum[j][1]-sum[j][0]

sum[i][2]-sum[i][0] = sum[j][2]-sum[j][0]

......

sum[i][k-1]-sum[i][0] = sum[j][k-1]-sum[j][0]

令C[i][y]=sum[i][y]-sum[i][0] (0<y<k)

初始条件C[0][0~k-1]=0

所以只需求满足C[i][]==C[j][] 中最大的i-j，其中0<=j<i<=n。

C[i][]==C[j][] 即二维数组C[][]第i行与第j行对应列的值相等，

那么原题就转化为求C数组中 相等且相隔最远的两行的距离i-j

以样例为例

7 3

7

6

7

2

1

4

2

先把7个十进制特征数转换为二进制，并逆序存放到特征数组feature[ ][ ]，得到：

7 ->1 1 1

6 ->0 1 1

7 ->1 1 1

2 ->0 1 0

1 ->1 0 0

4 ->0 0 1

2 ->0 1 0

（行数为cow编号，自上而下从1开始；列数为特征编号，自左到右从0开始）

再求sum数组，逐行累加得，sum数组为

1 1 1

1 2 2

2 3 3

2 4 3

3 4 3

3 4 4

3 5 4

再利用C[i][y]=sum[i][y]-sum[i][0]求C数组，即所有列都减去第一列

注意C数组有第0行，为全0

0 0 0 -> 第0行

0 0 0

0 1 1  <------

0 1 1

0 2 1

0 1 0

0 1 1 <-------

0 2 1

显然第2行与第6行相等，均为011，且距离最远，距离为6-2=4，这就是所求。

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

int num[][];

int N,K;

struct hash{

    int ind;

    hash* next;

};

hash hashtable[];

int gethash(int id){

    num[id][]=;

    for(int i=;i<K;i++){

        num[id][]+=num[id][i]*i;

    }

    num[id][]=(num[id][]&0x7fffffff)%;

return num[id][];

}

int isEqual(int id1,int id2){

    int flag=;

    for(int i=;i<K;i++){

        if(num[id1][i]!=num[id2][i]){

            flag=;

        }

    }

    return flag;

}

int main(){

    scanf("%d %d",&N,&K);

    for(int j=;j<K;j++){

        num[][j]=;

    }

    memset(hashtable,,sizeof(hash)*);

    for(int i=;i<=N;i++){

        int t;

        scanf("%d",&t);

        for(int j=;j<K;j++){

            num[i][j]=t%;

            t=t>>;

            num[i][j]+=num[i-][j];

        }

    }

    int result=;

    for(int i=;i<=N;i++){

        for(int j=;j<K;j++){

            num[i][j]-=num[i][];

        }

        int h=gethash(i);

        hash *t=(hash*)malloc(sizeof(hash));

        t->next=hashtable[h].next;

        t->ind=i;

        hashtable[h].next=t;

        while(t!=NULL){

            if(isEqual(t->ind,i)){

                int tmp=i-t->ind;

                result=result>tmp?result:tmp;

            }

            t=t->next;

        }

    }

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

    return ;

}

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.