POJ 3274 Gold Balanced Lineup

解题思路：hash

统计到当前位置为止，各位上出现1的个数，相对于最小次数的差值

比较出现相同差值序列的最长距离即可

#include<iostream>
using namespace std;
#define MAXN 100001
#define PRIME 50101
int feature[MAXN][30],next[MAXN],first[PRIME];
int main()
{
    int i,j,N,K,F,p,_min, _max=0;
    long long t,t2;
    int bit[30];
    for(i=0,j=1;i<30;i++,j*=2)bit[i]=j;
    for(i=0;i<MAXN;i++)next[i]=-1;
    for(i=0;i<PRIME;i++)first[i]=-1;
    scanf("%d %d", &N, &K);
    for(i=0;i<N;i++)
    {
        scanf("%d",&F);
        for(_min=MAXN,j=0;j<K;j++)
        {
            if(i>0)feature[i][j]=feature[i-1][j];
            if(F&bit[j])feature[i][j]++;
            if(_min > feature[i][j])_min=feature[i][j];
        }
        for(t=j=0;j<K;j++){feature[i][j]-=_min;t2=feature[i][j];t+=(t2*t2);}
        if(t==0&&_max<(i+1))_max=i+1;
        t%=PRIME;
        if(-1==(p=first[t])){first[t]=i;continue;}
        while(1)
        {
            for(j=0;j<K;j++)if(feature[p][j]!=feature[i][j])break;
            if(j==K&&((i-p)>_max))_max=i-p;
            if(next[p]==-1){next[p]=i;break;}
            p=next[p];
        }
    }
    printf("%d\n",_max);
    return 0;
}