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 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.
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.
1 /* 2 数组sum[i][j]表示从第1到第i头cow属性j的出现次数。 3 4 所以题目要求等价为: 5 6 求满足 7 8 sum[i][0]-sum[j][0]=sum[i][1]-sum[j][1]=.....=sum[i][k-1]-sum[j][k-1] (j<i) 9 10 中最大的i-j 11 12 13 14 将上式变换可得到 15 16 sum[i][1]-sum[i][0] = sum[j][1]-sum[j][0] 17 18 sum[i][2]-sum[i][0] = sum[j][2]-sum[j][0] 19 20 ...... 21 22 sum[i][k-1]-sum[i][0] = sum[j][k-1]-sum[j][0] 23 24 25 26 令C[i][y]=sum[i][y]-sum[i][0] (0<y<k) 27 28 初始条件C[0][0~k-1]=0 29 30 31 32 所以只需求满足C[i][]==C[j][] 中最大的i-j，其中0<=j<i<=n。 33 34 C[i][]==C[j][] 即二维数组C[][]第i行与第j行对应列的值相等， 35 36 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。 37 */ 38 #include<iostream> 39 #include<string> 40 #include<cstring> 41 #include<cmath> 42 #include<cstdio> 43 using namespace std; 44 const int maxn=107777; 45 int hash[maxn],c[maxn][50],sum[maxn]; 46 int i,j,k,n,m; 47 //min_index[]数组代表相等的上一行的下标 48 int min_index[maxn],next[maxn]; 49 bool judge(int a,int b)//判断a行和b行是否相等 50 { 51 for(int it=0;it<k;it++) 52 { 53 if(c[a][it]!=c[b][it]) 54 return false; 55 } 56 return true; 57 } 58 int hashcode(int *v)//hash函数 59 { 60 int s=0; 61 for(int it=0;it<k;it++) 62 { 63 s=((s<<2)+(v[it]>>4))^(v[it]<<10); 64 } 65 s=s%maxn; 66 if(s<0) 67 s=s+maxn; 68 return s; 69 } 70 int all_0(int index)//判断一行是否为0 71 { 72 for(int it=0;it<k;it++) 73 { 74 if(c[index][i]!=0) 75 return false; 76 } 77 return true; 78 } 79 void insert(int index) 80 { 81 int h=hashcode(c[index]); 82 if(!h) 83 { 84 if(all_0(index)) 85 { 86 min_index[index]=0; 87 return; 88 } 89 } 90 int u=hash[h]; 91 if(!u)//如果hash数组为0，则赋值 92 { 93 min_index[index]=index; 94 hash[h]=index; 95 return; 96 } 97 while(u) 98 { 99 if(judge(index,u))//找到相等的，则返回 100 { 101 min_index[index]=min_index[u]; 102 return; 103 } 104 u=next[u]; 105 } 106 min_index[index]=index; 107 next[index]=hash[h]; 108 hash[h]=index; 109 110 } 111 int main() 112 { 113 while(~scanf("%d%d",&n,&k)) 114 { 115 memset(c,0,sizeof(c)); 116 memset(min_index,0,sizeof(min_index)); 117 memset(next,0,sizeof(next)); 118 memset(hash,0,sizeof(hash)); 119 int sum[39]={0},num,max1=0; 120 for(i=1;i<=n;i++) 121 { 122 scanf("%d",&num); 123 for(j=0;j<k;j++) 124 { 125 sum[j]+=((1<<j)&num)?1:0; 126 c[i][j]=sum[j]-sum[0]; 127 } 128 insert(i); 129 } 130 for(i=1;i<=n;i++)//求最大值 131 { 132 if(max1<i-min_index[i]) 133 max1=i-min_index[i]; 134 } 135 printf("%d ",max1); 136 } 137 return 0; 138 }