Gold Balanced Lineup POJ - 3274
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
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
Sample Input
7 3 7 6 7 2 1 4 2
Sample Output
4
Hint
Source
题解：
其实这个题是想到怎么hash比较难。
假设我们去统计这样一组数据
1 0 1 0
0 1 0 1
求前缀和是
0 0 0 0
1 0 1 0
1 1 1 1
观察这组数据，我们可以发现每个数据减去第一位就会得到相同的
#include <cstdio>
#include<cstring>
#include <queue>
#include <vector>
#include <iostream>
using namespace std;
const int MAXN=1e5+10;
int sum[MAXN][40];
vector<int>Hash[MAXN];
int n,k;
const int pri=100005;
int abs(int x)
{
if(x<0) return -x;
return x;
}
queue<int>st;
void reslove(int x,int i)
{
while(x)
{
int z=x%2;
st.push(z);
x/=2;
}
while(st.size()<k)
{
st.push(0);
}
int num=0;
while(!st.empty())
{
if(st.front()==1)
{
sum[i][num]=sum[i-1][num]+1;
} else{
sum[i][num]=sum[i-1][num];
}
num++;
st.pop();
}
}
inline bool scan_d(int &num)//输入挂
{
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<'0'||in>'9')) in=getchar();
if(in=='-'){ IsN=true;num=0;}
else num=in-'0';
while(in=getchar(),in>='0'&&in<='9'){
num*=10,num+=in-'0';
}
if(IsN) num=-num;
return true;
}
int main()
{
scanf("%d%d",&n,&k);
int x;
memset(sum,0, sizeof(sum));
for (int i = 1; i <=n ; ++i) {
scan_d(x);
reslove(x,i);
}
for (int i = 0; i <=n ; ++i) {
int key=0;
for (int j = 1; j <k ; ++j) {
sum[i][j]-=sum[i][0];
key+=sum[i][j]*j;
}
key=abs(key)%pri;
Hash[key].push_back(i);
}
int MAX=0;
for (int i = 0; i <100005 ; ++i) {
if(Hash[i].size()>1)
{
for (int j = 0; j <Hash[i].size()-1 ; ++j) {
for (int l = j+1; l <Hash[i].size() ; ++l) {
int flag=0;
for (int m = 1; m <k ; ++m) {
if(sum[Hash[i][j]][m]!=sum[Hash[i][l]][m])
{
flag=1;
break;
}
}
if(flag==0)
{
MAX=max(MAX,abs(Hash[i][j]-Hash[i][l]));
}
}
}
}
}
printf("%d\n",MAX);
return 0;
}
/*
7 3
7
6
7
2
1
4
2
-----------------------
11 5
30
28
24
16
1
3
7
15
16
24
31
------------
6 30
123456789
234567890
345678901
456789012
567890123
678901234
------------
250 3
6
3
1
3
6
7
0
7
6
0
1
3
3
4
3
3
2
2
4
7
3
2
3
0
0
6
2
1
7
2
3
3
5
2
5
2
2
5
0
2
4
5
5
1
4
4
6
3
5
7
7
5
7
7
4
0
0
7
1
3
4
7
2
3
7
5
2
6
2
3
2
0
3
2
4
6
1
1
2
7
1
6
0
5
3
4
6
1
7
5
7
4
2
2
7
1
7
0
7
6
2
3
0
3
5
5
3
1
7
1
7
6
5
3
1
5
6
3
4
0
1
1
2
7
6
1
2
4
1
3
0
4
6
2
6
6
7
6
0
1
7
1
7
5
7
7
3
7
7
5
0
2
6
4
1
3
1
3
7
4
0
6
1
1
4
1
6
7
2
5
4
7
1
7
4
2
2
3
4
0
1
1
0
1
6
1
7
4
6
3
3
7
7
5
3
6
5
1
4
4
1
3
6
4
4
2
0
4
3
7
7
7
3
6
7
3
1
0
2
2
6
1
6
4
7
4
4
6
5
6
5
6
7
4
3
7
1
5
4
0
2
7
1
7
4
4
7
2
2
4
------------
1 3
0
=============================================
下面是答案：
4
--------------
8
------------
0
------------
205
----------
1
*/
