BZOJ1702. [Usaco2007 Mar] Gold Balanced Lineup balance queue

Portal

It seems impossible to start, to write out the equation calm down:

Conditions for the interval $ [i, j] $, to meet the requirements of:

$sum[i][1]-sum[j-1][1]=sum[i][2]-sum[j-1][2]$

$sum[i][1]-sum[j-1][1]=sum[i][3]-sum[j-1][3]$

...

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

Wherein $ sum [i] [j] $ represents the number of $ [1, i] within the interval characteristic $ $ $ appears J

Was put together under the same label:

$sum[j-1][2]-sum[j-1][1]=sum[i][2]-sum[i][1]$

$sum[j-1][3]-sum[j-1][1]=sum[i][3]-sum[i][1]$

...

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

Found satisfied as long as the $ sum [i] [p] -sum [i] [1] = sum [j] [p] -sum [j] [1], \ forall p \ in [1, k] $ 

So directly engage in a $ map $ maintenance on the line, I feel the state $ hash $ is not stable ah, so a direct enumeration judge put forward a structure like

Note that the prefix and so remember to put $ 0 $ Join $ map $!

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
    return x*f;
}
const int N=1e5+7;
int n,K,ans;
struct dat{
    int a[31];
    dat () { memset(a,0,sizeof(a)); }
    inline dat F()
    {
        dat res;
        for(int i=1;i<=K;i++) res.a[i]=a[i]-a[1];
        return res;
    }
    inline bool operator < (const dat &tmp) const {
        for(int i=1;i<=K;i++)
            if(a[i]!=tmp.a[i]) return a[i]<tmp.a[i];
        return 0;
    }
}C[N],sum;
map <dat,int> mp;
int main()
{
    n=read(),K=read(); int a;
    for(int i=1;i<=n;i++)
    {
        a=read();
        for(int j=1;j<=K;a>>=1,j++) C[i].a[j]=a&1;
    }
    mp[sum.F()]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=K;j++) sum.a[j]+=C[i].a[j];
        dat p=sum.F();
        if(mp[p]) ans=max(ans,i-(mp[p]-1));
        else mp[p]=i+1;
    }
    printf("%d\n",ans);
    return 0;
}