POJ-3261-Milk Patterns-二分+哈希

Milk Patterns

题意：

在一串数字中，求至少连续k次的最大子序列长度；

思路：

二分加哈希；

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

const int maxn = ;
const int p = ;

int n,k,a[maxn+];
unsigned long long hash1[maxn+],F[maxn+]={};        //F数组用来保存p的次方，可能是pow（）不稳定，慎用
unsigned long long re[maxn+];

bool judge(int x)
{
    memset(re,,sizeof(re));
    int tot = ;
    for(int i=;i<=n-x+;i++)
    {
        int r = i+x-;
        re[++tot] = (hash1[r]-hash1[i-]*F[r-i+]);
    }
    sort(re+,re++tot);
    int cnt=,mmax=;
    unsigned long long last = re[];
    for(int i=;i<=tot;i++)
    {
        if(re[i]==last)
        {
            cnt++;
            if(cnt>mmax)mmax = cnt;
        }
        else
        {
            cnt=;
        }
        last = re[i];
    }
    return mmax >= k;
}

int main(){
    scanf("%d%d",&n,&k);
    hash1[]=;
    F[] = ;
    for(int i=;i<=n;i++)
    {
        F[i] = F[i-]*p;
    }
    for(int i=;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(i==)hash1[i]=a[i];
        else
            hash1[i] = hash1[i-]*p +a[i];
    }
    int le = 1,ri = n;
    while(le+<ri)
    {
        int mid = le + (ri-le)/;
        if(judge(mid)) le = mid;
        else ri = mid;
    }
    printf("%d\n",le);
    return ;
}