PIPIOJ 1476: PIPI's string problem X binary string hash

http://www.pipioj.online/problem.php?id=1476

Idea: Through string hashing, it can be in O (n) O (n)Calculate the length within$O$$($$n$$)$ complexity to be$len$ all$s_1$The hash value of the substring of is stored in a hash table, then it can be in $O\left(n\right)$ Judges 1, s 2 s_1, s_2within complexity$s_1、s_2$Whether there is a common substring. Length of bipartite common substring, complexity $O\left(nlgn\right)$。

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

using ull=unsigned long long;

const int maxn=1e5+5;
const int bs=127;

int n;
char s1[maxn],s2[maxn];
ull s1hash[maxn],s2hash[maxn],base[maxn];

inline bool check(int len)
{
    
    
    unordered_map<ull,bool> hs;
    ull tmp=s1hash[len];
    for(int i=len;i<n;i++)
    {
    
    
        hs[tmp]=1;
        tmp=(tmp-s1[i-len]*base[len-1])*bs+s1[i];
    }
    hs[tmp]=1;
    tmp=s2hash[len];
    for(int i=len;i<n;i++)
    {
    
    
        if(hs.count(tmp))
            return 1;
        tmp=(tmp-s2[i-len]*base[len-1])*bs+s2[i];
    }
    return hs.count(tmp);
}

int main()
{
    
    
    base[0]=1;
    scanf("%d%s%s",&n,s1,s2);
    for(int i=1;i<=n;i++)
        s1hash[i]=s1hash[i-1]*bs+s1[i-1];
    for(int i=1;i<=n;i++)
    {
    
    
        s2hash[i]=s2hash[i-1]*bs+s2[i-1];
        base[i]=base[i-1]*bs;
    }
    int l=0,r=n,mid;
    while(l<=r)
    {
    
    
        mid=(l+r)>>1;
        if(check(mid))
            l=mid+1;
        else
            r=mid-1;
    }
    printf("%d\n",r);
    return 0;
}