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Give a string s, and specify some characters are legal and some are illegal

Find the maximum value of ∣ a ∣ f(a) ゗ * f(a) ∣ * a ∣ * f(a), aaa is the substring of s, f(a)f(a)f(a) f (a) is the number of occurrences ending with legal characters

Analysis

Build suffix array, merge according to height array from big to small, and check and maintain

Because the substring ending with illegal characters cannot be counted, and the suffix array cannot easily determine whether the end of the substring is legal or not, we first flip the string s, so that we can turn the problem into whether the beginning of the substring is legal or not

When initializing and querying the set, we set the suffix size at the beginning of the legal character to 1, and the suffix at the beginning of the illegal character to 0. Then we sort the height array, merge it from large to small, and update the answer every time

In addition, only once substring needs special judgment

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define debug(x) cerr<<#x<<' '<<x<<'\n'
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
const int maxn=2e5+10;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxbit=20;
struct SuffixArray
{
    int sa[maxn], rank[maxn], ws[maxn], wv[maxn], wa[maxn], wb[maxn], height[maxn], st[maxbit][maxn], N;
    bool cmp(int *r, int a, int b, int l){return r[a]==r[b] and r[a+l]==r[b+l];}
    void build(int *r, int n, int m)
    {
        N=n;
        n++;
        int i, j, k=0, p, *x=wa, *y=wb, *t;
        for(i=0;i<m;i++)ws[i]=0;
        for(i=0;i<n;i++)ws[x[i]=r[i]]++;
        for(i=1;i<m;i++)ws[i]+=ws[i-1];
        for(i=n-1;i>=0;i--)sa[--ws[x[i]]]=i;
        for(p=j=1;p<n;j<<=1,m=p)
        {
            for(p=0,i=n-j;i<n;i++)y[p++]=i;
            for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
            for(i=0;i<n;i++)wv[i]=x[y[i]];
            for(i=0;i<m;i++)ws[i]=0;
            for(i=0;i<n;i++)ws[wv[i]]++;
            for(i=1;i<m;i++)ws[i]+=ws[i-1];
            for(i=n-1;i>=0;i--)sa[--ws[wv[i]]]=y[i];
            for(t=x,x=y,y=t,p=1,i=1,x[sa[0]]=0;i<n;i++)
                x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
        }

        for(i=0;i<n;i++)rank[sa[i]]=i;

        for(i=0;i<n-1;height[rank[i++]]=k)
            for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
    }
    void build_st()     //st table
    {
        int i, k;
        for(i=1;i<=N;i++)st[0][i]=height[i];
        for(k=1;k<=maxbit;k++)
            for(i=1;i+(1<<k)-1<=N;i++)
                st[k][i]=min(st[k-1][i],st[k-1][i+(1<<k)-1]);
    }
    int lcp(int x, int y)   //Longest Common Prefix 
    {
        int l=rank[x], r=rank[y];
        if(l>r)swap(l,r);
        if(l==r)return N-sa[l];
        int t=log2(r-l);
        return min(st[t][l+1],st[t][r-(1<<t)+1]);
    }
}SA;
int strcp[maxn],fa[maxn];
ll siz[maxn];
bool can[maxn];
pii pos[maxn];
int Find(int x) {
    if(x==fa[x]) return x;
    else return fa[x]=Find(fa[x]);
}
void Union(int a,int b) {
    int faa=Find(a),fab=Find(b);
    siz[faa]+=siz[fab];
    fa[fab]=faa;
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);
    int len;
    cin>>len;
    string a,b;
    cin>>a;
    cin>>b;
    rep(i,0,len-1) {
        strcp[len-i-1]=a[i];
        if(b[i]=='0') can[len-i-1]=true;
        else can[len-i-1]=false;
    } 
    ll ans=0;
    rep(i,0,len) {
        fa[i]=i;
        if(can[i]) {
            siz[i]=1;
            ans=max(ans,(ll)len-i);
        } 
    }
    strcp[len]=0;
    SA.build(strcp,len,300);
    int num=0;
    rep(i,2,len) {
        pos[num].fi=SA.height[i];
        pos[num++].se=i;
    }
    sort(pos,pos+num);
    per(i,num-1,0) {
        int l=SA.sa[pos[i].se-1],r=SA.sa[pos[i].se];
        l=Find(l),r=Find(r);
        ans=max(ans,(siz[l]+siz[r])*pos[i].fi);
        Union(l,r);
    }
    cout<<ans<<'\n';
    return 0;
}