什么屑题啊……
统计每个状态的 \(\rm endpos\) 内不被禁止的位置的个数，乘该状态的 \(\rm len\)，取最大值即可。
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using namespace std;
const int N = 2e5;
int n;
char s[N + 5],t[N + 5];
long long ans;
namespace SAM
{
struct node
{
int ch[26];
int fa,len;
} sam[(N << 1) + 5];
int sz[(N << 1) + 5];
int las = 1,tot = 1;
int c[N + 5],a[(N << 1) + 5];
inline void insert(int x)
{
int cur = las,p = ++tot;
sam[p].len = sam[cur].len + 1,sz[p] = t[sam[p].len] ^ '1';
for(;cur && !sam[cur].ch[x];cur = sam[cur].fa)
sam[cur].ch[x] = p;
if(!cur)
sam[p].fa = 1;
else
{
int q = sam[cur].ch[x];
if(sam[cur].len + 1 == sam[q].len)
sam[p].fa = q;
else
{
int nxt = ++tot;
sam[nxt] = sam[q],sam[nxt].len = sam[cur].len + 1,sam[p].fa = sam[q].fa = nxt;
for(;cur && sam[cur].ch[x] == q;cur = sam[cur].fa)
sam[cur].ch[x] = nxt;
}
}
las = p;
}
inline void build()
{
for(register int i = 1;i <= n;++i)
insert(s[i] - 'a');
for(register int i = 1;i <= tot;++i)
++c[sam[i].len];
for(register int i = 1;i <= n;++i)
c[i] += c[i - 1];
for(register int i = tot;i > 1;--i)
a[c[sam[i].len]--] = i;
for(register int i = tot;i > 1;--i)
sz[sam[a[i]].fa] += sz[a[i]];
}
}
int main()
{
scanf("%d%s%s",&n,s + 1,t + 1);
SAM::build();
for(register int i = 2;i <= SAM::tot;++i)
ans = max(ans,(long long)SAM::sam[i].len * SAM::sz[i]);
printf("%lld\n",ans);
}