假设最长串为aaa…，其长度为x，子串从a到aaa…，总用长即为x*(x+1)/2<=1e5，x<=500。

最长串<=500，所以对于所有长度小于500的串，O(n^2)枚举所有子串，O(1)哈希判断是否子串存在，设串长分别为x1,x2…xk，则总时间复杂度x1^2+x2^2+…+xk^2<=500*(x1+x2+…+xk)<=500*1e5<=5e7，所以可以保证时间复杂度。

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define ull unsigned long long
#define endl '\n'
#define deb(n) cout << #n << "=" << n << "    "
#define debug(n) cout << #n << "=" << n << endl
#define div() cout << "_______________" << endl

const int N = 1e5 + 10, P = 131;

int n;
unordered_map<int, int> mp;
string A[N];
ull h[N], p[N];

ull geths(int l, int r) {
    return h[r] - h[l - 1] * p[r - l + 1];
}

void oper() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> A[i];
    for (int i = 1; i <= n; i++) {
        ull hs = 0;
        for (int j = 0; j < A[i].size(); j++) hs = hs * P + A[i][j];
        mp[hs] = 1;
    }
    int maxa = 0;
    for (int i = 1; i <= n; i++) {
        if (A[i].size() > 500) continue;
        int m = A[i].size();
        string tmp = '0' + A[i];
        for (int j = 1; j <= m; j++) h[j] = h[j - 1] * P + tmp[j];
        int flag = 1;
        for (int j = 1; j <= m; j++) {
            for (int k = j; k <= m; k++) {
                int tar = geths(j, k);
                if (!mp.count(tar)) flag = 0;
            }
        }
        if (flag) maxa = max(maxa, m);
    }
    cout << maxa << endl;
}

signed main() {
    p[0] = 1;
    for (int i = 1; i < N; i++) p[i] = p[i - 1] * P;
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t = 1; //cin >> t;
    while (t--) oper();
    return 0;
}