[ABC287E] Karuta

考虑二分 LCP，显然 LCP 是有单调性的。

然后维护一下哈希，用 multiset 判断存在性，就可以做到两只 $\log$。

ABC 还卡了 $10^9+7$ 的单模哈希，所以赛时用了双模。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <cstdlib>
#include <string>
using namespace std;

#define ll long long

const int N = 5e5 + 5, INF = 2e9, MOD = 1e9 + 7, MOD2 = 998244353;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, t, maxn = 0;
string s[N];
multiset<pair<long long, long long> > st;
vector<pair<long long, long long> > hs[N];

long long qpow(long long a, long long b)
{
	long long res = 1, base = a;
	while (b)
	{
		if (b & 1)
		{
			res = res * base;
			res %= MOD;
		}
		base *= base;
		base %= MOD;
		b >>= 1;
	}
	return res;
}

long long qpow2(long long a, long long b)
{
	long long res = 1, base = a;
	while (b)
	{
		if (b & 1)
		{
			res = res * base;
			res %= MOD2;
		}
		base *= base;
		base %= MOD2;
		b >>= 1;
	}
	return res;
}

int main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> s[i];
		long long ph = 0, ph2 = 0;
		for (int j = 0; j < s[i].size(); j++)
		{
			ph += s[i][j] * qpow(26, j * 1LL) % MOD;
			ph %= MOD;
			ph2 += s[i][j] * qpow2(26, j * 1LL) % MOD2;
			ph2 %= MOD2;
			hs[i].emplace_back(make_pair(ph, ph2));
			st.insert(make_pair(ph, ph2));
		}
	}
	for (int i = 1; i <= n; i++)
	{
		for (auto j : hs[i])
		{
			st.erase(st.find(j));
		}
		maxn = 0;
		int l = 1, r = s[i].size();
		while (l <= r)
		{
			int mid = l + r >> 1;
			auto g = hs[i][mid - 1];
			if (st.find(g) != st.end())
			{
				maxn = mid;
				l = mid + 1;
			}
			else r = mid - 1;
		}
		for (auto j : hs[i])
		{
			st.insert(j);
		}
		cout << maxn << "\n";
	}
	return 0;
}