Submission #1002721

#	Time	Username	Problem	Language	Result	Execution time	Memory
1002721		TheSahib	Dabbeh (INOI20_dabbeh)	C++14	100 / 100	1500 ms	132580 KiB

Main

// #pragma optimize("O3")
#include <bits/stdc++.h>
 
using namespace std;

#define ll long long
#define pii pair<int, int>
#define all(v) v.begin(), v.end()
#define oo 1e9

const int MAX = 5e5 + 5, B = 61, LOGMAX = 18;
const ll MOD = (1ll << 61) - 1;

struct ST{
    pii tree[4 * MAX];
    void init(){
        for(int i = 0; i < 4 * MAX; i++) tree[i] = {0, 0};
    }
    void update(int node, int l, int r, int pos, pii val){
        if(l == r){
            tree[node] = val;
            return;
        }
        int mid = (l + r) / 2;
        if(pos <= mid) update(2 * node, l, mid, pos, val);
        else update(2 * node + 1, mid + 1, r, pos, val);
        tree[node] = max(tree[2 * node], tree[2 * node + 1]);
    }
    pii ask(int node, int l, int r, int ql, int qr){
        if(qr < l || r < ql) return {0, 0};
        if(ql <= l && r <= qr) return tree[node];
        int mid = (l + r) / 2;
        return max(ask(2 * node, l, mid, ql, qr), ask(2 * node + 1, mid + 1, r, ql, qr));
    }
};
ll binpow(ll a, ll b){
    if(b == 0) return 1;
    if(b & 1) return (__int128_t)binpow(a, b - 1) % MOD * a % MOD;
    return binpow((__int128_t)a * a % MOD, b / 2) % MOD;
}
ll inv(ll a){
    return binpow(a, MOD - 2);
}
int n, q;
string s;
string t[MAX];
 
ll pre[MAX];
ll P[(int)5e5 + 5];
ll invP[(int)5e5 + 5];
int par[LOGMAX][MAX];
int nxt[MAX];
ST st;
unordered_set<ll> mp[MAX];
 
ll calc(int l, int r){
    return (__int128_t)(pre[r] - pre[l - 1] + MOD) % MOD * invP[l - 1] % MOD;
}
 
int sz[MAX];
 
void solve(){
    P[0]= 1;
    invP[0] = 1;
    for(int i = 1; i <= 5e5; i++){
        P[i] = (__int128_t)P[i - 1] * B % MOD;
        invP[i] = (__int128_t)invP[i - 1] * inv(B) % MOD;
    }
    cin >> n >> q;

    for(int i = 1; i <= n; i++) sz[t[i].size()]++;
    for(int i = 1; i < MAX; i++) mp[i].reserve(4 * sz[i]);

    for(int i = 1; i <= n; i++){
        cin >> t[i];
        ll H = 0;
        for(int j = 0; j < t[i].size(); j++){
            H = (__int128_t)(H + (__int128_t)(t[i][j] - 'a' + 1) * P[j] % MOD) % MOD;
            mp[j + 1].insert(H);
        }
    }
    cin >> s;
    n = s.size();
    for(int i = 1; i <= n; i++){
        pre[i] = (__int128_t)(pre[i - 1] + (__int128_t)(s[i - 1] - 'a' + 1) * P[i - 1] % MOD) % MOD;
    }
    st.init();
    st.update(1, 1, n + 1, n + 1, {n + 1, n + 1});
    par[0][n + 1] = n + 1;
    nxt[n + 1] = n + 1;
    for(int i = n; i >= 1; i--){
        if(!mp[1].count(calc(i, i))){
            par[0][i] = i;
            nxt[i] = i;
            st.update(1, 1, n + 1, i, {i, i});
            continue;
        }
        int l = i, r = n;
        while(l < r){
            int mid = (l + r + 1) / 2;
            if(mp[mid - i + 1].count(calc(i, mid))) l = mid;
            else r = mid - 1;
        }
        par[0][i] = st.ask(1, 1, n + 1, i + 1, l + 1).second;
        nxt[i] = l + 1;
        st.update(1, 1, n + 1, i, {l + 1, i});
    }
    for(int j = 1; j < LOGMAX; j++){
        for(int i = 1; i <= n + 1; i++){
            par[j][i] = par[j - 1][par[j - 1][i]];
        }
    }
    while(q--){
        int l, r; cin >> l >> r;
        l++;
        if(nxt[l] > r){
            cout << "1\n";
            continue;
        }
        ll ans = 0;
        for(int i = LOGMAX - 1; i >= 0; i--){
            if(nxt[par[i][l]] <= r){
                ans += (1ll << i);
                l = par[i][l];
            }
        }
        if(nxt[par[0][l]] <= r) cout << "-1\n";
        else cout << ans + 2 << '\n';
    }
}
 
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while(t--){
        solve();
    }
}

Compilation message (stderr)

Main.cpp: In function 'void solve()':
Main.cpp:77:26: warning: comparison of integer expressions of different signedness: 'int' and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   77 |         for(int j = 0; j < t[i].size(); j++){
      |                        ~~^~~~~~~~~~~~~

• Subtask 1: 25 / 25
• Subtask 2: 25 / 25
• Subtask 3: 50 / 50