Submission #61870

#TimeUsernameProblemLanguageResultExecution timeMemory
61870BenqMatch (CEOI16_match)C++11
100 / 100
49 ms18860 KiB

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

string x;
int st[MX][17], pre[MX][26];

bool ok(int l, int r) {
    r ++;
    F0Rd(i,17) if (st[r][i] >= l) r = st[r][i];
    return l == r;
}

vi v;

bool bad(int ind) {
    return !ok(ind+1,(sz(v)?v.back():sz(x))-1);
}

void ad(char c) {
    v.pb(pre[sz(v) ? v.back() : sz(x)][c-'a']);
}

void rem() {
    v.pop_back(), v.pop_back();
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> x;
    F0R(i,sz(x)+1) {
        if (i) st[i][0] = pre[i-1][x[i-1]-'a'];
        else st[i][0] = -1;
        
        // if (i) cout << i << " " << st[i][0] << " " << x[i-1] << "\n";
        F0R(j,26) pre[i][j] = (st[i][0] == -1 ? -1 : pre[st[i][0]][j]);
        if (i) {
            pre[i][x[i-1]-'a'] = i-1;
            // cout << "AH " << i << " " << x[i-1] << "\n";
        }
    }
    FOR(j,1,17) F0R(i,sz(x)+1) {
        if (st[i][j-1] == -1) st[i][j] = -1;
        else st[i][j] = st[st[i][j-1]][j-1];
    }
    if (!ok(0,sz(x)-1)) {
        cout << -1;
        exit(0);
    }
    // cout << pre[6][0]; exit(0);
    string ans;
    F0R(i,sz(x)) {
        ad(x[i]);
        //cout << v.back() << " " << bad(i) << "\n";
        //exit(0);
        if (bad(i)) {
            rem();
            ans += ')';
        } else {
            ans += '(';
        }
    }
    cout << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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