Submission #1019332

#	Time	Username	Problem	Language	Result	Execution time	Memory
1019332		stefanopulos	Palinilap (COI16_palinilap)	C++17	100 / 100	60 ms	24716 KiB

palinilap

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 200005; 

int n;
string s;

ll cnt[mxN];
ll sum[mxN];
ll kol[mxN][26];

int add(int a, int b){ a += b; if(a >= mod)a -= mod; return a; }
int sub(int a, int b){ a -= b; if(a < 0)a += mod; return a; }
int mul(int a, int b){ return (1ll * a * b) % mod; }
int power(int a, int b){
    if(!b)return 1;
    int p = power(a, b / 2);
    p = mul(p, p);
    if(b % 2 == 1)p = mul(p, a);
    return p;
}
int inv(int a){
    return power(a, mod - 2);
}

const int p1 = 37;
int pw[mxN];
int invz[mxN];

int pref[mxN];
int getP(int l, int r){
    return mul(sub(pref[r], (l == 0 ? 0 : pref[l - 1])), invz[l]); 
}

int sufi[mxN];
int getS(int l, int r){
    return mul(sub(sufi[l], sufi[r + 1]), invz[n - r - 1]); 
}

int calc(int a, int b){
    if(a < 0 || b >= n)return 0;

    int l = 1, r = n, ans = 0;
    while(l <= r){
        int mid = (l + r) / 2;
        if(a - mid + 1 >= 0 && b + mid - 1 < n && getP(a - mid + 1, a) == getS(b, b + mid - 1)){
            ans = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }
 
    return ans;
}

int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> s; n = sz(s);

    pw[0] = 1;
    ff(i,1,n)pw[i] = mul(pw[i - 1], p1);

    invz[n] = inv(pw[n]);
    fb(i,n - 1,0)invz[i] = mul(invz[i + 1], p1);

    ff(i,0,n - 1)pref[i] = add((i == 0 ? 0 : pref[i - 1]), mul(s[i] - 'a' + 1, pw[i]));
    fb(i,n - 1,0)sufi[i] = add(sufi[i + 1], mul(s[i] - 'a' + 1, pw[n - i - 1]));

    ll uk = 0;
    ff(i,0,n - 1){

        int l = 1, r = n, ans = 0;
        while(l <= r){
            int mid = (l + r) / 2;
            if(i - mid + 1 >= 0 && i + mid - 1 < n && getP(i - mid + 1, i) == getS(i, i + mid - 1)){
                ans = mid;
                l = mid + 1;
            }
            else r = mid - 1;
        }

        cnt[i - ans + 1] -= 1; cnt[i] += 1;
        sum[i - ans + 1] += (i - ans); sum[i] -= (i - ans); 

        cnt[i + 1] += 1; cnt[i + ans] -= 1;
        sum[i + 1] -= (i + ans); sum[i + ans] += (i + ans);

        uk += ans;

        int a = i - ans;
        int b = i + ans;
        if(a < 0 || b >= n)continue;

        int ima = 1 + calc(a - 1, b + 1);
        kol[a][s[b] - 'a'] += ima;
        kol[b][s[a] - 'a'] += ima;
    } 

    ff(i,1,n - 1){

        int l = 1, r = n, ans = 0;
        while(l <= r){
            int mid = (l + r) / 2;
            if(i - mid + 1 >= 0 && i + mid - 1 < n && getP(i - mid, i - 1) == getS(i, i + mid - 1)){
                ans = mid;
                l = mid + 1;
            }
            else r = mid - 1;
        }

        uk += ans;

        cnt[i - ans] -= 1; cnt[i] += 1;
        sum[i - ans] += (i - ans - 1); sum[i] -= (i - ans - 1);

        cnt[i] += 1; cnt[i + ans] -= 1;
        sum[i] -= (i + ans); sum[i + ans] += (i + ans);

        int a = i - ans - 1;
        int b = i + ans;
        if(a < 0 || b >= n)continue;

        int ima = 1 + calc(a - 1, b + 1);
        kol[a][s[b] - 'a'] += ima;
        kol[b][s[a] - 'a'] += ima;

    }

    ff(i,1,n - 1){
        cnt[i] += cnt[i - 1];
        sum[i] += sum[i - 1];
    }

    ll rez = uk;
    ff(i,0,n - 1){
        ll mx = 0; 
        ff(j,0,25)mx = max(mx, kol[i][j]);

        ll ans = uk + mx + 1ll * i * cnt[i] + sum[i];
        rez = max(rez, ans);
    }

    cout << rez << '\n';

    return 0;
}
/*



// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 17 / 17
• Subtask 2: 37 / 37
• Subtask 3: 46 / 46