제출 #1132790

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1132790		altern23	구경하기 (JOI13_watching)	C++20	100 / 100	105 ms	31884 KiB

watching

#include <bits/stdc++.h>
using namespace std;
 
#pragma GCC optimize ("O2")
#pragma GCC optimize ("unroll-loops")
 
#define ll long long
#define pii pair<ll,ll>
#define fi first
#define sec second
#define ld long double

template <typename T>
ostream& operator << (ostream& os, vector<T>tmp){
    os << "[";
    for(auto x : tmp) os << " " << x;
    os << " ]";
 
    return os;
}
 
template <typename T>
ostream& operator << (ostream& os, set<T>tmp){
    os << "[";
    for(auto x : tmp) os << " " << x;
    os << " ]";
 
    return os;
}
 
template <typename T>
ostream& operator << (ostream& os, multiset<T>tmp){
    os << "[";
    for(auto x : tmp) os << " " << x;
    os << " ]";
 
    return os;
}
 
ostream& operator << (ostream& os, pii x){
    os << "[";
    os << " " << x.fi << " " << x.sec;
    os << " ]";
 
    return os;
}

const ll MOD = 998244353;
const ll N = 2e5 + 5;
const ll INF = 2e18;

// modulo operations
void add(ll &a, ll b) { a = (a + b) % MOD; }
void sub(ll &a, ll b) { a -= b; while(a < 0) a += MOD; a %= MOD; }
void mul(ll &a, ll b) { a = (a * b) % MOD; }

ll expo(ll a, ll b) {
    ll ans = 1;
    while(b > 0){
        if(b & 1) mul(ans, a);
        mul(a, a);
        b /= 2;
    }

    return ans;
}

ll dp[2005][2005];
ll l1[2005], l2[2005];

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int tc = 1;
    // cin >> tc;
    for(;tc--;){
        ll n, p, q; cin >> n >> p >> q;
        vector<ll> a(n + 5);
        for(int i = 1; i <= n; i++) cin >> a[i];

        if(p + q >= n){
            cout << 1 << "\n";
            continue;
        }
        
        sort(a.begin() + 1, a.begin() + 1 + n);

        ll l = 1, r = 1e9, ans = -1;
        for(;l <= r;){
            ll mid = (l + r) / 2;

            ll pt = 1, pt2 = 1;
            dp[0][0] = 0;
            ll mn = INF;

            for(int i = 1; i <= n; i++){
                while(pt <= n && a[i] - a[pt] + 1 > mid) pt++;
                while(pt2 <= n && a[i] - a[pt2] + 1 > 2 * mid) pt2++;
                l1[i] = pt, l2[i] = pt2;
                for(int j = 0; j <= p; j++){
                    dp[i][j] = INF;
                    dp[i][j] = dp[l2[i] - 1][j] + 1;
                    if(j) dp[i][j] = min(dp[i][j], dp[l1[i] - 1][j - 1]);
                    if(i == n) mn = min(mn, dp[i][j]);
                }
            }

            if(mn <= q){
                ans = mid;
                r = mid - 1;
            }
            else l = mid + 1;
        }

        cout << ans << "\n";
    }   

}
 
/*

*/

• Subtask 1: 50 / 50
• Subtask 2: 50 / 50