oj.uz

답안 #552342

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
552342 2022-04-23T03:20:36 Z pokmui9909 Krov (COCI17_krov) C++17
84 / 140
 1500 ms 12352 KB 
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
  
ll N;
ll A[100005];
ll B[100005];
ll C[100005];
struct BIT{
    ll T[300015] = {};
    void update(ll k, ll v){while(k <= 300010) T[k] += v, k += k & -k;}
    ll sum(ll k){ll r = 0;while(k >= 1) r += T[k], k -= k & -k; return r;}
    ll query(ll l, ll r) {return sum(r) - sum(l - 1);}
};
BIT pfs, pfc, sfs, sfc;
vector<ll> V1, V2;
const ll INF = 1e10;
ll Idx1(ll k){
    return lower_bound(V1.begin(), V1.end(), k) - V1.begin();
}
ll Idx2(ll k){
    return upper_bound(V1.begin(), V1.end(), k) - V1.begin() - 1;
}
ll Idx3(ll k){
    return lower_bound(V2.begin(), V2.end(), k) - V2.begin();
}
ll Idx4(ll k){
    return upper_bound(V2.begin(), V2.end(), k) - V2.begin() - 1;
}
  
ll f(ll X, ll K){
    ll r = 0;
    K -= X;
    r += K * pfc.query(Idx1(-INF), Idx2(K)) - pfs.query(Idx1(-INF), Idx2(K));
    r += pfs.query(Idx1(K + 1), Idx2(INF)) - K * pfc.query(Idx1(K + 1), Idx2(INF));
  
    K += 2 * X;
    r += K * sfc.query(Idx3(-INF), Idx4(K)) - sfs.query(Idx3(-INF), Idx4(K));
    r += sfs.query(Idx3(K + 1), Idx4(INF)) - K * sfc.query(Idx3(K + 1), Idx4(INF));
    return r;
}
  
int main(){
    cin.tie(0) -> sync_with_stdio(false);
  
    V1.push_back(-INF - 1);
    V1.push_back(-INF);
    V1.push_back(INF);
    V2.push_back(-INF - 1);
    V2.push_back(-INF);
    V2.push_back(INF);
    cin >> N;
    for(ll i = 1; i <= N; i++){
        cin >> A[i];
        B[i] = A[i] + i;
        C[i] = A[i] - i;
        V1.push_back(A[i]); V1.push_back(C[i]);
        V2.push_back(A[i]); V2.push_back(B[i]);
    }
    sort(V1.begin(), V1.end());
    V1.erase(unique(V1.begin(), V1.end()), V1.end());
    sort(V2.begin(), V2.end());
    V2.erase(unique(V2.begin(), V2.end()), V2.end());
    ll ans = 1e18;
    for(ll i = 1; i <= N; i++){
        sfs.update(Idx3(B[i]), B[i]);
        sfc.update(Idx3(B[i]), 1);
    }
    for(ll i = 1; i <= N; i++){
        pfs.update(Idx1(C[i]), C[i]);
        pfc.update(Idx1(C[i]), 1);
        sfs.update(Idx3(B[i]), -B[i]);
        sfc.update(Idx3(B[i]), -1);
        ll L = 0, R = 2e9;
        while(R - L > 10){
            ll p = (L * 2 + R) / 3, q = (L + R * 2) / 3;
            if(f(i, p) <= f(i, q)) R = q;
            else L = p;
        }
        L = max(L, i);
        L = max(L, N + 1 - i);
        for(ll j = L; j <= R; j++){
            ans = min(ans, f(i, j));
        }
    }
    cout << ans;
}

Subtask #1
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 33 ms 596 KB Output is correct
2 Correct 39 ms 632 KB Output is correct

Subtask #2
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 32 ms 596 KB Output is correct
2 Correct 41 ms 640 KB Output is correct

Subtask #3
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 67 ms 716 KB Output is correct
2 Correct 70 ms 692 KB Output is correct

Subtask #4
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 69 ms 852 KB Output is correct
2 Correct 100 ms 768 KB Output is correct

Subtask #5
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 108 ms 724 KB Output is correct
2 Correct 127 ms 828 KB Output is correct

Subtask #6
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 177 ms 1108 KB Output is correct
2 Correct 225 ms 1016 KB Output is correct
3 Correct 108 ms 844 KB Output is correct

Subtask #7
0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1417 ms 3300 KB Output is correct
2 Correct 1290 ms 2864 KB Output is correct
3 Correct 1435 ms 3432 KB Output is correct
4 Execution timed out 1577 ms 4056 KB Time limit exceeded
5 Halted 0 ms 0 KB -

Subtask #8
0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1598 ms 4564 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #9
0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1591 ms 6040 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #10
0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1568 ms 12352 KB Time limit exceeded
2 Halted 0 ms 0 KB -